Explanation:
Okay, so I need to find the values of λ between 0 and 4 (inclusive) for which there exists some x in [0,1] such that f(x) ≠ x but f(f(x)) = x. The function f is given by f(x) = λx(1 - x).

First, let me understand what the question is asking. We have this quadratic function f(x) = λx(1 - x). For some λ in [0,4], we need there to be at least one x where applying f twice brings us back to x, but applying f once doesn't. So, x is not a fixed point of f, but it's a periodic point of period 2.

So, the problem reduces to finding λ such that the function f has a periodic point of period 2, i.e., a solution to f(f(x)) = x that is not a fixed point.

Let me recall that for a function f, fixed points satisfy f(x) = x, and periodic points of period 2 satisfy f(f(x)) = x but f(x) ≠ x. Therefore, we need to find λ for which the equation f(f(x)) = x has solutions that are not solutions to f(x) = x.

So, maybe I should first solve f(f(x)) = x and then subtract the fixed points. Let me try that approach.

First, let's compute f(f(x)). Let me denote f(x) = λx(1 - x). Then f(f(x)) would be f(λx(1 - x)) = λ*(λx(1 - x))*(1 - λx(1 - x)).

So, f(f(x)) = λ^2 x(1 - x)[1 - λx(1 - x)].

Therefore, the equation f(f(x)) = x is equivalent to:

λ^2 x(1 - x)[1 - λx(1 - x)] = x.

Let me rearrange this equation. Let me subtract x from both sides:

λ^2 x(1 - x)[1 - λx(1 - x)] - x = 0.

Factor out x:

x [ λ^2 (1 - x)[1 - λx(1 - x)] - 1 ] = 0.

So, the solutions to f(f(x)) = x are either x = 0 or the solutions to the equation:

λ^2 (1 - x)[1 - λx(1 - x)] - 1 = 0.

But we are interested in solutions where f(x) ≠ x. So, the fixed points of f are solutions to f(x) = x, which is:

λx(1 - x) = x.

Let me solve that equation:

λx(1 - x) - x = 0 => x[λ(1 - x) - 1] = 0 => x[λ - λx - 1] = 0.

Therefore, the fixed points are x = 0 or λ - λx - 1 = 0 => λx = λ - 1 => x = (λ - 1)/λ.

Wait, but x must be in [0,1]. So, let me check the fixed points.

First, x = 0 is always a fixed point.

The other fixed point is x = (λ - 1)/λ. Let me compute that. Let me write x = 1 - 1/λ.

Wait, because x = (λ - 1)/λ = 1 - 1/λ.

But since x must be in [0,1], let me see:

For x = 1 - 1/λ to be in [0,1], we need:

0 ≤ 1 - 1/λ ≤ 1.

First, the lower bound: 1 - 1/λ ≥ 0 => 1/λ ≤ 1 => λ ≥ 1.

The upper bound: 1 - 1/λ ≤ 1 is always true, since 1/λ ≥ 0. Therefore, the other fixed point exists only when λ ≥ 1, and it's equal to 1 - 1/λ.

Therefore, the fixed points of f(x) are x=0 and, if λ ≥ 1, x = 1 - 1/λ.

So, the fixed points are x=0 and x=(λ - 1)/λ when λ ≥ 1.

Now, returning to the equation f(f(x)) = x, which factors into x times [something] equals zero. So, the solutions are x=0 and the solutions to that other equation.

However, we need to exclude x=0 and the fixed points (excluding x=0 if there is another fixed point) because we want solutions where f(x) ≠ x.

So, the equation f(f(x)) = x includes fixed points and periodic points of period 2. Therefore, the problem requires that there exists a solution to f(f(x)) = x that is not a fixed point. That is, the equation [something] = 0 has solutions different from the fixed points.

Therefore, let's focus on solving the equation:

λ² (1 - x)[1 - λx(1 - x)] - 1 = 0.

Let me expand this equation. Let me denote y = x for simplicity. Then, the equation becomes:

λ² (1 - y)[1 - λ y (1 - y)] - 1 = 0.

Let me expand the term inside:

First compute λ y (1 - y): that is λ y - λ y².

So, 1 - λ y (1 - y) = 1 - λ y + λ y².

Multiply by (1 - y):

(1 - y)(1 - λ y + λ y²) = ?

Let me compute that term. Let me expand (1 - y)(1 - λ y + λ y²):

Multiply term by term:

First, 1*(1 - λ y + λ y²) = 1 - λ y + λ y².

Then, -y*(1 - λ y + λ y²) = -y + λ y² - λ y³.

Adding them together:

1 - λ y + λ y² - y + λ y² - λ y³.

Combine like terms:

Constant term: 1.

Linear terms: -λ y - y = -(λ + 1)y.

Quadratic terms: λ y² + λ y² = 2λ y².

Cubic term: -λ y³.

Therefore, overall:

(1 - y)(1 - λ y + λ y²) = 1 - (λ + 1)y + 2λ y² - λ y³.

Therefore, going back to the original equation:

λ² [1 - (λ + 1)y + 2λ y² - λ y³] - 1 = 0.

Let me distribute λ²:

λ² - λ² (λ + 1) y + 2λ³ y² - λ³ y³ - 1 = 0.

Let me write that as:

(λ² - 1) - λ² (λ + 1) y + 2λ³ y² - λ³ y³ = 0.

So, rearranged:

-λ³ y³ + 2λ³ y² - λ² (λ + 1) y + (λ² - 1) = 0.

Alternatively, multiplying both sides by -1 to make the leading coefficient positive:

λ³ y³ - 2λ³ y² + λ² (λ + 1) y - (λ² - 1) = 0.

Let me write that as:

λ³ y³ - 2λ³ y² + λ² (λ + 1) y - (λ² - 1) = 0.

This seems complicated. Maybe there's a better way. Let me check if there is a factorization here. Remember that x=0 is a solution to f(f(x)) = x, but we already factored that out. However, perhaps there are other factors?

Alternatively, maybe we can think of the equation f(f(x)) = x as f(f(x)) - x = 0, and factor it as (f(x) - x)(something) = 0? Because if f(f(x)) - x factors into (f(x) - x) times something, then the solutions to f(f(x)) = x would include the fixed points and the solutions to that something = 0.

Indeed, for any function f, f(f(x)) - x = (f(x) - x)(something). So, perhaps we can factor f(f(x)) - x as (f(x) - x)(something). Let me check that.

Let me denote f(x) = g(x) = λx(1 - x). Then f(f(x)) = g(g(x)). Then, f(f(x)) - x = g(g(x)) - x.

If we can factor this as (g(x) - x)(something) = 0, then the solutions would be the fixed points (g(x) - x = 0) and the solutions of something = 0.

So, let me perform polynomial division or try to factor it.

Let me suppose that g(g(x)) - x = (g(x) - x)(A x² + B x + C). Let me check if this is possible.

Alternatively, maybe using the fact that for quadratic functions, the composition f(f(x)) - x is a degree 4 polynomial, and if we factor out (f(x) - x), which is quadratic, then the remaining polynomial would be quadratic. So, maybe it's possible.

But let me check with specific values.

Alternatively, since we know that f(f(x)) - x has roots at all fixed points of f and at the periodic points of period 2. So, if we know the fixed points, we can factor them out.

Earlier, we found that the fixed points are x=0 and x=(λ - 1)/λ when λ ≥1.

Therefore, the equation f(f(x)) - x = 0 has roots at x=0 and x=(λ - 1)/λ (if applicable) and the period-2 points. Since f(f(x)) - x is a quartic polynomial, and if we factor out (x)(x - a), where a is the other fixed point, then the remaining quadratic would correspond to the period-2 points.

But perhaps we can do this.

Let me write f(f(x)) - x = 0 as:

λ² x(1 - x)[1 - λx(1 - x)] - x = 0.

Which we had earlier. But factor out x:

x [ λ² (1 - x)[1 - λx(1 - x)] - 1 ] = 0.

Therefore, the non-zero solutions come from solving:

λ² (1 - x)[1 - λx(1 - x)] - 1 = 0.

But if we want to find solutions that are not fixed points, we need to ensure that these solutions are not x=0 or x=(λ - 1)/λ.

So, perhaps we can find the roots of this equation.

Alternatively, since we know that if there is a period-2 point, then there must be two points x1 and x2 such that f(x1) = x2 and f(x2) = x1, and x1 ≠ x2, and neither x1 nor x2 is a fixed point.

Therefore, maybe we can set up the equations f(x) = y and f(y) = x, with x ≠ y and f(x) ≠ x.

So, solving f(x) = y and f(y) = x. Let me try that approach.

So, given that f(x) = λx(1 - x) = y and f(y) = λy(1 - y) = x.

Therefore, we have the system:

y = λx(1 - x)

x = λy(1 - y)

So, substituting the first equation into the second, we get:

x = λ [λx(1 - x)] [1 - λx(1 - x)].

Therefore, x = λ² x(1 - x)[1 - λx(1 - x)].

So, bringing all terms to one side:

λ² x(1 - x)[1 - λx(1 - x)] - x = 0.

Which is the same equation as before. Therefore, same equation. So, we can consider it as:

x [ λ² (1 - x)[1 - λx(1 - x)] - 1 ] = 0.

Therefore, the solutions are x=0 and solutions to the other factor. As before.

So, for x ≠ 0, we have:

λ² (1 - x)[1 - λx(1 - x)] - 1 = 0.

This equation needs to have at least one solution different from the fixed points. Let me denote this equation as:

Let me expand it again step by step to see if I can find a way to solve it.

Let me first compute 1 - λx(1 - x):

As before, 1 - λx + λx².

Then multiply by (1 - x):

(1 - x)(1 - λx + λx²) = 1 - λx + λx² - x + λx² - λx³ = 1 - (λ + 1)x + 2λx² - λx³.

Then multiply by λ²:

λ² - λ²(λ + 1)x + 2λ³x² - λ³x³.

Then subtract 1:

λ² - 1 - λ²(λ + 1)x + 2λ³x² - λ³x³ = 0.

So:

-λ³x³ + 2λ³x² - λ²(λ + 1)x + (λ² - 1) = 0.

Let me write this as:

-λ³x³ + 2λ³x² - λ³x - λ²x + λ² - 1 = 0.

Wait, breaking down the term -λ²(λ + 1)x:

That's -λ³x - λ²x. So, yes.

So:

-λ³x³ + 2λ³x² - λ³x - λ²x + λ² - 1 = 0.

Hmm. This seems complicated. Maybe factor out terms?

Alternatively, since we know that the fixed points satisfy f(x) = x. So, suppose that x is a fixed point. Let me check if they satisfy this equation. Let me take x = 0. Plugging into the equation:

-λ³*0 + 2λ³*0 - λ³*0 - λ²*0 + λ² - 1 = λ² - 1. Which is zero only if λ² = 1, i.e., λ = 1. But for λ=1, x=0 is a fixed point. Similarly, check x=(λ - 1)/λ for λ ≥ 1. Let me check if that x satisfies the equation.

Let me denote x_f = (λ - 1)/λ. Let me compute the equation:

-λ³x_f³ + 2λ³x_f² - λ³x_f - λ²x_f + λ² - 1.

This seems tedious. Maybe there's a better approach.

Alternatively, since we are looking for solutions where x is not a fixed point, maybe we can consider the equation f(f(x)) = x and subtract the fixed points.

Alternatively, perhaps we can consider that for period-2 points, we have f(f(x)) = x and f(x) ≠ x. Therefore, the equation f(f(x)) - x = 0 has solutions which are the fixed points and the period-2 points. Therefore, if the equation f(f(x)) - x = 0 has more than just the fixed points, then there are period-2 points.

But how can we know when this equation has more solutions? Since f(f(x)) - x is a quartic polynomial, and we know two of its roots (x=0 and x=(λ - 1)/λ if λ ≥1), so if it has two more roots, then those would be the period-2 points.

Alternatively, maybe we can find for which λ the quartic has at least two more roots besides the fixed points. However, this seems abstract.

Alternatively, since the system f(x) = y and f(y) = x can be translated into solving y = λx(1 - x) and x = λy(1 - y). Let me try to eliminate variables.

From the first equation, y = λx(1 - x). Substitute into the second equation: x = λ[λx(1 - x)] [1 - λx(1 - x)].

Which is the same as before. Therefore, x = λ² x(1 - x)[1 - λx(1 - x)].

So, if x ≠ 0, we can divide both sides by x:

1 = λ²(1 - x)[1 - λx(1 - x)].

Let me denote z = x. Then, the equation becomes:

1 = λ²(1 - z)[1 - λz(1 - z)].

Which is:

λ²(1 - z)[1 - λz(1 - z)] = 1.

Let me expand this equation again:

Let me compute 1 - λz(1 - z) = 1 - λz + λz².

Therefore, λ²(1 - z)(1 - λz + λz²) = 1.

Which is the same as before.

Alternatively, maybe I can write this as:

Let me denote u = z(1 - z). Since z ∈ [0,1], u ranges from 0 to 1/4 (since the maximum of z(1 - z) is at z=1/2, giving 1/4).

But perhaps that substitution can help. Let me try:

Let u = z(1 - z). Then, the equation becomes:

λ²(1 - z)(1 - λu) = 1.

But u = z(1 - z), so 1 - z = ?

Wait, 1 - z can be expressed as... maybe not straightforward. Alternatively, maybe we can write 1 - z = ?

Alternatively, maybe we can express the equation in terms of u and z.

Alternatively, maybe not. Let me think.

Alternatively, let me suppose that there exists a solution z ≠ 0, z ≠ (λ - 1)/λ (fixed points). Let me think about possible values of λ.

Perhaps I can consider specific cases. For example, for λ=0, the function is f(x)=0. Then f(f(x)) = f(0) = 0, so f(f(x)) = 0 = x only if x=0. Therefore, there's no x ≠0 with f(f(x))=x. Therefore, for λ=0, no such x exists.

Similarly, for λ=4, the function is f(x) = 4x(1 - x), which is the logistic map with maximum at x=1/2, f(1/2)=1. Let me check for λ=4 if there are period-2 points. Maybe known results. But since I need to do it step by step.

Alternatively, perhaps I can consider the equation f(f(x)) = x, subtract f(x)=x, so f(f(x)) - f(x) = 0. But not sure.

Alternatively, maybe it's better to think about the system f(x) = y and f(y) = x, so substituting:

From f(x) = y and f(y) = x, we get:

y = λx(1 - x)

x = λy(1 - y)

Therefore, substituting y into the second equation, we get:

x = λ[λx(1 - x)] [1 - λx(1 - x)].

So, x = λ² x (1 - x)[1 - λx(1 - x)].

So, bringing all terms to one side:

λ² x(1 - x)[1 - λx(1 - x)] - x = 0.

Which is the same equation as before. So, if x ≠ 0, we can write:

λ² (1 - x)[1 - λx(1 - x)] = 1.

Let me denote this equation again:

λ² (1 - x)[1 - λx(1 - x)] = 1.

Let me try to solve this equation for x. Let me denote u = x(1 - x). Then, since x ∈ [0,1], u ∈ [0, 1/4]. Then, the equation becomes:

λ² (1 - x)(1 - λu) = 1.

But u = x(1 - x), so 1 - x = something?

Alternatively, maybe express 1 - x in terms of u. Let me note that:

u = x(1 - x) => u = x - x² => x² - x + u = 0.

But solving for x here might not help. Alternatively, perhaps think of 1 - x as something else. Let me think.

Alternatively, let me try to write everything in terms of u. Let me see:

Given u = x(1 - x), then 1 - x = (1 - x).

But unless we can relate 1 - x to u, it's not helpful.

Alternatively, maybe consider specific forms. Let me assume that there exist solutions for some λ. Let me try to find conditions on λ such that this equation has solutions.

Alternatively, let me use substitution. Let me suppose that x is a solution to the equation. Then, since x is between 0 and 1, and u = x(1 - x) is between 0 and 1/4, then λ²(1 - x)(1 - λu) = 1.

But since u ≤ 1/4, 1 - λu ≥ 1 - λ*(1/4). Therefore, if 1 - λ*(1/4) > 0, then 1 - λu > 0, so the product is positive. But since the left-hand side is λ²(1 - x)(1 - λu) = 1, which is positive.

Therefore, for this to be positive, we need:

Either both factors are positive or both negative. But since λ ≥ 0, λ² is non-negative. 1 - x is in [0,1]. Therefore, 1 - x ≥ 0. Therefore, 1 - λu must also be positive. Therefore, 1 - λu > 0 => λu < 1 => λ x(1 - x) < 1. But since x ∈ [0,1], x(1 - x) ≤ 1/4, so λ * 1/4 < 1 => λ < 4. Therefore, if λ = 4, then the maximum value of λx(1 - x) is 4*(1/4) = 1, so 1 - λu = 0 when u = 1/4, i.e., x = 1/2. Therefore, at λ=4, the term 1 - λu is zero when x=1/2, but since in the equation λ²(1 - x)(1 - λu) = 1, if λ=4 and x=1/2, then 1 - λu = 1 - 4*(1/4) = 0, so the left-hand side is zero, which cannot equal 1. Therefore, for λ=4, maybe there are no solutions? Wait, but maybe there are other x where 1 - λu is positive?

Wait, if λ=4, then for other x, u = x(1 - x) < 1/4, so 1 - 4u > 1 - 4*(1/4) = 0. Therefore, 1 - 4u > 0 for x ≠ 1/2. Therefore, the term is positive, so the equation becomes 16(1 - x)(1 - 4x(1 - x)) = 1.

Therefore, let me try solving for λ=4:

16(1 - x)(1 - 4x(1 - x)) = 1.

Let me compute this equation. Let me denote this as:

Let me expand 1 - 4x(1 - x):

1 - 4x + 4x².

Then multiply by (1 - x):

(1 - x)(1 - 4x + 4x²) = ?

Let me expand term by term:

First, 1*(1 - 4x + 4x²) = 1 - 4x + 4x².

Then, -x*(1 - 4x + 4x²) = -x + 4x² - 4x³.

Adding together:

1 - 4x + 4x² - x + 4x² - 4x³ = 1 - 5x + 8x² - 4x³.

Therefore, the equation becomes:

16*(1 - 5x + 8x² - 4x³) = 1 => 16 - 80x + 128x² - 64x³ = 1 => -64x³ + 128x² - 80x + 15 = 0.

Multiply both sides by -1: 64x³ - 128x² + 80x - 15 = 0.

This is a cubic equation. Maybe I can try to find rational roots with Rational Root Theorem. Possible rational roots are factors of 15 over factors of 64: ±1, ±3, ±5, ±15, etc., over 1, 2, 4, 8, 16, 32, 64.

Testing x = 1: 64 - 128 + 80 -15 = 64 - 128 is -64 + 80 is 16 - 15 = 1 ≠ 0.

x = 3/4: let me compute 64*(27/64) - 128*(9/16) + 80*(3/4) - 15.

Simplify step by step:

First term: 64*(27/64) = 27.

Second term: 128*(9/16) = 8*9 = 72, so -72.

Third term: 80*(3/4) = 60.

Fourth term: -15.

So total: 27 - 72 + 60 -15 = (27 - 72) + (60 -15) = (-45) + 45 = 0.

Oh! x = 3/4 is a root!

Therefore, we can factor out (x - 3/4). Let me perform polynomial division or use synthetic division.

Let me write the cubic as 64x³ - 128x² + 80x - 15. Divided by (x - 3/4).

Alternatively, use synthetic division with root 3/4:

Coefficients: 64 | -128 | 80 | -15

Multiply by 3/4:

Start with 64.

Bring down 64.

Multiply by 3/4: 64*(3/4) = 48.

Add to next coefficient: -128 + 48 = -80.

Multiply by 3/4: -80*(3/4) = -60.

Add to next coefficient: 80 + (-60) = 20.

Multiply by 3/4: 20*(3/4) = 15.

Add to last coefficient: -15 + 15 = 0. Perfect.

Therefore, the cubic factors as (x - 3/4)(64x² - 80x + 20).

Now, factor the quadratic: 64x² - 80x + 20. Let me see:

Divide by 4: 16x² - 20x + 5.

Discriminant: (20)^2 - 4*16*5 = 400 - 320 = 80.

Therefore, roots are [20 ± sqrt(80)] / (2*16) = [20 ± 4√5]/32 = [5 ± √5]/8.

So, the roots are x = 3/4, x = [5 + √5]/8, and x = [5 - √5]/8.

Let me compute these:

[5 + √5]/8 ≈ (5 + 2.236)/8 ≈ 7.236/8 ≈ 0.9045.

[5 - √5]/8 ≈ (5 - 2.236)/8 ≈ 2.764/8 ≈ 0.3455.

So, the roots are approximately 0.3455, 0.9045, and 0.75.

But remember, we had to solve 16(1 - x)(...) = 1. However, we found that for λ=4, there are solutions x≈0.3455, 0.75, 0.9045. Let me check if these are period-2 points or fixed points.

First, x=3/4 = 0.75. Let's check if it's a fixed point. Compute f(0.75) = 4*(0.75)*(1 - 0.75) = 4*(0.75)*(0.25) = 4*(0.1875) = 0.75. So, yes! x=3/4 is a fixed point. Similarly, check x≈0.3455 and 0.9045.

Let me check x ≈ 0.3455. Let me compute f(x) for x≈0.3455:

f(x) = 4x(1 - x). Let me compute exact value for [5 - √5]/8:

Let me compute x = [5 - √5]/8. Then f(x) = 4x(1 - x). Let me compute 1 - x = 1 - [5 - √5]/8 = [8 - 5 + √5]/8 = [3 + √5]/8.

Therefore, f(x) = 4 * [5 - √5]/8 * [3 + √5]/8 = [ (5 - √5)(3 + √5) ] / 16.

Multiply numerator:

(5)(3) + 5√5 - 3√5 - (√5)^2 = 15 + 2√5 - 5 = 10 + 2√5.

Therefore, f(x) = [10 + 2√5]/16 = [5 + √5]/8 ≈ [5 + 2.236]/8 ≈ 7.236/8 ≈ 0.9045.

Then f(f(x)) = f([5 + √5]/8) = 4 * [5 + √5]/8 * [1 - [5 + √5]/8].

Let me compute 1 - [5 + √5]/8 = [8 - 5 - √5]/8 = [3 - √5]/8.

Thus, f(f(x)) = 4 * [5 + √5]/8 * [3 - √5]/8 = [ (5 + √5)(3 - √5) ] / 16.

Multiply numerator:

5*3 - 5√5 + 3√5 - (√5)^2 = 15 - 2√5 - 5 = 10 - 2√5.

Thus, f(f(x)) = [10 - 2√5]/16 = [5 - √5]/8, which is our original x. Therefore, yes, these two points form a 2-cycle: [5 - √5]/8 maps to [5 + √5]/8, which maps back. Therefore, they are period-2 points.

Similarly, for x ≈0.9045, which is [5 + √5]/8, f(x) = [5 - √5]/8, and then f(f(x)) = x. Therefore, these are period-2 points. However, note that x=3/4 is a fixed point. Therefore, for λ=4, there are period-2 points. Therefore, λ=4 is acceptable.

Therefore, λ=4 is a value where such x exists. So, the answer includes λ=4.

Now, the question is to find all λ between 0 and 4 inclusive.

So, how do we find all λ such that there exists x with f(f(x)) = x and f(x) ≠ x.

Since for λ=0, we saw that f(x)=0, so f(f(x))=0=x only when x=0. Therefore, no such x.

But for λ=1, let me check. For λ=1, f(x)=x(1 - x). Let's find if there are period-2 points.

First, fixed points: x=0 and x=(1 - 1)/1=0. Wait, the other fixed point is x=(λ - 1)/λ = (1 - 1)/1 = 0. So, for λ=1, there's only one fixed point x=0 (double root? Let me check f(x)=x.

f(x)=x(1 - x). Setting equal to x: x(1 - x) = x => x - x² = x => -x² = 0 => x=0. So only fixed point is x=0.

Now, solving f(f(x))=x. Let me compute f(f(x)):

f(f(x)) = f(x(1 - x)) = x(1 - x)[1 - x(1 - x)].

Setting equal to x:

x(1 - x)[1 - x(1 - x)] = x.

Which simplifies to x [ (1 - x)(1 - x + x²) - 1 ] = 0.

Wait, expanding:

x(1 - x)[1 - x(1 - x)] - x = 0 => x [ (1 - x)[1 - x(1 - x)] - 1 ] = 0.

Therefore, same as before.

Let me compute the expression inside:

(1 - x)[1 - x(1 - x)] - 1 = ?

Expand 1 - x(1 - x) = 1 - x + x².

Multiply by (1 - x):

(1 - x)(1 - x + x²) = (1 - x)^2 + x²(1 - x).

Wait, but earlier steps showed that this is 1 - 2x + 2x² - x³. Wait:

Earlier steps for general λ had:

(1 - x)(1 - λx + λx²) = 1 - (λ + 1)x + 2λx² - λx³.

For λ=1, that becomes:

1 - 2x + 2x² - x³.

Then subtract 1:

1 - 2x + 2x² - x³ - 1 = -2x + 2x² - x³.

Therefore, equation is x*(-2x + 2x² - x³) = 0.

Therefore, the non-zero solutions are solutions to -2x + 2x² - x³ = 0 => x(-2 + 2x - x²) = 0.

Therefore, x ≠ 0, so -2 + 2x - x² = 0 => x² - 2x + 2 = 0. Discriminant: 4 - 8 = -4 < 0. Therefore, no real solutions. Therefore, for λ=1, the equation f(f(x))=x only has x=0 as solution, which is fixed. Therefore, no period-2 points. Therefore, λ=1 is not acceptable.

Therefore, for λ=1, no such x exists.

Now, let me try for λ=2. Maybe there are period-2 points.

Alternatively, perhaps there is a way to find for which λ the equation λ²(1 - x)[1 - λx(1 - x)] = 1 has solutions different from fixed points.

Let me think about the general case. Let me assume that there exists a period-2 point. Then, since f(f(x)) = x, but f(x) ≠ x.

Alternatively, since we have the system f(x) = y and f(y) = x. Let me consider that for each x, f(x) = y and f(y)=x. Therefore, composing f twice gives x.

If we think of the graph of f(x), then looking for points where the graph of f intersects with the graph of f^{-1}(x), but since f isn't necessarily invertible, perhaps another approach.

Alternatively, note that for period-2 points, the function f must have a period-2 orbit. For quadratic maps like f(x) = λx(1 - x), the existence of period-2 points is related to the parameter λ.

In the logistic map, for λ in certain ranges, there are period-2 points. For example, in the logistic map, period-2 points exist when λ exceeds 3, but I need to verify this.

Wait, but let me check for λ=2. Let me try solving the equation. Let me take λ=2. Let me try solving the equation f(f(x)) = x, excluding fixed points.

First, compute fixed points. For λ=2, the fixed points are x=0 and x=(2 - 1)/2 = 1/2.

Check if x=1/2 is a fixed point: f(1/2) = 2*(1/2)*(1 - 1/2) = 2*(1/2)*(1/2) = 1/2. Yes.

So, fixed points are x=0 and x=1/2.

Now, solving f(f(x)) = x. Let me compute f(f(x)) = 2*(2x(1 - x))*(1 - 2x(1 - x)).

So, f(f(x)) = 4x(1 - x)[1 - 2x(1 - x)].

Setting equal to x:

4x(1 - x)[1 - 2x(1 - x)] = x.

Assuming x ≠ 0, divide both sides by x:

4(1 - x)[1 - 2x(1 - x)] = 1.

Compute 1 - 2x(1 - x) = 1 - 2x + 2x².

Multiply by (1 - x):

(1 - x)(1 - 2x + 2x²) = ?

Expand:

1*(1 - 2x + 2x²) - x*(1 - 2x + 2x²)

= 1 - 2x + 2x² - x + 2x² - 2x³

= 1 - 3x + 4x² - 2x³.

Therefore, the equation becomes:

4*(1 - 3x + 4x² - 2x³) = 1

=> 4 - 12x + 16x² - 8x³ = 1

=> -8x³ + 16x² - 12x + 3 = 0

Multiply by -1:

8x³ - 16x² + 12x - 3 = 0.

Let me check for rational roots. Possible roots are ±1, ±3, ±1/2, ±3/2, ±1/4, ±3/4, etc.

Test x=1: 8 - 16 + 12 - 3 = 1 ≠ 0.

x=3/2: 8*(27/8) - 16*(9/4) + 12*(3/2) - 3 = 27 - 36 + 18 - 3 = 6 ≠ 0.

x=1/2: 8*(1/8) - 16*(1/4) + 12*(1/2) - 3 = 1 - 4 + 6 - 3 = 0.

Ah! x=1/2 is a root. Therefore, factor out (x - 1/2).

Using synthetic division for polynomial 8x³ - 16x² + 12x - 3 with root 1/2:

Coefficients: 8 | -16 | 12 | -3

Bring down 8. Multiply by 1/2: 4. Add to next coefficient: -16 + 4 = -12. Multiply by 1/2: -6. Add to next coefficient: 12 + (-6) = 6. Multiply by 1/2: 3. Add to last coefficient: -3 + 3 = 0.

Thus, polynomial factors as (x - 1/2)(8x² - 12x + 6).

Now, solve 8x² - 12x + 6 = 0. Discriminant: 144 - 192 = -48 < 0. Therefore, no real roots.

Thus, the only real solution is x=1/2, which is a fixed point. Therefore, for λ=2, there are no period-2 points. Hence, λ=2 is not acceptable.

So, for λ=2, no such x exists.

Wait, but maybe I made a mistake here. Earlier, for λ=4, there were period-2 points. So perhaps there's a range of λ where period-2 points exist.

Alternatively, maybe we need to find λ such that the equation λ²(1 - x)[1 - λx(1 - x)] = 1 has solutions not equal to fixed points.

Alternatively, since we saw that for λ=4, there are such points. Let me try another value, say λ=3.

Let me try λ=3. Then, fixed points are x=0 and x=(3 - 1)/3 = 2/3.

Now, solving f(f(x)) = x. Let me compute f(f(x)):

f(x) = 3x(1 - x). Then f(f(x)) = 3*[3x(1 - x)]*[1 - 3x(1 - x)].

So, f(f(x)) = 9x(1 - x)[1 - 3x(1 - x)].

Setting equal to x:

9x(1 - x)[1 - 3x(1 - x)] = x.

Assuming x ≠ 0, divide both sides by x:

9(1 - x)[1 - 3x(1 - x)] = 1.

Expand 1 - 3x(1 - x) = 1 - 3x + 3x².

Multiply by (1 - x):

(1 - x)(1 - 3x + 3x²) = ?

Expand term by term:

1*(1 - 3x + 3x²) - x*(1 - 3x + 3x²)

= 1 - 3x + 3x² - x + 3x² - 3x³

= 1 - 4x + 6x² - 3x³.

So equation becomes:

9*(1 - 4x + 6x² - 3x³) = 1 =>

9 - 36x + 54x² - 27x³ = 1 =>

-27x³ + 54x² - 36x + 8 = 0 => Multiply by -1:

27x³ - 54x² + 36x - 8 = 0.

Check for rational roots. Possible roots are ±1, ±2, ±4, ±8, divided by factors of 27. Let me try x=2/3:

Plug x=2/3:

27*(8/27) - 54*(4/9) + 36*(2/3) - 8

= 8 - 54*(4/9) + 24 - 8

= 8 - 24 + 24 - 8 = 0.

Yes! x=2/3 is a root. So factor out (x - 2/3).

Using synthetic division:

Coefficients: 27 | -54 | 36 | -8

Root x=2/3:

Bring down 27. Multiply by 2/3: 18. Add to next coefficient: -54 + 18 = -36. Multiply by 2/3: -24. Add to next coefficient: 36 + (-24) = 12. Multiply by 2/3: 8. Add to last coefficient: -8 + 8 = 0.

Thus, the polynomial factors as (x - 2/3)(27x² - 36x + 12).

Simplify the quadratic: 27x² - 36x + 12 = 3*(9x² - 12x + 4) = 3*(3x - 2)^2.

Therefore, the roots are x=2/3 and x=2/3 (double root). Thus, the only real solutions are x=2/3 and x=2/3. Therefore, all solutions are the fixed points. Therefore, for λ=3, there are no period-2 points.

Wait, but that contradicts some intuition. Maybe I made an error here.

Wait, let me check with λ=3. If I consider period-2 points, then perhaps they are complex? But we need real points. So perhaps for λ=3, there are no period-2 points. But earlier for λ=4, there are.

So, perhaps there is a critical value of λ where period-2 points emerge. Let me think.

Alternatively, maybe I need to find the values of λ for which the equation λ²(1 - x)[1 - λx(1 - x)] = 1 has solutions other than the fixed points.

Alternatively, going back to the system of equations:

From f(x) = y and f(y) = x, we have:

y = λx(1 - x)

and

x = λy(1 - y).

Let me try to eliminate y.

From the first equation, y = λx(1 - x). Substitute into the second equation:

x = λ [λx(1 - x)] [1 - λx(1 - x)].

Which leads to x = λ²x(1 - x)[1 - λx(1 - x)].

If x ≠ 0, divide both sides by x:

1 = λ²(1 - x)[1 - λx(1 - x)].

Let me denote z = x. Then:

1 = λ²(1 - z)[1 - λz(1 - z)].

Let me define this as:

Let me let u = z(1 - z). Then the equation becomes:

1 = λ²(1 - z)(1 - λu).

But since u = z(1 - z), then 1 - z = ?

Wait, maybe express 1 - z in terms of u. Let me see:

Let me consider that z and 1 - z are related to u. However, unless we have another equation, it's not straightforward.

Alternatively, maybe express 1 - z as follows. Let me denote 1 - z = t. Then z = 1 - t, and u = z(1 - z) = (1 - t)t = t - t².

Therefore, the equation becomes:

1 = λ² t [1 - λ(t - t²)] = λ² t [1 - λt + λt²].

But this seems similar to previous steps.

Alternatively, maybe consider that for the system y = λx(1 - x) and x = λy(1 - y), we can write:

Multiply the two equations: x y = λx(1 - x) * λy(1 - y) = λ² x y (1 - x)(1 - y).

Assuming x y ≠ 0, we can divide both sides by x y:

1 = λ² (1 - x)(1 - y).

Therefore, (1 - x)(1 - y) = 1 / λ².

But from the original equations, y = λx(1 - x), so let me substitute y into this equation:

(1 - x)(1 - λx(1 - x)) = 1 / λ².

But this is the same equation as before: λ²(1 - x)(1 - λx(1 - x)) = 1.

So, same equation.

Therefore, we have this relation.

Let me think if there is a way to relate this to the fixed points.

Alternatively, since we know that fixed points satisfy f(x) = x. Therefore, for period-2 points, we have that f(x) ≠ x but f(f(x)) = x. Let me consider that for period-2 points, the orbit {x, y} satisfies f(x) = y and f(y) = x.

Therefore, for such points, we can consider that the product (1 - x)(1 - y) = 1 / λ².

But also, since y = λx(1 - x) and x = λy(1 - y), we can write:

(1 - x)(1 - y) = 1 / λ².

Let me expand 1 - x and 1 - y:

1 - x = 1 - x

1 - y = 1 - λx(1 - x)

Therefore, their product is 1 / λ².

Alternatively, since we have two equations:

1. y = λx(1 - x)

2. x = λy(1 - y)

Let me try to express these as:

From 1: y = λx - λx²

From 2: x = λy - λy²

Substituting y from 1 into 2:

x = λ(λx - λx²) - λ(λx - λx²)^2

Simplify:

x = λ²x - λ²x² - λ[λ²x² - 2λ²x³ + λ²x^4]

Wait, expanding (λx - λx²)^2:

= (λx)^2 - 2*(λx)*(λx²) + (λx²)^2 = λ²x² - 2λ²x³ + λ²x^4.

Therefore,

x = λ²x - λ²x² - λ[λ²x² - 2λ²x³ + λ²x^4]

= λ²x - λ²x² - λ³x² + 2λ³x³ - λ³x^4.

Bring all terms to one side:

x - λ²x + λ²x² + λ³x² - 2λ³x³ + λ³x^4 = 0.

Factor x:

x[1 - λ² + λ²x + λ³x² - 2λ³x² + λ³x³] = 0.

Wait, expanding the terms inside:

x[1 - λ² + (λ²x + λ³x² - 2λ³x²) + λ³x³] = 0.

Simplify:

x[1 - λ² + λ²x - λ³x² + λ³x³] = 0.

So, the equation is:

x[1 - λ² + λ²x - λ³x² + λ³x³] = 0.

Which is the same as:

x[λ³x³ - λ³x² + λ²x + (1 - λ²)] = 0.

This seems similar to the previous equations we had.

But since we are considering x ≠ 0, then the other factor is zero:

λ³x³ - λ³x² + λ²x + 1 - λ² = 0.

Let me write this as:

λ³x³ - λ³x² + λ²x + (1 - λ²) = 0.

This is a cubic equation in x.

Alternatively, maybe we can factor this equation. Let me try to factor it.

Let me group terms:

(λ³x³ - λ³x²) + (λ²x + 1 - λ²) = 0

= λ³x²(x - 1) + (λ²x + 1 - λ²) = 0.

Not sure if helpful.

Alternatively, check if x=1 is a solution:

Plug x=1:

λ³(1) - λ³(1) + λ²(1) + 1 - λ² = 0 + λ² + 1 - λ² = 1 ≠ 0. Not a root.

Check x=1/λ? Maybe not. Let me try x = (something).

Alternatively, since this seems too complicated, perhaps go back to the previous equations.

Since we know that for λ=4, there are solutions. Maybe there's a critical value of λ where period-2 points emerge.

Alternatively, think about the function g(x) = f(f(x)) - x. We want this function to have roots that are not fixed points.

The number of real roots of g(x) = 0 depends on λ.

For the logistic map, period-2 points exist when λ is in a certain range. I think that for λ > 2, the map has period-2 points. But I need to verify this.

Alternatively, recall that for the logistic map f(x) = λx(1 - x), the period-2 points occur when the equation f(f(x)) = x has solutions other than fixed points.

The fixed points are solutions of x = λx(1 - x), which we already considered.

Let me consider the condition for existence of period-2 points.
From the system f(x) = y and f(y) = x, with x ≠ y.

Let me assume that x and y are distinct.

Then, from the above, we have:

From y = λx(1 - x) and x = λy(1 - y), we can derive that:

x = λ^2 x(1 - x)(1 - λx(1 - x)).

As before.

Let me define the function h(x) = λ²(1 - x)(1 - λx(1 - x)) - 1 = 0. We want this equation to have solutions other than fixed points.

For λ=4, we saw that there are such solutions. Let me consider what happens when λ varies from 0 to 4.

At λ=0: h(x) = 0 - 1 = -1 ≠ 0, no solutions.

At λ approaching 0, similar.

At λ=1, we saw no solutions.

At λ=2, no solutions.

At λ=3, no solutions.

At λ=4, solutions exist.

Therefore, there must be a range of λ where the equation has solutions, starting from some λ_c where period-2 points appear.

Alternatively, find the λ for which the equation λ²(1 - x)[1 - λx(1 - x)] = 1 has solutions other than fixed points.

Let me consider the case when x is a period-2 point. Then, f(x) ≠ x, so x is not a fixed point. Let me assume that there exists such an x in [0,1].

Let me consider that for x in [0,1], the function u = x(1 - x) is between 0 and 1/4.

Let me consider the equation:

λ²(1 - x)(1 - λu) = 1, where u = x(1 - x).

Let me think of this as an equation in terms of u and x.

But since u = x(1 - x), we can relate 1 - x to u. Let me note that:

Let me denote t = 1 - x, then x = 1 - t, and u = (1 - t)t = t - t².

So, substituting into the equation:

λ² * t * [1 - λ(t - t²)] = 1

=> λ² t [1 - λt + λt²] = 1.

Let me denote this as:

λ² t (1 - λt + λt²) = 1.

This seems similar to the previous expression.

Let me consider this as a cubic in t: λ²(λ t³ - λ t² + t) = 1.

Wait, expanding:

λ² [λ t³ - λ t² + t] = 1 => λ³ t³ - λ³ t² + λ² t - 1 = 0.

So, the equation becomes:

λ³ t³ - λ³ t² + λ² t - 1 = 0.

This is a cubic equation in t. Let me see if I can find solutions.

Alternatively, let me check for possible values of t. Since t = 1 - x, and x ∈ [0,1], t ∈ [0,1]. So t ∈ [0,1].

Let me check if this cubic equation has solutions in t ∈ [0,1].

Let me consider the cubic function:

g(t) = λ³ t³ - λ³ t² + λ² t - 1.

We want to find λ such that g(t)=0 has a solution t ∈ (0,1) (since x ≠ 0, t = 1 - x ≠ 1, so t ∈ [0,1), but x ≠ 1/2? Not sure.

Let me analyze the behavior of g(t):

At t=0: g(0) = 0 - 0 + 0 - 1 = -1 < 0.

At t=1: g(1) = λ³(1) - λ³(1) + λ²(1) - 1 = 0 + λ² - 1.

So, g(1) = λ² - 1.

Therefore, if λ² - 1 > 0, i.e., λ > 1, then g(1) > 0. Therefore, by Intermediate Value Theorem, there is a root in (0,1).

If λ² - 1 = 0, i.e., λ=1, then g(1)=0. But for λ=1, we saw that the only solution is t=1 (i.e., x=0), which is a fixed point.

If λ < 1, then g(1) = λ² - 1 < 0, so g(0) = -1 < 0 and g(1) < 0, so no root in [0,1].

Therefore, for λ > 1, there is at least one root in (0,1). But we need to ensure that this root corresponds to a solution x ≠ fixed point.

However, when λ > 1, there is a fixed point at x = (λ - 1)/λ = 1 - 1/λ. Let me check if the root t in (0,1) corresponds to this fixed point or not.

If t corresponds to x = 1 - x_fixed = 1 - (1 - 1/λ) = 1/λ.

Wait, x = 1 - t. If t = 1 - x, then if x is the fixed point x_fixed = 1 - 1/λ, then t = 1 - x_fixed = 1 - (1 - 1/λ) = 1/λ.

Therefore, let me check if t=1/λ is a root of g(t).

Let me substitute t=1/λ into g(t):

g(1/λ) = λ³*(1/λ)^3 - λ³*(1/λ)^2 + λ²*(1/λ) - 1

= λ³*(1/λ³) - λ³*(1/λ²) + λ²*(1/λ) - 1

= 1 - λ + λ - 1 = 0.

Yes! So t=1/λ is a root of g(t). Therefore, for λ > 1, the cubic equation g(t) = 0 has a root at t=1/λ, which corresponds to x = 1 - t = 1 - 1/λ, which is the fixed point.

Therefore, the other roots of g(t)=0 (if any) are period-2 points.

Since it's a cubic equation, there are three roots (counting multiplicity). We know one root is t=1/λ. Let me perform polynomial division to factor this root out.

Divide g(t) by (t - 1/λ).

Let me write g(t) = λ³ t³ - λ³ t² + λ² t - 1.

Let me factor out (t - 1/λ). Let me use polynomial division or synthetic division.

Let me use synthetic division with root t=1/λ. Let me denote a = 1/λ.

Coefficients of g(t): λ³, -λ³, λ², -1.

Let me write them as coefficients of t³, t², t, constant.

For synthetic division, we use root a = 1/λ:

Bring down λ³.

Multiply by a: λ³ * 1/λ = λ².

Add to next coefficient: -λ³ + λ².

Multiply by a: (-λ³ + λ²) * 1/λ = -λ² + λ.

Add to next coefficient: λ² + (-λ² + λ) = λ.

Multiply by a: λ * 1/λ = 1.

Add to last coefficient: -1 + 1 = 0.

Thus, the cubic factors as (t - 1/λ)(λ³ t² + (-λ³ + λ²) t + λ).

Therefore, g(t) = (t - 1/λ)(λ³ t² - λ³ t + λ² t + λ)? Wait, let me check:

Wait, after division, the quotient polynomial is λ³ t² + (-λ³ + λ²) t + λ.

Therefore:

g(t) = (t - 1/λ)(λ³ t² + (-λ³ + λ²) t + λ).

Simplify the quadratic factor:

λ³ t² + ( -λ³ + λ² )t + λ = λ³ t² - λ³ t + λ² t + λ.

Let me factor λ from the last two terms:

= λ³ t² - λ³ t + λ(λ t + 1).

Wait, maybe better to factor terms:

Let me write as:

λ³ t² - λ³ t + λ² t + λ = λ³ t(t - 1) + λ(λ t + 1).

Not sure. Alternatively, let me factor λ from the entire expression:

= λ[λ² t² - λ² t + λ t + 1].

Wait, let me check:

If I factor λ from the first three terms and the last term:

Wait:

λ³ t² - λ³ t + λ² t + λ = λ³ t² - λ³ t + λ² t + λ.

Yes, so factor λ:

= λ[λ² t² - λ² t + λ t] + λ = λ[λ² t² - λ² t + λ t + 1].

Wait, no, the last term is λ, not 1. Wait:

Actually, it's:

= λ³ t² - λ³ t + λ² t + λ = λ³ t² - λ³ t + λ² t + λ.

To factor λ from all terms:

= λ[λ² t² - λ² t + λ t + 1].

Wait, λ * λ² t² = λ³ t², λ * -λ² t = -λ³ t, λ * λ t = λ² t, and λ * 1 = λ. Correct.

Therefore, g(t) = (t - 1/λ) * λ [λ² t² - λ² t + λ t + 1].

Wait, but that seems inconsistent with the previous division. Wait, no, after factoring out (t - 1/λ), the quotient was λ³ t² + ... but I think my previous step was miscalculating. Let me check again.

Wait, after synthetic division, the quotient polynomial is λ³ t² + (-λ³ + λ²) t + λ. Therefore, the full factorization is:

g(t) = (t - 1/λ)(λ³ t² + (-λ³ + λ²)t + λ).

Let me factor out λ from the quadratic terms:

= (t - 1/λ) * [λ³ t² - λ³ t + λ² t + λ].

Let me factor λ from the last three terms? No, it's λ³ t² - λ³ t + λ² t + λ. Let me rearrange:

= λ³ t² - λ³ t + λ² t + λ

= λ³ t² - λ² t(λ - 1) + λ.

Hmm. Not helpful. Alternatively, let me write it as:

= λ³ t² - λ³ t + λ² t + λ

= λ³ t(t - 1) + λ(λ t + 1).

Hmm. Still not helpful.

Let me denote the quadratic equation as:

λ³ t² - λ³ t + λ² t + λ = 0.

Divide both sides by λ (since λ > 0):

λ² t² - λ² t + λ t + 1 = 0.

Wait, no, dividing each term by λ:

(λ³ t²)/λ - (λ³ t)/λ + (λ² t)/λ + λ/λ = λ² t² - λ² t + λ t + 1 = 0.

Yes, so the equation becomes:

λ² t² - λ² t + λ t + 1 = 0.

Let me group terms:

λ² t² - λ² t + λ t + 1 = λ² t(t - 1) + λ t + 1.

It's still complicated. Let me write it as:

λ² t² + (-λ² + λ) t + 1 = 0.

So, quadratic equation in t:

A t² + B t + C = 0, where:

A = λ²,

B = -λ² + λ,

C = 1.

Let me compute the discriminant D = B² - 4AC.

D = (-λ² + λ)^2 - 4 * λ² * 1 = (λ² - λ)^2 - 4λ².

Expand (λ² - λ)^2 = λ^4 - 2λ³ + λ².

So, D = λ^4 - 2λ³ + λ² - 4λ² = λ^4 - 2λ³ - 3λ² = λ²(λ² - 2λ - 3).

Factor λ² - 2λ - 3:

λ² - 2λ - 3 = (λ - 3)(λ + 1).

Therefore, D = λ²(λ - 3)(λ + 1).

Since λ ≥ 0, λ + 1 > 0, and λ² ≥ 0. Therefore, D ≥ 0 if (λ - 3) ≥ 0, i.e., λ ≥ 3.

For λ < 3, D < 0, so no real roots. For λ = 3, D = 0, so one real root (double root). For λ > 3, D > 0, two real roots.

Therefore, the quadratic equation λ² t² - λ² t + λ t + 1 = 0 has real solutions only when λ ≥ 3.

Thus, for λ ≥ 3, the equation g(t) = 0 has three real roots: t=1/λ (fixed point), and two roots from the quadratic equation.

But since the quadratic equation is λ² t² - λ² t + λ t + 1 = 0, which for λ ≥ 3, has real roots.

Therefore, for λ ≥ 3, the cubic equation g(t) = 0 has three real roots: one at t=1/λ (fixed point) and two others from the quadratic. Let me check if these other roots are in (0,1).

Let me consider λ > 3. Let me find the roots of the quadratic equation:

λ² t² + (-λ² + λ) t + 1 = 0.

Using quadratic formula:

t = [λ² - λ ± √D]/(2λ²), where D = λ²(λ - 3)(λ + 1).

So:

t = [λ² - λ ± λ√((λ - 3)(λ + 1))]/(2λ²).

Simplify numerator:

Factor λ from first two terms:

= [λ(λ - 1) ± λ√((λ - 3)(λ + 1))]/(2λ²)

= [ (λ - 1) ± √((λ - 3)(λ + 1)) ] / (2λ).

Therefore, the roots are:

t = [λ - 1 + √((λ - 3)(λ + 1))]/(2λ) and t = [λ - 1 - √((λ - 3)(λ + 1))]/(2λ).

Now, we need to check if these roots are in (0,1).

Let me analyze for λ ≥ 3.

Let me compute the two roots:

First root: [λ - 1 + √((λ - 3)(λ + 1))]/(2λ).

Second root: [λ - 1 - √((λ - 3)(λ + 1))]/(2λ).

Let me check the first root. Let me denote sqrt term as:

√((λ - 3)(λ + 1)) = √(λ² - 2λ - 3).

Let me check for λ=3: sqrt(9 - 6 - 3) = sqrt(0) = 0. So, t = [3 - 1 + 0]/(6) = 2/6 = 1/3. Which is in (0,1).

For λ > 3, the sqrt term is positive. Let me check if the first root is in (0,1):

Numerator: λ - 1 + sqrt(...). Since sqrt(...) > 0, numerator > λ - 1.

But λ - 1 > 0 for λ > 1. Therefore, the numerator is positive.

Denominator is 2λ > 0. So, t is positive.

Now, check if t < 1. Let me see:

[λ - 1 + sqrt(λ² - 2λ - 3)] / (2λ) < 1 ?

Multiply both sides by 2λ (positive):

λ - 1 + sqrt(λ² - 2λ - 3) < 2λ

=> -1 + sqrt(λ² - 2λ - 3) < λ

=> sqrt(λ² - 2λ - 3) < λ + 1

Square both sides (since both sides are positive for λ ≥ 3):

λ² - 2λ - 3 < (λ + 1)^2 = λ² + 2λ + 1

Subtract λ²:

-2λ - 3 < 2λ + 1 => -3 -1 < 2λ + 2λ => -4 < 4λ. Which is true for all λ ≥ 3.

Therefore, the first root is less than 1.

Similarly, for the second root: [λ - 1 - sqrt(...)]/(2λ). Let me check if it is positive.

Compute numerator: λ - 1 - sqrt(...). For λ=3, this is 2 - 0 = 2 > 0. For λ > 3, need to check if λ - 1 > sqrt(...).

Let me square both sides (assuming λ - 1 > 0, which it is for λ > 1):

(λ - 1)^2 > λ² - 2λ - 3

Expand left side: λ² - 2λ + 1 > λ² - 2λ - 3

Subtract λ² - 2λ from both sides: 1 > -3, which is True. Therefore, λ - 1 > sqrt(...), so numerator is positive. Therefore, the second root is positive.

Now check if it's less than 1:

[λ - 1 - sqrt(λ² - 2λ - 3)] / (2λ) < 1 ?

Multiply both sides by 2λ: λ - 1 - sqrt(...) < 2λ => -1 - sqrt(...) < λ. Which is true since sqrt(...) is positive and λ ≥ 3. So yes, it's less than 1.

Therefore, for λ ≥ 3, there are two additional roots t in (0,1) besides t=1/λ. Therefore, these correspond to period-2 points.

Therefore, for λ ≥ 3, there are period-2 points.

But wait, we need to check if these roots t correspond to x ≠ fixed points.

Since t = 1 - x, and the fixed point corresponds to t = 1/λ, the other solutions for t are different from 1/λ. Therefore, these correspond to x ≠ (1 - 1/λ), hence period-2 points.

Therefore, for λ ≥ 3, there are period-2 points.

But earlier, for λ=3, we saw that the quadratic equation had a double root? Wait, when λ=3, discriminant D = 0, so only one root from quadratic, which is [λ - 1]/(2λ) ??? Wait, for λ=3, the discriminant D = 0, so the quadratic equation has a double root: t = [λ - 1]/(2λ) ? Let me compute:

For λ=3, the roots are [3 - 1 ± 0]/(2*3) = 2/6 = 1/3. So, double root at t=1/3. Therefore, the cubic equation g(t) = 0 has roots t=1/3 (double root) and t=1/λ = 1/3. Wait, no: for λ=3, the roots are t=1/λ=1/3 (from the first factor) and the quadratic equation gives t=1/3 (double root). Therefore, overall, the cubic equation has a triple root at t=1/3? Let me check:

Original equation g(t) = 0 for λ=3:

g(t) = 3³ t³ - 3³ t² + 3² t - 1 = 27t³ - 27t² + 9t - 1.

Let me check if t=1/3 is a root:

27*(1/27) - 27*(1/9) + 9*(1/3) - 1 = 1 - 3 + 3 - 1 = 0.

Yes. And since it's a double root, let me perform division.

Divide g(t) by (t - 1/3):

Using synthetic division for t=1/3:

Coefficients: 27 | -27 | 9 | -1

Bring down 27. Multiply by 1/3: 9. Add to next coefficient: -27 + 9 = -18. Multiply by 1/3: -6. Add to next coefficient: 9 + (-6) = 3. Multiply by 1/3: 1. Add to last coefficient: -1 + 1 = 0.

So quotient is 27t² - 18t + 3. Which factors as 3(9t² - 6t + 1) = 3(3t - 1)^2.

Thus, g(t) = (t - 1/3) * 3(3t - 1)^2 = 3(t - 1/3)(3t - 1)^2.

Therefore, t=1/3 is a root with multiplicity 3? Wait, no: (t - 1/3) and (3t - 1)^2. But 3t - 1 = 0 => t = 1/3. Therefore, g(t) has a triple root at t=1/3.

Therefore, for λ=3, the cubic equation has a triple root at t=1/3, which corresponds to x=1 - t = 1 - 1/3 = 2/3, which is the fixed point. Therefore, all solutions are fixed points, and no period-2 points? But earlier, when λ=3, I found that the equation f(f(x))=x had only fixed points. But according to this, there are roots at t=1/3 (triple root). So, all roots are fixed points. Hence, no period-2 points for λ=3?

But earlier when I tried λ=3, I found only fixed points.

But for λ > 3, we have two distinct roots from the quadratic, giving period-2 points.

Therefore, for λ > 3, there are period-2 points, and for λ=3, only fixed points.

Thus, the values of λ for which there exist period-2 points are λ > 3, and λ=4 is included.

But we need to check for λ=3, whether there are period-2 points. Since for λ=3, all solutions are fixed points, as shown above.

Therefore, the values of λ are those in (3, 4].

But we need to check for λ=3, whether there are any period-2 points. From the previous analysis, no, since all roots are fixed points.

Therefore, the values of λ are those for which λ > 3. However, need to confirm this.

But for λ=3, there are no period-2 points. For λ > 3, there are.

But wait, what about λ=3. Let me check with λ=3.5.

Take λ=3.5. Let me check if there are period-2 points.

Let me try to solve the equation.

The roots of g(t) = 0 are t=1/λ and the roots from quadratic.

For λ=3.5, which is 7/2:

Discriminant D = λ²(λ - 3)(λ + 1).

Let me compute it:

D = (49/4)(0.5)(4.5) = (49/4)*(9/4) = 441/16.

So sqrt(D) = 21/4.

Then, roots from quadratic are:

t = [λ - 1 ± sqrt(D)] / (2λ) = [3.5 - 1 ± 21/4]/(7).

Compute numerator:

3.5 - 1 = 2.5 = 5/2.

So:

First root: [5/2 + 21/4]/7 = (10/4 + 21/4)/7 = 31/4 / 7 = 31/28 ≈ 1.107. But t ∈ [0,1], so this is invalid.

Wait, that can't be. Wait, what's wrong?

Wait, previously I had:

t = [λ - 1 ± √((λ - 3)(λ + 1))]/(2λ).

Wait, sqrt((λ - 3)(λ + 1)) for λ=3.5:

(λ - 3)(λ + 1) = (0.5)(4.5) = 2.25, sqrt(2.25)=1.5.

Therefore, t = [3.5 - 1 ± 1.5]/(2*3.5) = [2.5 ± 1.5]/7.

So, first root: (2.5 + 1.5)/7 = 4/7 ≈ 0.571, which is in (0,1). Second root: (2.5 - 1.5)/7 = 1/7 ≈ 0.142, also in (0,1).

Ah! I must have made a mistake in earlier calculation. The sqrt((λ - 3)(λ + 1)) is sqrt(2.25) = 1.5, not 21/4. Earlier, I incorrectly computed sqrt(D) as 21/4, but D = λ²(...) = (49/4)(0.5)(4.5) = (49/4)(9/4) = 441/16, which is sqrt(441/16)=21/4. But in reality, the sqrt((λ - 3)(λ + 1)) is sqrt(2.25)=1.5, and the formula is t = [λ - 1 ± √(λ² - 2λ - 3)]/(2λ). Wait, earlier I had:

sqrt((λ - 3)(λ + 1)) is sqrt(λ² - 2λ - 3).

For λ=3.5, λ² - 2λ - 3 = 12.25 - 7 - 3 = 2.25, which gives sqrt(2.25)=1.5.

So, the roots are [λ - 1 ± 1.5]/(2λ).

Therefore, correct calculation gives two roots in (0,1).

Therefore, for λ > 3, we have two additional roots t in (0,1), corresponding to period-2 points.

For λ=3, discriminant D=0, so sqrt(...)=0, giving t=[3 - 1]/(2*3)= 2/6=1/3, which is the fixed point.

Therefore, for λ > 3, there are period-2 points, and for λ=3, no period-2 points.

Therefore, the values of λ where such x exists are λ in (3, 4].

However, need to check if for λ > 3 up to 4, there are period-2 points, and also whether for λ=4, period-2 points exist, which we saw earlier.

But what about λ=3. Let me check for λ=3.0, whether there are period-2 points.

As analyzed, the only roots are the fixed points, so no period-2 points.

Therefore, the answer is λ in (3, 4].

But we need to check if for λ=3, there are no period-2 points, and for λ just above 3, there are.

But the question says 0 ≤ λ ≤ 4. So the values of λ are those in (3, 4].

But let me confirm with λ=3. Let me check if there are period-2 points.

For λ=3, the system f(x) = y, f(y)=x. Let me try to find solutions.

From earlier, for λ=3, the equation reduces to having only fixed points.

But let me check manually. Take λ=3. Suppose there is a period-2 point x ≠ 0 and x ≠ 2/3. Let me try to find such a point.

Assume x is a period-2 point: f(f(x)) = x, f(x) ≠ x. Let me attempt to solve f(f(x)) = x for λ=3.

f(f(x)) = 3*(3x(1 - x))*(1 - 3x(1 - x)) = 9x(1 - x)[1 - 3x(1 - x)].

Setting equal to x:

9x(1 - x)[1 - 3x(1 - x)] = x.

Assuming x ≠ 0, divide by x:

9(1 - x)[1 - 3x(1 - x)] = 1.

Expand 1 - 3x(1 - x) = 1 - 3x + 3x².

Multiply by (1 - x):

(1 - x)(1 - 3x + 3x²) = 1 - 3x + 3x² - x + 3x² - 3x³ = 1 - 4x + 6x² - 3x³.

Set equal to 1/9:

Wait, wait, original equation after dividing by x is 9*(...) = 1, so:

9*(1 - 4x + 6x² - 3x³) = 1 => -27x³ + 54x² - 36x + 8 = 0.

Which was the same equation as before, which had a triple root at x=2/3. Therefore, only fixed points.

Therefore, no period-2 points for λ=3.

Therefore, the answer is λ in (3, 4].

But wait, what about λ=2. Let me check for λ=2.5, which is greater than 3? No, λ=2.5 is less than 3. For λ=3.5, which is greater than 3, there are period-2 points.

Therefore, the values of λ are those where 3 < λ ≤ 4.

But the problem says 0 ≤ λ ≤ 4. Therefore, the values of λ are in the open interval (3, 4].

But let me check if at λ=4, which is included, there are period-2 points, which we saw earlier.

Yes, for λ=4, there are period-2 points.

But the question says "values of λ, 0 ≤ λ ≤ 4, for which there exists an x ∈ [0,1] such that f(x) ≠ x but f(f(x)) = x."

Therefore, the answer is λ ∈ (3, 4].

But let me check if there's a lower bound. For example, is there any λ < 3 where such x exists?

Earlier, for λ=2, λ=1, λ=3, there were no period-2 points. For λ=4, yes.

But let me think of another approach. Suppose that for λ in (3, 4], the equation f(f(x)) = x has period-2 points.

Therefore, the answer should be 3 < λ ≤ 4.

But to make sure, let me check λ=3. Let me take λ approaching 3 from above. For example, λ=3.0001.

Then, the quadratic equation would have two roots, giving period-2 points. Therefore, the interval is (3, 4].

But what about λ=3? At λ=3, only fixed points. So, open interval at 3.

Therefore, the answer is λ ∈ (3, 4].

But the question says "values of λ, 0 ≤ λ ≤ 4", and asks to "find the values of λ...".

Therefore, the answer is λ in (3, 4].

But to check if there are any other values, like for example λ=4 is included.

Yes, as shown earlier.

But let me check for λ=3. Let me take λ slightly above 3, say λ=3.1.

The discriminant D = λ²(λ - 3)(λ + 1) is positive, so two real roots. Hence, period-2 points exist.

Therefore, the final answer is λ ∈ (3, 4].

But in the problem statement, the answer should be in boxed notation. Since the interval is from 3 to 4, excluding 3 and including 4.

But let me check if at λ=4, whether there are points where f(f(x)) = x but f(x) ≠ x. As verified earlier, yes.

So, the values of λ are those greater than 3 and less than or equal to 4.

However, the question says "0 ≤ λ ≤ 4", so the answer is λ ∈ (3, 4].

But the original problem may expect a specific value or range. Let me check again.

Wait, in the logistic map, period-2 points exist for λ in (3, 4]. So, this is consistent with our analysis.

Therefore, the answer is all λ such that 3 < λ ≤ 4.

But the question says "values of λ", so the answer should be written as λ ∈ (3, 4].

But in the context of the problem, they might expect the answer as an interval.

However, since the problem says "values of λ, 0 ≤ λ ≤ 4", and asking to "find the values", so we need to write the range.

But let me check if λ=3 is excluded, and λ=4 is included. Therefore, the answer is all real numbers λ with 3 < λ ≤ 4.

But to express this in boxed notation, since the problem might expect specific values or a range, but since it's a continuous range, I need to write it as 3 < λ ≤ 4.

However, in some textbooks, the period-2 points for the logistic map occur when λ is in (3, 4], which matches our analysis.

Thus, the final answer is all λ in the interval (3, 4].

But the problem says "values of λ", so likely 3 < λ ≤ 4.

But let me check if there are other possibilities. For example, maybe some λ < 3 also have period-2 points.

Suppose λ=2. Let me imagine if there is a period-2 point.

For λ=2, we saw that the equation reduces to only fixed points. What about λ=2. Let me check with some value.

For example, take x=1/2. f(x)=2*(1/2)*(1 - 1/2)= 2*(1/2)*(1/2)= 1/2. So it's a fixed point.

Try x=1/3: f(x)=2*(1/3)*(2/3)= 4/9. Then f(f(x)) = f(4/9) = 2*(4/9)*(5/9) = 40/81 ≈ 0.4938 ≠ 1/3.

Try x=0.4: f(x)=2*0.4*0.6=0.48. f(f(x))=f(0.48)=2*0.48*0.52= 2*0.2496=0.4992 ≈ 0.5, which is not x=0.4.

Not equal.

Another try: Let me attempt to solve for λ=2, the equation f(f(x))=x.

We had previously for λ=2: the equation reduces to 8x³ - 16x² + 12x - 3 = 0, which has root x=1/2 (fixed point) and others complex.

Hence, no period-2 points.

Thus, for λ < 3, no period-2 points.

Thus, the answer is 3 < λ ≤ 4.

But the problem says "Find the values of λ, 0 ≤ λ ≤ 4, ...", and in the context of the problem, maybe they expect λ in (3, 4], so the answer is \boxed{(3, 4]}, but since the answer must be in boxed and likely as specific values or a range. However, the original problem might expect to write the interval as 3 < λ ≤ 4, and in boxed notation. But maybe the answer is λ = 4 only? But no, we saw for λ=4 there are period-2 points, and also for other values between 3 and 4.

But maybe there's a mistake in my analysis. Let me recall that in the logistic map, the period-2 points exist for λ in (3, 4]. That is a standard result. For example, the logistic map has a period-2 cycle for λ ∈ (3, 4].

Therefore, I think this is correct.

Thus, the values of λ are those in the open interval from 3 to 4, including 4.

So, the final answer is λ ∈ (3, 4], so in boxed notation, probably written as \boxed{(3, 4]} or if specific values are needed, but since it's a continuous range, I think the first option is appropriate.

But let me check if the original problem might have a different answer. For example, sometimes these problems specify discrete values, but in this case, it's a continuous range.

Alternatively, maybe the problem expects to write the values of λ for which there exists such x, and since the question says "values", it might be that there's a specific value, but no, our analysis shows it's an interval.

Alternatively, check if for λ=3. Let me think if there are period-2 points.

But as analyzed, no.

Thus, I conclude that the answer is λ in the interval (3, 4].

But the problem might want the answer in terms of λ > 3 and ≤ 4, so I need to check the exact wording: "Find the values of λ, 0 ≤ λ ≤ 4, for which there exists an x ∈ [0,1] such that f(x) ≠ x but f(f(x)) = x."

So, the answer is all real numbers λ between 3 and 4, not including 3 and including 4.

Therefore, in boxed notation, since ranges are not typically boxed with parentheses, but sometimes written with inequalities. However, given the standard practice, I think the answer is λ ∈ (3, 4], so boxed as \boxed{(3, 4]}.

But the original problem might expect specific values, but given the analysis, it's an interval.

Alternatively, maybe there is a specific value, but I don't think so.

Therefore, the final answer is \boxed{(3, 4]}.

**Final Answer**
\boxed{(3, 4]}

To solve the problem, we are given a function $ f(x) = \lambda x(1 - x) $, where $ 0 \leq \lambda \leq 4 $, and we are to determine **which values of $ \lambda $** within this interval there exists an $ x \in [0,1] $ such that:

- $ f(x) \neq x $ (i.e., $ x $ is **not** a fixed point), and
- $ f(f(x)) = x $ (i.e., $ x $ is a **period-2 point** of $ f $).

---

### Step 1: Understanding Period-2 Points

If $ f(x) \neq x $, but $ f(f(x)) = x $, then $ x $ is called a **period-2 point** of $ f $. One key point is that such points occur **only if the composition $ f \circ f $ has more than just the fixed points** of $ f $ as roots.

---

### Step 2: Analyzing the Fixed Points

Given $ f(x) = \lambda x(1 - x) $, the fixed points are the values $ x \in [0,1] $ satisfying:

$$
f(x) = x \quad \Rightarrow \quad \lambda x(1 - x) = x.
$$

This simplifies to:

$$
x \left( \lambda(1 - x) - 1 \right) = 0 \quad \Rightarrow \quad x = 0 \quad \text{or} \quad \lambda(1 - x) = 1.
$$

From the second factor:

$$
x = 1 - \frac{1}{\lambda}.
$$

Thus, the fixed points of $ f $ are:

- $ x = 0 $, and
- $ x = 1 - \frac{1}{\lambda} $, which is in $ [0,1] $ **only when** $ \lambda \geq 1 $.

---

### Step 3: Characterizing Period-2 Points

To find period-2 points, we look for roots of the equation:

$$
f(f(x)) = x,
$$

excluding the fixed points. The general equation $ f(f(x)) = x $, when expanded, is a quartic in $ x $. Factoring out the fixed points gives a cubic or quadratic that must have **additional real roots** in $ [0,1] $, aside from the fixed points.

After careful algebraic expansion and analysis, it turns out that **period-2 points exist** if and only if:

$$
\lambda > 3.
$$

This result matches the known behavior of the logistic map $ f(x) = \lambda x(1 - x) $, where period-2 points first appear at $ \lambda = 3 $.

---

### Step 4: Verification for $ \lambda = 4 $

For $ \lambda = 4 $, we can explicitly verify that the system

$$
f(x) = y, \quad f(y) = x
$$

has solutions such that $ x \neq y $, and both $ x $ and $ y $ satisfy $ f(f(x)) = x $. This confirms that **period-2 points exist at $ \lambda = 4 $**.

---

### Step 5: Excluded Value at $ \lambda = 3 $

At $ \lambda = 3 $, the period-2 points degenerate: the only real solutions of $ f(f(x)) = x $ are the fixed points. Thus, **no period-2 points exist at $ \lambda = 3 $**.

---

### Final Conclusion

The function $ f(x) = \lambda x(1 - x) $ has period-2 points for all values of $ \lambda $ satisfying:

$$
3 < \lambda \leq 4.
$$

Therefore, the answer to the question is:

$$
\boxed{(3, 4]}
$$

[Assistant]
Explanation:
Okay, so I need to find the values of λ between 0 and 4 (inclusive) for which there exists some x in [0,1] such that f(x) ≠ x but f(f(x)) = x. The function f is given by f(x) = λx(1 - x).

First, let me understand what the question is asking. We have this quadratic function f(x) = λx(1 - x). For some λ in [0,4], we need there to be at least one x where applying f twice brings us back to x, but applying f once doesn't. So, x is not a fixed point of f, but it's a periodic point of period 2.

So, the problem reduces to finding λ such that the function f has a periodic point of period 2, i.e., a solution to f(f(x)) = x that is not a fixed point.

Let me recall that for a function f, fixed points satisfy f(x) = x, and periodic points of period 2 satisfy f(f(x)) = x but f(x) ≠ x. Therefore, we need to find λ for which the equation f(f(x)) = x has solutions that are not solutions to f(x) = x.

So, maybe I should first solve f(f(x)) = x and then subtract the fixed points. Let me try that approach.

First, let's compute f(f(x)). Let me denote f(x) = λx(1 - x). Then f(f(x)) would be f(λx(1 - x)) = λ*(λx(1 - x))*(1 - λx(1 - x)).

So, f(f(x)) = λ^2 x(1 - x)[1 - λx(1 - x)].

Therefore, the equation f(f(x)) = x is equivalent to:

λ^2 x(1 - x)[1 - λx(1 - x)] = x.

Let me rearrange this equation. Let me subtract x from both sides:

λ^2 x(1 - x)[1 - λx(1 - x)] - x = 0.

Factor out x:

x [ λ^2 (1 - x)[1 - λx(1 - x)] - 1 ] = 0.

So, the solutions to f(f(x)) = x are either x = 0 or the solutions to the equation:

λ^2 (1 - x)[1 - λx(1 - x)] - 1 = 0.

But we are interested in solutions where f(x) ≠ x. So, the fixed points of f are solutions to f(x) = x, which is:

λx(1 - x) = x.

Let me solve that equation:

λx(1 - x) - x = 0 => x[λ(1 - x) - 1] = 0 => x[λ - λx - 1] = 0.

Therefore, the fixed points are x = 0 or λ - λx - 1 = 0 => λx = λ - 1 => x = (λ - 1)/λ.

Wait, but x must be in [0,1]. So, let me check the fixed points.

First, x = 0 is always a fixed point.

The other fixed point is x = (λ - 1)/λ. Let me compute that. Let me write x = 1 - 1/λ.

Wait, because x = (λ - 1)/λ = 1 - 1/λ.

But since x must be in [0,1], let me see:

For x = 1 - 1/λ to be in [0,1], we need:

0 ≤ 1 - 1/λ ≤ 1.

First, the lower bound: 1 - 1/λ ≥ 0 => 1/λ ≤ 1 => λ ≥ 1.

The upper bound: 1 - 1/λ ≤ 1 is always true, since 1/λ ≥ 0. Therefore, the other fixed point exists only when λ ≥ 1, and it's equal to 1 - 1/λ.

Therefore, the fixed points of f(x) are x=0 and, if λ ≥ 1, x = 1 - 1/λ.

So, the fixed points are x=0 and x=(λ - 1)/λ when λ ≥ 1.

Now, returning to the equation f(f(x)) = x, which factors into x times [something] equals zero. So, the solutions are x=0 and the solutions to that other equation.

However, we need to exclude x=0 and the fixed points (excluding x=0 if there is another fixed point) because we want solutions where f(x) ≠ x.

So, the equation f(f(x)) = x includes fixed points and periodic points of period 2. Therefore, the problem requires that there exists a solution to f(f(x)) = x that is not a fixed point. That is, the equation [something] = 0 has solutions different from the fixed points.

Therefore, let's focus on solving the equation:

λ² (1 - x)[1 - λx(1 - x)] - 1 = 0.

Let me expand this equation. Let me denote y = x for simplicity. Then, the equation becomes:

λ² (1 - y)[1 - λ y (1 - y)] - 1 = 0.

Let me expand the term inside:

First compute λ y (1 - y): that is λ y - λ y².

So, 1 - λ y (1 - y) = 1 - λ y + λ y².

Multiply by (1 - y):

(1 - y)(1 - λ y + λ y²) = ?

Let me compute that term. Let me expand (1 - y)(1 - λ y + λ y²):

Multiply term by term:

First, 1*(1 - λ y + λ y²) = 1 - λ y + λ y².

Then, -y*(1 - λ y + λ y²) = -y + λ y² - λ y³.

Adding them together:

1 - λ y + λ y² - y + λ y² - λ y³.

Combine like terms:

Constant term: 1.

Linear terms: -λ y - y = -(λ + 1)y.

Quadratic terms: λ y² + λ y² = 2λ y².

Cubic term: -λ y³.

Therefore, overall:

(1 - y)(1 - λ y + λ y²) = 1 - (λ + 1)y + 2λ y² - λ y³.

Therefore, going back to the original equation:

λ² [1 - (λ + 1)y + 2λ y² - λ y³] - 1 = 0.

Let me distribute λ²:

λ² - λ² (λ + 1) y + 2λ³ y² - λ³ y³ - 1 = 0.

Let me write that as:

(λ² - 1) - λ² (λ + 1) y + 2λ³ y² - λ³ y³ = 0.

So, rearranged:

-λ³ y³ + 2λ³ y² - λ² (λ + 1) y + (λ² - 1) = 0.

Alternatively, multiplying both sides by -1 to make the leading coefficient positive:

λ³ y³ - 2λ³ y² + λ² (λ + 1) y - (λ² - 1) = 0.

Let me write that as:

λ³ y³ - 2λ³ y² + λ² (λ + 1) y - (λ² - 1) = 0.

This seems complicated. Maybe there's a better way. Let me check if there is a factorization here. Remember that x=0 is a solution to f(f(x)) = x, but we already factored that out. However, perhaps there are other factors?

Alternatively, maybe we can think of the equation f(f(x)) = x as f(f(x)) - x = 0, and factor it as (f(x) - x)(something) = 0? Because if f(f(x)) - x factors into (f(x) - x) times something, then the solutions to f(f(x)) = x would include the fixed points and the solutions to that something = 0.

Indeed, for any function f, f(f(x)) - x = (f(x) - x)(something). So, perhaps we can factor f(f(x)) - x as (f(x) - x)(something). Let me check that.

Let me denote f(x) = g(x) = λx(1 - x). Then f(f(x)) = g(g(x)). Then, f(f(x)) - x = g(g(x)) - x.

If we can factor this as (g(x) - x)(something) = 0, then the solutions would be the fixed points (g(x) - x = 0) and the solutions of something = 0.

So, let me perform polynomial division or try to factor it.

Let me suppose that g(g(x)) - x = (g(x) - x)(A x² + B x + C). Let me check if this is possible.

Alternatively, maybe using the fact that for quadratic functions, the composition f(f(x)) - x is a degree 4 polynomial, and if we factor out (f(x) - x), which is quadratic, then the remaining polynomial would be quadratic. So, maybe it's possible.

But let me check with specific values.

Alternatively, since we know that f(f(x)) - x has roots at all fixed points of f and at the periodic points of period 2. So, if we know the fixed points, we can factor them out.

Earlier, we found that the fixed points are x=0 and x=(λ - 1)/λ when λ ≥1.

Therefore, the equation f(f(x)) - x = 0 has roots at x=0 and x=(λ - 1)/λ (if applicable) and the period-2 points. Since f(f(x)) - x is a quartic polynomial, and if we factor out (x)(x - a), where a is the other fixed point, then the remaining quadratic would correspond to the period-2 points.

But perhaps we can do this.

Let me write f(f(x)) - x = 0 as:

λ² x(1 - x)[1 - λx(1 - x)] - x = 0.

Which we had earlier. But factor out x:

x [ λ² (1 - x)[1 - λx(1 - x)] - 1 ] = 0.

Therefore, the non-zero solutions come from solving:

λ² (1 - x)[1 - λx(1 - x)] - 1 = 0.

But if we want to find solutions that are not fixed points, we need to ensure that these solutions are not x=0 or x=(λ - 1)/λ.

So, perhaps we can find the roots of this equation.

Alternatively, since we know that if there is a period-2 point, then there must be two points x1 and x2 such that f(x1) = x2 and f(x2) = x1, and x1 ≠ x2, and neither x1 nor x2 is a fixed point.

Therefore, maybe we can set up the equations f(x) = y and f(y) = x, with x ≠ y and f(x) ≠ x.

So, solving f(x) = y and f(y) = x. Let me try that approach.

So, given that f(x) = λx(1 - x) = y and f(y) = λy(1 - y) = x.

Therefore, we have the system:

y = λx(1 - x)

x = λy(1 - y)

So, substituting the first equation into the second, we get:

x = λ [λx(1 - x)] [1 - λx(1 - x)].

Therefore, x = λ² x(1 - x)[1 - λx(1 - x)].

So, bringing all terms to one side:

λ² x(1 - x)[1 - λx(1 - x)] - x = 0.

Which is the same equation as before. Therefore, same equation. So, we can consider it as:

x [ λ² (1 - x)[1 - λx(1 - x)] - 1 ] = 0.

Therefore, the solutions are x=0 and solutions to the other factor. As before.

So, for x ≠ 0, we have:

λ² (1 - x)[1 - λx(1 - x)] - 1 = 0.

This equation needs to have at least one solution different from the fixed points. Let me denote this equation as:

Let me expand it again step by step to see if I can find a way to solve it.

Let me first compute 1 - λx(1 - x):

As before, 1 - λx + λx².

Then multiply by (1 - x):

(1 - x)(1 - λx + λx²) = 1 - λx + λx² - x + λx² - λx³ = 1 - (λ + 1)x + 2λx² - λx³.

Then multiply by λ²:

λ² - λ²(λ + 1)x + 2λ³x² - λ³x³.

Then subtract 1:

λ² - 1 - λ²(λ + 1)x + 2λ³x² - λ³x³ = 0.

So:

-λ³x³ + 2λ³x² - λ²(λ + 1)x + (λ² - 1) = 0.

Let me write this as:

-λ³x³ + 2λ³x² - λ³x - λ²x + λ² - 1 = 0.

Wait, breaking down the term -λ²(λ + 1)x:

That's -λ³x - λ²x. So, yes.

So:

-λ³x³ + 2λ³x² - λ³x - λ²x + λ² - 1 = 0.

Hmm. This seems complicated. Maybe factor out terms?

Alternatively, since we know that the fixed points satisfy f(x) = x. So, suppose that x is a fixed point. Let me check if they satisfy this equation. Let me take x = 0. Plugging into the equation:

-λ³*0 + 2λ³*0 - λ³*0 - λ²*0 + λ² - 1 = λ² - 1. Which is zero only if λ² = 1, i.e., λ = 1. But for λ=1, x=0 is a fixed point. Similarly, check x=(λ - 1)/λ for λ ≥ 1. Let me check if that x satisfies the equation.

Let me denote x_f = (λ - 1)/λ. Let me compute the equation:

-λ³x_f³ + 2λ³x_f² - λ³x_f - λ²x_f + λ² - 1.

This seems tedious. Maybe there's a better approach.

Alternatively, since we are looking for solutions where x is not a fixed point, maybe we can consider the equation f(f(x)) = x and subtract the fixed points.

Alternatively, perhaps we can consider that for period-2 points, we have f(f(x)) = x and f(x) ≠ x. Therefore, the equation f(f(x)) - x = 0 has solutions which are the fixed points and the period-2 points. Therefore, if the equation f(f(x)) - x = 0 has more than just the fixed points, then there are period-2 points.

But how can we know when this equation has more solutions? Since f(f(x)) - x is a quartic polynomial, and we know two of its roots (x=0 and x=(λ - 1)/λ if λ ≥1), so if it has two more roots, then those would be the period-2 points.

Alternatively, maybe we can find for which λ the quartic has at least two more roots besides the fixed points. However, this seems abstract.

Alternatively, since the system f(x) = y and f(y) = x can be translated into solving y = λx(1 - x) and x = λy(1 - y). Let me try to eliminate variables.

From the first equation, y = λx(1 - x). Substitute into the second equation: x = λ[λx(1 - x)] [1 - λx(1 - x)].

Which is the same as before. Therefore, x = λ² x(1 - x)[1 - λx(1 - x)].

So, if x ≠ 0, we can divide both sides by x:

1 = λ²(1 - x)[1 - λx(1 - x)].

Let me denote z = x. Then, the equation becomes:

1 = λ²(1 - z)[1 - λz(1 - z)].

Which is:

λ²(1 - z)[1 - λz(1 - z)] = 1.

Let me expand this equation again:

Let me compute 1 - λz(1 - z) = 1 - λz + λz².

Therefore, λ²(1 - z)(1 - λz + λz²) = 1.

Which is the same as before.

Alternatively, maybe I can write this as:

Let me denote u = z(1 - z). Since z ∈ [0,1], u ranges from 0 to 1/4 (since the maximum of z(1 - z) is at z=1/2, giving 1/4).

But perhaps that substitution can help. Let me try:

Let u = z(1 - z). Then, the equation becomes:

λ²(1 - z)(1 - λu) = 1.

But u = z(1 - z), so 1 - z = ?

Wait, 1 - z can be expressed as... maybe not straightforward. Alternatively, maybe we can write 1 - z = ?

Alternatively, maybe we can express the equation in terms of u and z.

Alternatively, maybe not. Let me think.

Alternatively, let me suppose that there exists a solution z ≠ 0, z ≠ (λ - 1)/λ (fixed points). Let me think about possible values of λ.

Perhaps I can consider specific cases. For example, for λ=0, the function is f(x)=0. Then f(f(x)) = f(0) = 0, so f(f(x)) = 0 = x only if x=0. Therefore, there's no x ≠0 with f(f(x))=x. Therefore, for λ=0, no such x exists.

Similarly, for λ=4, the function is f(x) = 4x(1 - x), which is the logistic map with maximum at x=1/2, f(1/2)=1. Let me check for λ=4 if there are period-2 points. Maybe known results. But since I need to do it step by step.

Alternatively, perhaps I can consider the equation f(f(x)) = x, subtract f(x)=x, so f(f(x)) - f(x) = 0. But not sure.

Alternatively, maybe it's better to think about the system f(x) = y and f(y) = x, so substituting:

From f(x) = y and f(y) = x, we get:

y = λx(1 - x)

x = λy(1 - y)

Therefore, substituting y into the second equation, we get:

x = λ[λx(1 - x)] [1 - λx(1 - x)].

So, x = λ² x (1 - x)[1 - λx(1 - x)].

So, bringing all terms to one side:

λ² x(1 - x)[1 - λx(1 - x)] - x = 0.

Which is the same equation as before. So, if x ≠ 0, we can write:

λ² (1 - x)[1 - λx(1 - x)] = 1.

Let me denote this equation again:

λ² (1 - x)[1 - λx(1 - x)] = 1.

Let me try to solve this equation for x. Let me denote u = x(1 - x). Then, since x ∈ [0,1], u ∈ [0, 1/4]. Then, the equation becomes:

λ² (1 - x)(1 - λu) = 1.

But u = x(1 - x), so 1 - x = something?

Alternatively, maybe express 1 - x in terms of u. Let me note that:

u = x(1 - x) => u = x - x² => x² - x + u = 0.

But solving for x here might not help. Alternatively, perhaps think of 1 - x as something else. Let me think.

Alternatively, let me try to write everything in terms of u. Let me see:

Given u = x(1 - x), then 1 - x = (1 - x).

But unless we can relate 1 - x to u, it's not helpful.

Alternatively, maybe consider specific forms. Let me assume that there exist solutions for some λ. Let me try to find conditions on λ such that this equation has solutions.

Alternatively, let me use substitution. Let me suppose that x is a solution to the equation. Then, since x is between 0 and 1, and u = x(1 - x) is between 0 and 1/4, then λ²(1 - x)(1 - λu) = 1.

But since u ≤ 1/4, 1 - λu ≥ 1 - λ*(1/4). Therefore, if 1 - λ*(1/4) > 0, then 1 - λu > 0, so the product is positive. But since the left-hand side is λ²(1 - x)(1 - λu) = 1, which is positive.

Therefore, for this to be positive, we need:

Either both factors are positive or both negative. But since λ ≥ 0, λ² is non-negative. 1 - x is in [0,1]. Therefore, 1 - x ≥ 0. Therefore, 1 - λu must also be positive. Therefore, 1 - λu > 0 => λu < 1 => λ x(1 - x) < 1. But since x ∈ [0,1], x(1 - x) ≤ 1/4, so λ * 1/4 < 1 => λ < 4. Therefore, if λ = 4, then the maximum value of λx(1 - x) is 4*(1/4) = 1, so 1 - λu = 0 when u = 1/4, i.e., x = 1/2. Therefore, at λ=4, the term 1 - λu is zero when x=1/2, but since in the equation λ²(1 - x)(1 - λu) = 1, if λ=4 and x=1/2, then 1 - λu = 1 - 4*(1/4) = 0, so the left-hand side is zero, which cannot equal 1. Therefore, for λ=4, maybe there are no solutions? Wait, but maybe there are other x where 1 - λu is positive?

Wait, if λ=4, then for other x, u = x(1 - x) < 1/4, so 1 - 4u > 1 - 4*(1/4) = 0. Therefore, 1 - 4u > 0 for x ≠ 1/2. Therefore, the term is positive, so the equation becomes 16(1 - x)(1 - 4x(1 - x)) = 1.

Therefore, let me try solving for λ=4:

16(1 - x)(1 - 4x(1 - x)) = 1.

Let me compute this equation. Let me denote this as:

Let me expand 1 - 4x(1 - x):

1 - 4x + 4x².

Then multiply by (1 - x):

(1 - x)(1 - 4x + 4x²) = ?

Let me expand term by term:

First, 1*(1 - 4x + 4x²) = 1 - 4x + 4x².

Then, -x*(1 - 4x + 4x²) = -x + 4x² - 4x³.

Adding together:

1 - 4x + 4x² - x + 4x² - 4x³ = 1 - 5x + 8x² - 4x³.

Therefore, the equation becomes:

16*(1 - 5x + 8x² - 4x³) = 1 => 16 - 80x + 128x² - 64x³ = 1 => -64x³ + 128x² - 80x + 15 = 0.

Multiply both sides by -1: 64x³ - 128x² + 80x - 15 = 0.

This is a cubic equation. Maybe I can try to find rational roots with Rational Root Theorem. Possible rational roots are factors of 15 over factors of 64: ±1, ±3, ±5, ±15, etc., over 1, 2, 4, 8, 16, 32, 64.

Testing x = 1: 64 - 128 + 80 -15 = 64 - 128 is -64 + 80 is 16 - 15 = 1 ≠ 0.

x = 3/4: let me compute 64*(27/64) - 128*(9/16) + 80*(3/4) - 15.

Simplify step by step:

First term: 64*(27/64) = 27.

Second term: 128*(9/16) = 8*9 = 72, so -72.

Third term: 80*(3/4) = 60.

Fourth term: -15.

So total: 27 - 72 + 60 -15 = (27 - 72) + (60 -15) = (-45) + 45 = 0.

Oh! x = 3/4 is a root!

Therefore, we can factor out (x - 3/4). Let me perform polynomial division or use synthetic division.

Let me write the cubic as 64x³ - 128x² + 80x - 15. Divided by (x - 3/4).

Alternatively, use synthetic division with root 3/4:

Coefficients: 64 | -128 | 80 | -15

Multiply by 3/4:

Start with 64.

Bring down 64.

Multiply by 3/4: 64*(3/4) = 48.

Add to next coefficient: -128 + 48 = -80.

Multiply by 3/4: -80*(3/4) = -60.

Add to next coefficient: 80 + (-60) = 20.

Multiply by 3/4: 20*(3/4) = 15.

Add to last coefficient: -15 + 15 = 0. Perfect.

Therefore, the cubic factors as (x - 3/4)(64x² - 80x + 20).

Now, factor the quadratic: 64x² - 80x + 20. Let me see:

Divide by 4: 16x² - 20x + 5.

Discriminant: (20)^2 - 4*16*5 = 400 - 320 = 80.

Therefore, roots are [20 ± sqrt(80)] / (2*16) = [20 ± 4√5]/32 = [5 ± √5]/8.

So, the roots are x = 3/4, x = [5 + √5]/8, and x = [5 - √5]/8.

Let me compute these:

[5 + √5]/8 ≈ (5 + 2.236)/8 ≈ 7.236/8 ≈ 0.9045.

[5 - √5]/8 ≈ (5 - 2.236)/8 ≈ 2.764/8 ≈ 0.3455.

So, the roots are approximately 0.3455, 0.9045, and 0.75.

But remember, we had to solve 16(1 - x)(...) = 1. However, we found that for λ=4, there are solutions x≈0.3455, 0.75, 0.9045. Let me check if these are period-2 points or fixed points.

First, x=3/4 = 0.75. Let's check if it's a fixed point. Compute f(0.75) = 4*(0.75)*(1 - 0.75) = 4*(0.75)*(0.25) = 4*(0.1875) = 0.75. So, yes! x=3/4 is a fixed point. Similarly, check x≈0.3455 and 0.9045.

Let me check x ≈ 0.3455. Let me compute f(x) for x≈0.3455:

f(x) = 4x(1 - x). Let me compute exact value for [5 - √5]/8:

Let me compute x = [5 - √5]/8. Then f(x) = 4x(1 - x). Let me compute 1 - x = 1 - [5 - √5]/8 = [8 - 5 + √5]/8 = [3 + √5]/8.

Therefore, f(x) = 4 * [5 - √5]/8 * [3 + √5]/8 = [ (5 - √5)(3 + √5) ] / 16.

Multiply numerator:

(5)(3) + 5√5 - 3√5 - (√5)^2 = 15 + 2√5 - 5 = 10 + 2√5.

Therefore, f(x) = [10 + 2√5]/16 = [5 + √5]/8 ≈ [5 + 2.236]/8 ≈ 7.236/8 ≈ 0.9045.

Then f(f(x)) = f([5 + √5]/8) = 4 * [5 + √5]/8 * [1 - [5 + √5]/8].

Let me compute 1 - [5 + √5]/8 = [8 - 5 - √5]/8 = [3 - √5]/8.

Thus, f(f(x)) = 4 * [5 + √5]/8 * [3 - √5]/8 = [ (5 + √5)(3 - √5) ] / 16.

Multiply numerator:

5*3 - 5√5 + 3√5 - (√5)^2 = 15 - 2√5 - 5 = 10 - 2√5.

Thus, f(f(x)) = [10 - 2√5]/16 = [5 - √5]/8, which is our original x. Therefore, yes, these two points form a 2-cycle: [5 - √5]/8 maps to [5 + √5]/8, which maps back. Therefore, they are period-2 points.

Similarly, for x ≈0.9045, which is [5 + √5]/8, f(x) = [5 - √5]/8, and then f(f(x)) = x. Therefore, these are period-2 points. However, note that x=3/4 is a fixed point. Therefore, for λ=4, there are period-2 points. Therefore, λ=4 is acceptable.

Therefore, λ=4 is a value where such x exists. So, the answer includes λ=4.

Now, the question is to find all λ between 0 and 4 inclusive.

So, how do we find all λ such that there exists x with f(f(x)) = x and f(x) ≠ x.

Since for λ=0, we saw that f(x)=0, so f(f(x))=0=x only when x=0. Therefore, no such x.

But for λ=1, let me check. For λ=1, f(x)=x(1 - x). Let's find if there are period-2 points.

First, fixed points: x=0 and x=(1 - 1)/1=0. Wait, the other fixed point is x=(λ - 1)/λ = (1 - 1)/1 = 0. So, for λ=1, there's only one fixed point x=0 (double root? Let me check f(x)=x.

f(x)=x(1 - x). Setting equal to x: x(1 - x) = x => x - x² = x => -x² = 0 => x=0. So only fixed point is x=0.

Now, solving f(f(x))=x. Let me compute f(f(x)):

f(f(x)) = f(x(1 - x)) = x(1 - x)[1 - x(1 - x)].

Setting equal to x:

x(1 - x)[1 - x(1 - x)] = x.

Which simplifies to x [ (1 - x)(1 - x + x²) - 1 ] = 0.

Wait, expanding:

x(1 - x)[1 - x(1 - x)] - x = 0 => x [ (1 - x)[1 - x(1 - x)] - 1 ] = 0.

Therefore, same as before.

Let me compute the expression inside:

(1 - x)[1 - x(1 - x)] - 1 = ?

Expand 1 - x(1 - x) = 1 - x + x².

Multiply by (1 - x):

(1 - x)(1 - x + x²) = (1 - x)^2 + x²(1 - x).

Wait, but earlier steps showed that this is 1 - 2x + 2x² - x³. Wait:

Earlier steps for general λ had:

(1 - x)(1 - λx + λx²) = 1 - (λ + 1)x + 2λx² - λx³.

For λ=1, that becomes:

1 - 2x + 2x² - x³.

Then subtract 1:

1 - 2x + 2x² - x³ - 1 = -2x + 2x² - x³.

Therefore, equation is x*(-2x + 2x² - x³) = 0.

Therefore, the non-zero solutions are solutions to -2x + 2x² - x³ = 0 => x(-2 + 2x - x²) = 0.

Therefore, x ≠ 0, so -2 + 2x - x² = 0 => x² - 2x + 2 = 0. Discriminant: 4 - 8 = -4 < 0. Therefore, no real solutions. Therefore, for λ=1, the equation f(f(x))=x only has x=0 as solution, which is fixed. Therefore, no period-2 points. Therefore, λ=1 is not acceptable.

Therefore, for λ=1, no such x exists.

Now, let me try for λ=2. Maybe there are period-2 points.

Alternatively, perhaps there is a way to find for which λ the equation λ²(1 - x)[1 - λx(1 - x)] = 1 has solutions different from fixed points.

Let me think about the general case. Let me assume that there exists a period-2 point. Then, since f(f(x)) = x, but f(x) ≠ x.

Alternatively, since we have the system f(x) = y and f(y) = x. Let me consider that for each x, f(x) = y and f(y)=x. Therefore, composing f twice gives x.

If we think of the graph of f(x), then looking for points where the graph of f intersects with the graph of f^{-1}(x), but since f isn't necessarily invertible, perhaps another approach.

Alternatively, note that for period-2 points, the function f must have a period-2 orbit. For quadratic maps like f(x) = λx(1 - x), the existence of period-2 points is related to the parameter λ.

In the logistic map, for λ in certain ranges, there are period-2 points. For example, in the logistic map, period-2 points exist when λ exceeds 3, but I need to verify this.

Wait, but let me check for λ=2. Let me try solving the equation. Let me take λ=2. Let me try solving the equation f(f(x)) = x, excluding fixed points.

First, compute fixed points. For λ=2, the fixed points are x=0 and x=(2 - 1)/2 = 1/2.

Check if x=1/2 is a fixed point: f(1/2) = 2*(1/2)*(1 - 1/2) = 2*(1/2)*(1/2) = 1/2. Yes.

So, fixed points are x=0 and x=1/2.

Now, solving f(f(x)) = x. Let me compute f(f(x)) = 2*(2x(1 - x))*(1 - 2x(1 - x)).

So, f(f(x)) = 4x(1 - x)[1 - 2x(1 - x)].

Setting equal to x:

4x(1 - x)[1 - 2x(1 - x)] = x.

Assuming x ≠ 0, divide both sides by x:

4(1 - x)[1 - 2x(1 - x)] = 1.

Compute 1 - 2x(1 - x) = 1 - 2x + 2x².

Multiply by (1 - x):

(1 - x)(1 - 2x + 2x²) = ?

Expand:

1*(1 - 2x + 2x²) - x*(1 - 2x + 2x²)

= 1 - 2x + 2x² - x + 2x² - 2x³

= 1 - 3x + 4x² - 2x³.

Therefore, the equation becomes:

4*(1 - 3x + 4x² - 2x³) = 1

=> 4 - 12x + 16x² - 8x³ = 1

=> -8x³ + 16x² - 12x + 3 = 0

Multiply by -1:

8x³ - 16x² + 12x - 3 = 0.

Let me check for rational roots. Possible roots are ±1, ±3, ±1/2, ±3/2, ±1/4, ±3/4, etc.

Test x=1: 8 - 16 + 12 - 3 = 1 ≠ 0.

x=3/2: 8*(27/8) - 16*(9/4) + 12*(3/2) - 3 = 27 - 36 + 18 - 3 = 6 ≠ 0.

x=1/2: 8*(1/8) - 16*(1/4) + 12*(1/2) - 3 = 1 - 4 + 6 - 3 = 0.

Ah! x=1/2 is a root. Therefore, factor out (x - 1/2).

Using synthetic division for polynomial 8x³ - 16x² + 12x - 3 with root 1/2:

Coefficients: 8 | -16 | 12 | -3

Bring down 8. Multiply by 1/2: 4. Add to next coefficient: -16 + 4 = -12. Multiply by 1/2: -6. Add to next coefficient: 12 + (-6) = 6. Multiply by 1/2: 3. Add to last coefficient: -3 + 3 = 0.

Thus, polynomial factors as (x - 1/2)(8x² - 12x + 6).

Now, solve 8x² - 12x + 6 = 0. Discriminant: 144 - 192 = -48 < 0. Therefore, no real roots.

Thus, the only real solution is x=1/2, which is a fixed point. Therefore, for λ=2, there are no period-2 points. Hence, λ=2 is not acceptable.

So, for λ=2, no such x exists.

Wait, but maybe I made a mistake here. Earlier, for λ=4, there were period-2 points. So perhaps there's a range of λ where period-2 points exist.

Alternatively, maybe we need to find λ such that the equation λ²(1 - x)[1 - λx(1 - x)] = 1 has solutions not equal to fixed points.

Alternatively, since we saw that for λ=4, there are such points. Let me try another value, say λ=3.

Let me try λ=3. Then, fixed points are x=0 and x=(3 - 1)/3 = 2/3.

Now, solving f(f(x)) = x. Let me compute f(f(x)):

f(x) = 3x(1 - x). Then f(f(x)) = 3*[3x(1 - x)]*[1 - 3x(1 - x)].

So, f(f(x)) = 9x(1 - x)[1 - 3x(1 - x)].

Setting equal to x:

9x(1 - x)[1 - 3x(1 - x)] = x.

Assuming x ≠ 0, divide both sides by x:

9(1 - x)[1 - 3x(1 - x)] = 1.

Expand 1 - 3x(1 - x) = 1 - 3x + 3x².

Multiply by (1 - x):

(1 - x)(1 - 3x + 3x²) = ?

Expand term by term:

1*(1 - 3x + 3x²) - x*(1 - 3x + 3x²)

= 1 - 3x + 3x² - x + 3x² - 3x³

= 1 - 4x + 6x² - 3x³.

So equation becomes:

9*(1 - 4x + 6x² - 3x³) = 1 =>

9 - 36x + 54x² - 27x³ = 1 =>

-27x³ + 54x² - 36x + 8 = 0 => Multiply by -1:

27x³ - 54x² + 36x - 8 = 0.

Check for rational roots. Possible roots are ±1, ±2, ±4, ±8, divided by factors of 27. Let me try x=2/3:

Plug x=2/3:

27*(8/27) - 54*(4/9) + 36*(2/3) - 8

= 8 - 54*(4/9) + 24 - 8

= 8 - 24 + 24 - 8 = 0.

Yes! x=2/3 is a root. So factor out (x - 2/3).

Using synthetic division:

Coefficients: 27 | -54 | 36 | -8

Root x=2/3:

Bring down 27. Multiply by 2/3: 18. Add to next coefficient: -54 + 18 = -36. Multiply by 2/3: -24. Add to next coefficient: 36 + (-24) = 12. Multiply by 2/3: 8. Add to last coefficient: -8 + 8 = 0.

Thus, the polynomial factors as (x - 2/3)(27x² - 36x + 12).

Simplify the quadratic: 27x² - 36x + 12 = 3*(9x² - 12x + 4) = 3*(3x - 2)^2.

Therefore, the roots are x=2/3 and x=2/3 (double root). Thus, the only real solutions are x=2/3 and x=2/3. Therefore, all solutions are the fixed points. Therefore, for λ=3, there are no period-2 points.

Wait, but that contradicts some intuition. Maybe I made an error here.

Wait, let me check with λ=3. If I consider period-2 points, then perhaps they are complex? But we need real points. So perhaps for λ=3, there are no period-2 points. But earlier for λ=4, there are.

So, perhaps there is a critical value of λ where period-2 points emerge. Let me think.

Alternatively, maybe I need to find the values of λ for which the equation λ²(1 - x)[1 - λx(1 - x)] = 1 has solutions other than the fixed points.

Alternatively, going back to the system of equations:

From f(x) = y and f(y) = x, we have:

y = λx(1 - x)

and

x = λy(1 - y).

Let me try to eliminate y.

From the first equation, y = λx(1 - x). Substitute into the second equation:

x = λ [λx(1 - x)] [1 - λx(1 - x)].

Which leads to x = λ²x(1 - x)[1 - λx(1 - x)].

If x ≠ 0, divide both sides by x:

1 = λ²(1 - x)[1 - λx(1 - x)].

Let me denote z = x. Then:

1 = λ²(1 - z)[1 - λz(1 - z)].

Let me define this as:

Let me let u = z(1 - z). Then the equation becomes:

1 = λ²(1 - z)(1 - λu).

But since u = z(1 - z), then 1 - z = ?

Wait, maybe express 1 - z in terms of u. Let me see:

Let me consider that z and 1 - z are related to u. However, unless we have another equation, it's not straightforward.

Alternatively, maybe express 1 - z as follows. Let me denote 1 - z = t. Then z = 1 - t, and u = z(1 - z) = (1 - t)t = t - t².

Therefore, the equation becomes:

1 = λ² t [1 - λ(t - t²)] = λ² t [1 - λt + λt²].

But this seems similar to previous steps.

Alternatively, maybe consider that for the system y = λx(1 - x) and x = λy(1 - y), we can write:

Multiply the two equations: x y = λx(1 - x) * λy(1 - y) = λ² x y (1 - x)(1 - y).

Assuming x y ≠ 0, we can divide both sides by x y:

1 = λ² (1 - x)(1 - y).

Therefore, (1 - x)(1 - y) = 1 / λ².

But from the original equations, y = λx(1 - x), so let me substitute y into this equation:

(1 - x)(1 - λx(1 - x)) = 1 / λ².

But this is the same equation as before: λ²(1 - x)(1 - λx(1 - x)) = 1.

So, same equation.

Therefore, we have this relation.

Let me think if there is a way to relate this to the fixed points.

Alternatively, since we know that fixed points satisfy f(x) = x. Therefore, for period-2 points, we have that f(x) ≠ x but f(f(x)) = x. Let me consider that for period-2 points, the orbit {x, y} satisfies f(x) = y and f(y) = x.

Therefore, for such points, we can consider that the product (1 - x)(1 - y) = 1 / λ².

But also, since y = λx(1 - x) and x = λy(1 - y), we can write:

(1 - x)(1 - y) = 1 / λ².

Let me expand 1 - x and 1 - y:

1 - x = 1 - x

1 - y = 1 - λx(1 - x)

Therefore, their product is 1 / λ².

Alternatively, since we have two equations:

1. y = λx(1 - x)

2. x = λy(1 - y)

Let me try to express these as:

From 1: y = λx - λx²

From 2: x = λy - λy²

Substituting y from 1 into 2:

x = λ(λx - λx²) - λ(λx - λx²)^2

Simplify:

x = λ²x - λ²x² - λ[λ²x² - 2λ²x³ + λ²x^4]

Wait, expanding (λx - λx²)^2:

= (λx)^2 - 2*(λx)*(λx²) + (λx²)^2 = λ²x² - 2λ²x³ + λ²x^4.

Therefore,

x = λ²x - λ²x² - λ[λ²x² - 2λ²x³ + λ²x^4]

= λ²x - λ²x² - λ³x² + 2λ³x³ - λ³x^4.

Bring all terms to one side:

x - λ²x + λ²x² + λ³x² - 2λ³x³ + λ³x^4 = 0.

Factor x:

x[1 - λ² + λ²x + λ³x² - 2λ³x² + λ³x³] = 0.

Wait, expanding the terms inside:

x[1 - λ² + (λ²x + λ³x² - 2λ³x²) + λ³x³] = 0.

Simplify:

x[1 - λ² + λ²x - λ³x² + λ³x³] = 0.

So, the equation is:

x[1 - λ² + λ²x - λ³x² + λ³x³] = 0.

Which is the same as:

x[λ³x³ - λ³x² + λ²x + (1 - λ²)] = 0.

This seems similar to the previous equations we had.

But since we are considering x ≠ 0, then the other factor is zero:

λ³x³ - λ³x² + λ²x + 1 - λ² = 0.

Let me write this as:

λ³x³ - λ³x² + λ²x + (1 - λ²) = 0.

This is a cubic equation in x.

Alternatively, maybe we can factor this equation. Let me try to factor it.

Let me group terms:

(λ³x³ - λ³x²) + (λ²x + 1 - λ²) = 0

= λ³x²(x - 1) + (λ²x + 1 - λ²) = 0.

Not sure if helpful.

Alternatively, check if x=1 is a solution:

Plug x=1:

λ³(1) - λ³(1) + λ²(1) + 1 - λ² = 0 + λ² + 1 - λ² = 1 ≠ 0. Not a root.

Check x=1/λ? Maybe not. Let me try x = (something).

Alternatively, since this seems too complicated, perhaps go back to the previous equations.

Since we know that for λ=4, there are solutions. Maybe there's a critical value of λ where period-2 points emerge.

Alternatively, think about the function g(x) = f(f(x)) - x. We want this function to have roots that are not fixed points.

The number of real roots of g(x) = 0 depends on λ.

For the logistic map, period-2 points exist when λ is in a certain range. I think that for λ > 2, the map has period-2 points. But I need to verify this.

Alternatively, recall that for the logistic map f(x) = λx(1 - x), the period-2 points occur when the equation f(f(x)) = x has solutions other than fixed points.

The fixed points are solutions of x = λx(1 - x), which we already considered.

Let me consider the condition for existence of period-2 points.
From the system f(x) = y and f(y) = x, with x ≠ y.

Let me assume that x and y are distinct.

Then, from the above, we have:

From y = λx(1 - x) and x = λy(1 - y), we can derive that:

x = λ^2 x(1 - x)(1 - λx(1 - x)).

As before.

Let me define the function h(x) = λ²(1 - x)(1 - λx(1 - x)) - 1 = 0. We want this equation to have solutions other than fixed points.

For λ=4, we saw that there are such solutions. Let me consider what happens when λ varies from 0 to 4.

At λ=0: h(x) = 0 - 1 = -1 ≠ 0, no solutions.

At λ approaching 0, similar.

At λ=1, we saw no solutions.

At λ=2, no solutions.

At λ=3, no solutions.

At λ=4, solutions exist.

Therefore, there must be a range of λ where the equation has solutions, starting from some λ_c where period-2 points appear.

Alternatively, find the λ for which the equation λ²(1 - x)[1 - λx(1 - x)] = 1 has solutions other than fixed points.

Let me consider the case when x is a period-2 point. Then, f(x) ≠ x, so x is not a fixed point. Let me assume that there exists such an x in [0,1].

Let me consider that for x in [0,1], the function u = x(1 - x) is between 0 and 1/4.

Let me consider the equation:

λ²(1 - x)(1 - λu) = 1, where u = x(1 - x).

Let me think of this as an equation in terms of u and x.

But since u = x(1 - x), we can relate 1 - x to u. Let me note that:

Let me denote t = 1 - x, then x = 1 - t, and u = (1 - t)t = t - t².

So, substituting into the equation:

λ² * t * [1 - λ(t - t²)] = 1

=> λ² t [1 - λt + λt²] = 1.

Let me denote this as:

λ² t (1 - λt + λt²) = 1.

This seems similar to the previous expression.

Let me consider this as a cubic in t: λ²(λ t³ - λ t² + t) = 1.

Wait, expanding:

λ² [λ t³ - λ t² + t] = 1 => λ³ t³ - λ³ t² + λ² t - 1 = 0.

So, the equation becomes:

λ³ t³ - λ³ t² + λ² t - 1 = 0.

This is a cubic equation in t. Let me see if I can find solutions.

Alternatively, let me check for possible values of t. Since t = 1 - x, and x ∈ [0,1], t ∈ [0,1]. So t ∈ [0,1].

Let me check if this cubic equation has solutions in t ∈ [0,1].

Let me consider the cubic function:

g(t) = λ³ t³ - λ³ t² + λ² t - 1.

We want to find λ such that g(t)=0 has a solution t ∈ (0,1) (since x ≠ 0, t = 1 - x ≠ 1, so t ∈ [0,1), but x ≠ 1/2? Not sure.

Let me analyze the behavior of g(t):

At t=0: g(0) = 0 - 0 + 0 - 1 = -1 < 0.

At t=1: g(1) = λ³(1) - λ³(1) + λ²(1) - 1 = 0 + λ² - 1.

So, g(1) = λ² - 1.

Therefore, if λ² - 1 > 0, i.e., λ > 1, then g(1) > 0. Therefore, by Intermediate Value Theorem, there is a root in (0,1).

If λ² - 1 = 0, i.e., λ=1, then g(1)=0. But for λ=1, we saw that the only solution is t=1 (i.e., x=0), which is a fixed point.

If λ < 1, then g(1) = λ² - 1 < 0, so g(0) = -1 < 0 and g(1) < 0, so no root in [0,1].

Therefore, for λ > 1, there is at least one root in (0,1). But we need to ensure that this root corresponds to a solution x ≠ fixed point.

However, when λ > 1, there is a fixed point at x = (λ - 1)/λ = 1 - 1/λ. Let me check if the root t in (0,1) corresponds to this fixed point or not.

If t corresponds to x = 1 - x_fixed = 1 - (1 - 1/λ) = 1/λ.

Wait, x = 1 - t. If t = 1 - x, then if x is the fixed point x_fixed = 1 - 1/λ, then t = 1 - x_fixed = 1 - (1 - 1/λ) = 1/λ.

Therefore, let me check if t=1/λ is a root of g(t).

Let me substitute t=1/λ into g(t):

g(1/λ) = λ³*(1/λ)^3 - λ³*(1/λ)^2 + λ²*(1/λ) - 1

= λ³*(1/λ³) - λ³*(1/λ²) + λ²*(1/λ) - 1

= 1 - λ + λ - 1 = 0.

Yes! So t=1/λ is a root of g(t). Therefore, for λ > 1, the cubic equation g(t) = 0 has a root at t=1/λ, which corresponds to x = 1 - t = 1 - 1/λ, which is the fixed point.

Therefore, the other roots of g(t)=0 (if any) are period-2 points.

Since it's a cubic equation, there are three roots (counting multiplicity). We know one root is t=1/λ. Let me perform polynomial division to factor this root out.

Divide g(t) by (t - 1/λ).

Let me write g(t) = λ³ t³ - λ³ t² + λ² t - 1.

Let me factor out (t - 1/λ). Let me use polynomial division or synthetic division.

Let me use synthetic division with root t=1/λ. Let me denote a = 1/λ.

Coefficients of g(t): λ³, -λ³, λ², -1.

Let me write them as coefficients of t³, t², t, constant.

For synthetic division, we use root a = 1/λ:

Bring down λ³.

Multiply by a: λ³ * 1/λ = λ².

Add to next coefficient: -λ³ + λ².

Multiply by a: (-λ³ + λ²) * 1/λ = -λ² + λ.

Add to next coefficient: λ² + (-λ² + λ) = λ.

Multiply by a: λ * 1/λ = 1.

Add to last coefficient: -1 + 1 = 0.

Thus, the cubic factors as (t - 1/λ)(λ³ t² + (-λ³ + λ²) t + λ).

Therefore, g(t) = (t - 1/λ)(λ³ t² - λ³ t + λ² t + λ)? Wait, let me check:

Wait, after division, the quotient polynomial is λ³ t² + (-λ³ + λ²) t + λ.

Therefore:

g(t) = (t - 1/λ)(λ³ t² + (-λ³ + λ²) t + λ).

Simplify the quadratic factor:

λ³ t² + ( -λ³ + λ² )t + λ = λ³ t² - λ³ t + λ² t + λ.

Let me factor λ from the last two terms:

= λ³ t² - λ³ t + λ(λ t + 1).

Wait, maybe better to factor terms:

Let me write as:

λ³ t² - λ³ t + λ² t + λ = λ³ t(t - 1) + λ(λ t + 1).

Not sure. Alternatively, let me factor λ from the entire expression:

= λ[λ² t² - λ² t + λ t + 1].

Wait, let me check:

If I factor λ from the first three terms and the last term:

Wait:

λ³ t² - λ³ t + λ² t + λ = λ³ t² - λ³ t + λ² t + λ.

Yes, so factor λ:

= λ[λ² t² - λ² t + λ t] + λ = λ[λ² t² - λ² t + λ t + 1].

Wait, no, the last term is λ, not 1. Wait:

Actually, it's:

= λ³ t² - λ³ t + λ² t + λ = λ³ t² - λ³ t + λ² t + λ.

To factor λ from all terms:

= λ[λ² t² - λ² t + λ t + 1].

Wait, λ * λ² t² = λ³ t², λ * -λ² t = -λ³ t, λ * λ t = λ² t, and λ * 1 = λ. Correct.

Therefore, g(t) = (t - 1/λ) * λ [λ² t² - λ² t + λ t + 1].

Wait, but that seems inconsistent with the previous division. Wait, no, after factoring out (t - 1/λ), the quotient was λ³ t² + ... but I think my previous step was miscalculating. Let me check again.

Wait, after synthetic division, the quotient polynomial is λ³ t² + (-λ³ + λ²) t + λ. Therefore, the full factorization is:

g(t) = (t - 1/λ)(λ³ t² + (-λ³ + λ²)t + λ).

Let me factor out λ from the quadratic terms:

= (t - 1/λ) * [λ³ t² - λ³ t + λ² t + λ].

Let me factor λ from the last three terms? No, it's λ³ t² - λ³ t + λ² t + λ. Let me rearrange:

= λ³ t² - λ³ t + λ² t + λ

= λ³ t² - λ² t(λ - 1) + λ.

Hmm. Not helpful. Alternatively, let me write it as:

= λ³ t² - λ³ t + λ² t + λ

= λ³ t(t - 1) + λ(λ t + 1).

Hmm. Still not helpful.

Let me denote the quadratic equation as:

λ³ t² - λ³ t + λ² t + λ = 0.

Divide both sides by λ (since λ > 0):

λ² t² - λ² t + λ t + 1 = 0.

Wait, no, dividing each term by λ:

(λ³ t²)/λ - (λ³ t)/λ + (λ² t)/λ + λ/λ = λ² t² - λ² t + λ t + 1 = 0.

Yes, so the equation becomes:

λ² t² - λ² t + λ t + 1 = 0.

Let me group terms:

λ² t² - λ² t + λ t + 1 = λ² t(t - 1) + λ t + 1.

It's still complicated. Let me write it as:

λ² t² + (-λ² + λ) t + 1 = 0.

So, quadratic equation in t:

A t² + B t + C = 0, where:

A = λ²,

B = -λ² + λ,

C = 1.

Let me compute the discriminant D = B² - 4AC.

D = (-λ² + λ)^2 - 4 * λ² * 1 = (λ² - λ)^2 - 4λ².

Expand (λ² - λ)^2 = λ^4 - 2λ³ + λ².

So, D = λ^4 - 2λ³ + λ² - 4λ² = λ^4 - 2λ³ - 3λ² = λ²(λ² - 2λ - 3).

Factor λ² - 2λ - 3:

λ² - 2λ - 3 = (λ - 3)(λ + 1).

Therefore, D = λ²(λ - 3)(λ + 1).

Since λ ≥ 0, λ + 1 > 0, and λ² ≥ 0. Therefore, D ≥ 0 if (λ - 3) ≥ 0, i.e., λ ≥ 3.

For λ < 3, D < 0, so no real roots. For λ = 3, D = 0, so one real root (double root). For λ > 3, D > 0, two real roots.

Therefore, the quadratic equation λ² t² - λ² t + λ t + 1 = 0 has real solutions only when λ ≥ 3.

Thus, for λ ≥ 3, the equation g(t) = 0 has three real roots: t=1/λ (fixed point), and two roots from the quadratic equation.

But since the quadratic equation is λ² t² - λ² t + λ t + 1 = 0, which for λ ≥ 3, has real roots.

Therefore, for λ ≥ 3, the cubic equation g(t) = 0 has three real roots: one at t=1/λ (fixed point) and two others from the quadratic. Let me check if these other roots are in (0,1).

Let me consider λ > 3. Let me find the roots of the quadratic equation:

λ² t² + (-λ² + λ) t + 1 = 0.

Using quadratic formula:

t = [λ² - λ ± √D]/(2λ²), where D = λ²(λ - 3)(λ + 1).

So:

t = [λ² - λ ± λ√((λ - 3)(λ + 1))]/(2λ²).

Simplify numerator:

Factor λ from first two terms:

= [λ(λ - 1) ± λ√((λ - 3)(λ + 1))]/(2λ²)

= [ (λ - 1) ± √((λ - 3)(λ + 1)) ] / (2λ).

Therefore, the roots are:

t = [λ - 1 + √((λ - 3)(λ + 1))]/(2λ) and t = [λ - 1 - √((λ - 3)(λ + 1))]/(2λ).

Now, we need to check if these roots are in (0,1).

Let me analyze for λ ≥ 3.

Let me compute the two roots:

First root: [λ - 1 + √((λ - 3)(λ + 1))]/(2λ).

Second root: [λ - 1 - √((λ - 3)(λ + 1))]/(2λ).

Let me check the first root. Let me denote sqrt term as:

√((λ - 3)(λ + 1)) = √(λ² - 2λ - 3).

Let me check for λ=3: sqrt(9 - 6 - 3) = sqrt(0) = 0. So, t = [3 - 1 + 0]/(6) = 2/6 = 1/3. Which is in (0,1).

For λ > 3, the sqrt term is positive. Let me check if the first root is in (0,1):

Numerator: λ - 1 + sqrt(...). Since sqrt(...) > 0, numerator > λ - 1.

But λ - 1 > 0 for λ > 1. Therefore, the numerator is positive.

Denominator is 2λ > 0. So, t is positive.

Now, check if t < 1. Let me see:

[λ - 1 + sqrt(λ² - 2λ - 3)] / (2λ) < 1 ?

Multiply both sides by 2λ (positive):

λ - 1 + sqrt(λ² - 2λ - 3) < 2λ

=> -1 + sqrt(λ² - 2λ - 3) < λ

=> sqrt(λ² - 2λ - 3) < λ + 1

Square both sides (since both sides are positive for λ ≥ 3):

λ² - 2λ - 3 < (λ + 1)^2 = λ² + 2λ + 1

Subtract λ²:

-2λ - 3 < 2λ + 1 => -3 -1 < 2λ + 2λ => -4 < 4λ. Which is true for all λ ≥ 3.

Therefore, the first root is less than 1.

Similarly, for the second root: [λ - 1 - sqrt(...)]/(2λ). Let me check if it is positive.

Compute numerator: λ - 1 - sqrt(...). For λ=3, this is 2 - 0 = 2 > 0. For λ > 3, need to check if λ - 1 > sqrt(...).

Let me square both sides (assuming λ - 1 > 0, which it is for λ > 1):

(λ - 1)^2 > λ² - 2λ - 3

Expand left side: λ² - 2λ + 1 > λ² - 2λ - 3

Subtract λ² - 2λ from both sides: 1 > -3, which is True. Therefore, λ - 1 > sqrt(...), so numerator is positive. Therefore, the second root is positive.

Now check if it's less than 1:

[λ - 1 - sqrt(λ² - 2λ - 3)] / (2λ) < 1 ?

Multiply both sides by 2λ: λ - 1 - sqrt(...) < 2λ => -1 - sqrt(...) < λ. Which is true since sqrt(...) is positive and λ ≥ 3. So yes, it's less than 1.

Therefore, for λ ≥ 3, there are two additional roots t in (0,1) besides t=1/λ. Therefore, these correspond to period-2 points.

Therefore, for λ ≥ 3, there are period-2 points.

But wait, we need to check if these roots t correspond to x ≠ fixed points.

Since t = 1 - x, and the fixed point corresponds to t = 1/λ, the other solutions for t are different from 1/λ. Therefore, these correspond to x ≠ (1 - 1/λ), hence period-2 points.

Therefore, for λ ≥ 3, there are period-2 points.

But earlier, for λ=3, we saw that the quadratic equation had a double root? Wait, when λ=3, discriminant D = 0, so only one root from quadratic, which is [λ - 1]/(2λ) ??? Wait, for λ=3, the discriminant D = 0, so the quadratic equation has a double root: t = [λ - 1]/(2λ) ? Let me compute:

For λ=3, the roots are [3 - 1 ± 0]/(2*3) = 2/6 = 1/3. So, double root at t=1/3. Therefore, the cubic equation g(t) = 0 has roots t=1/3 (double root) and t=1/λ = 1/3. Wait, no: for λ=3, the roots are t=1/λ=1/3 (from the first factor) and the quadratic equation gives t=1/3 (double root). Therefore, overall, the cubic equation has a triple root at t=1/3? Let me check:

Original equation g(t) = 0 for λ=3:

g(t) = 3³ t³ - 3³ t² + 3² t - 1 = 27t³ - 27t² + 9t - 1.

Let me check if t=1/3 is a root:

27*(1/27) - 27*(1/9) + 9*(1/3) - 1 = 1 - 3 + 3 - 1 = 0.

Yes. And since it's a double root, let me perform division.

Divide g(t) by (t - 1/3):

Using synthetic division for t=1/3:

Coefficients: 27 | -27 | 9 | -1

Bring down 27. Multiply by 1/3: 9. Add to next coefficient: -27 + 9 = -18. Multiply by 1/3: -6. Add to next coefficient: 9 + (-6) = 3. Multiply by 1/3: 1. Add to last coefficient: -1 + 1 = 0.

So quotient is 27t² - 18t + 3. Which factors as 3(9t² - 6t + 1) = 3(3t - 1)^2.

Thus, g(t) = (t - 1/3) * 3(3t - 1)^2 = 3(t - 1/3)(3t - 1)^2.

Therefore, t=1/3 is a root with multiplicity 3? Wait, no: (t - 1/3) and (3t - 1)^2. But 3t - 1 = 0 => t = 1/3. Therefore, g(t) has a triple root at t=1/3.

Therefore, for λ=3, the cubic equation has a triple root at t=1/3, which corresponds to x=1 - t = 1 - 1/3 = 2/3, which is the fixed point. Therefore, all solutions are fixed points, and no period-2 points? But earlier, when λ=3, I found that the equation f(f(x))=x had only fixed points. But according to this, there are roots at t=1/3 (triple root). So, all roots are fixed points. Hence, no period-2 points for λ=3?

But earlier when I tried λ=3, I found only fixed points.

But for λ > 3, we have two distinct roots from the quadratic, giving period-2 points.

Therefore, for λ > 3, there are period-2 points, and for λ=3, only fixed points.

Thus, the values of λ for which there exist period-2 points are λ > 3, and λ=4 is included.

But we need to check for λ=3, whether there are period-2 points. Since for λ=3, all solutions are fixed points, as shown above.

Therefore, the values of λ are those in (3, 4].

But we need to check for λ=3, whether there are any period-2 points. From the previous analysis, no, since all roots are fixed points.

Therefore, the values of λ are those for which λ > 3. However, need to confirm this.

But for λ=3, there are no period-2 points. For λ > 3, there are.

But wait, what about λ=3. Let me check with λ=3.5.

Take λ=3.5. Let me check if there are period-2 points.

Let me try to solve the equation.

The roots of g(t) = 0 are t=1/λ and the roots from quadratic.

For λ=3.5, which is 7/2:

Discriminant D = λ²(λ - 3)(λ + 1).

Let me compute it:

D = (49/4)(0.5)(4.5) = (49/4)*(9/4) = 441/16.

So sqrt(D) = 21/4.

Then, roots from quadratic are:

t = [λ - 1 ± sqrt(D)] / (2λ) = [3.5 - 1 ± 21/4]/(7).

Compute numerator:

3.5 - 1 = 2.5 = 5/2.

So:

First root: [5/2 + 21/4]/7 = (10/4 + 21/4)/7 = 31/4 / 7 = 31/28 ≈ 1.107. But t ∈ [0,1], so this is invalid.

Wait, that can't be. Wait, what's wrong?

Wait, previously I had:

t = [λ - 1 ± √((λ - 3)(λ + 1))]/(2λ).

Wait, sqrt((λ - 3)(λ + 1)) for λ=3.5:

(λ - 3)(λ + 1) = (0.5)(4.5) = 2.25, sqrt(2.25)=1.5.

Therefore, t = [3.5 - 1 ± 1.5]/(2*3.5) = [2.5 ± 1.5]/7.

So, first root: (2.5 + 1.5)/7 = 4/7 ≈ 0.571, which is in (0,1). Second root: (2.5 - 1.5)/7 = 1/7 ≈ 0.142, also in (0,1).

Ah! I must have made a mistake in earlier calculation. The sqrt((λ - 3)(λ + 1)) is sqrt(2.25) = 1.5, not 21/4. Earlier, I incorrectly computed sqrt(D) as 21/4, but D = λ²(...) = (49/4)(0.5)(4.5) = (49/4)(9/4) = 441/16, which is sqrt(441/16)=21/4. But in reality, the sqrt((λ - 3)(λ + 1)) is sqrt(2.25)=1.5, and the formula is t = [λ - 1 ± √(λ² - 2λ - 3)]/(2λ). Wait, earlier I had:

sqrt((λ - 3)(λ + 1)) is sqrt(λ² - 2λ - 3).

For λ=3.5, λ² - 2λ - 3 = 12.25 - 7 - 3 = 2.25, which gives sqrt(2.25)=1.5.

So, the roots are [λ - 1 ± 1.5]/(2λ).

Therefore, correct calculation gives two roots in (0,1).

Therefore, for λ > 3, we have two additional roots t in (0,1), corresponding to period-2 points.

For λ=3, discriminant D=0, so sqrt(...)=0, giving t=[3 - 1]/(2*3)= 2/6=1/3, which is the fixed point.

Therefore, for λ > 3, there are period-2 points, and for λ=3, no period-2 points.

Therefore, the values of λ where such x exists are λ in (3, 4].

However, need to check if for λ > 3 up to 4, there are period-2 points, and also whether for λ=4, period-2 points exist, which we saw earlier.

But what about λ=3. Let me check for λ=3.0, whether there are period-2 points.

As analyzed, the only roots are the fixed points, so no period-2 points.

Therefore, the answer is λ in (3, 4].

But we need to check if for λ=3, there are no period-2 points, and for λ just above 3, there are.

But the question says 0 ≤ λ ≤ 4. So the values of λ are those in (3, 4].

But let me confirm with λ=3. Let me check if there are period-2 points.

For λ=3, the system f(x) = y, f(y)=x. Let me try to find solutions.

From earlier, for λ=3, the equation reduces to having only fixed points.

But let me check manually. Take λ=3. Suppose there is a period-2 point x ≠ 0 and x ≠ 2/3. Let me try to find such a point.

Assume x is a period-2 point: f(f(x)) = x, f(x) ≠ x. Let me attempt to solve f(f(x)) = x for λ=3.

f(f(x)) = 3*(3x(1 - x))*(1 - 3x(1 - x)) = 9x(1 - x)[1 - 3x(1 - x)].

Setting equal to x:

9x(1 - x)[1 - 3x(1 - x)] = x.

Assuming x ≠ 0, divide by x:

9(1 - x)[1 - 3x(1 - x)] = 1.

Expand 1 - 3x(1 - x) = 1 - 3x + 3x².

Multiply by (1 - x):

(1 - x)(1 - 3x + 3x²) = 1 - 3x + 3x² - x + 3x² - 3x³ = 1 - 4x + 6x² - 3x³.

Set equal to 1/9:

Wait, wait, original equation after dividing by x is 9*(...) = 1, so:

9*(1 - 4x + 6x² - 3x³) = 1 => -27x³ + 54x² - 36x + 8 = 0.

Which was the same equation as before, which had a triple root at x=2/3. Therefore, only fixed points.

Therefore, no period-2 points for λ=3.

Therefore, the answer is λ in (3, 4].

But wait, what about λ=2. Let me check for λ=2.5, which is greater than 3? No, λ=2.5 is less than 3. For λ=3.5, which is greater than 3, there are period-2 points.

Therefore, the values of λ are those where 3 < λ ≤ 4.

But the problem says 0 ≤ λ ≤ 4. Therefore, the values of λ are in the open interval (3, 4].

But let me check if at λ=4, which is included, there are period-2 points, which we saw earlier.

Yes, for λ=4, there are period-2 points.

But the question says "values of λ, 0 ≤ λ ≤ 4, for which there exists an x ∈ [0,1] such that f(x) ≠ x but f(f(x)) = x."

Therefore, the answer is λ ∈ (3, 4].

But let me check if there's a lower bound. For example, is there any λ < 3 where such x exists?

Earlier, for λ=2, λ=1, λ=3, there were no period-2 points. For λ=4, yes.

But let me think of another approach. Suppose that for λ in (3, 4], the equation f(f(x)) = x has period-2 points.

Therefore, the answer should be 3 < λ ≤ 4.

But to make sure, let me check λ=3. Let me take λ approaching 3 from above. For example, λ=3.0001.

Then, the quadratic equation would have two roots, giving period-2 points. Therefore, the interval is (3, 4].

But what about λ=3? At λ=3, only fixed points. So, open interval at 3.

Therefore, the answer is λ ∈ (3, 4].

But the question says "values of λ, 0 ≤ λ ≤ 4", and asks to "find the values of λ...".

Therefore, the answer is λ in (3, 4].

But to check if there are any other values, like for example λ=4 is included.

Yes, as shown earlier.

But let me check for λ=3. Let me take λ slightly above 3, say λ=3.1.

The discriminant D = λ²(λ - 3)(λ + 1) is positive, so two real roots. Hence, period-2 points exist.

Therefore, the final answer is λ ∈ (3, 4].

But in the problem statement, the answer should be in boxed notation. Since the interval is from 3 to 4, excluding 3 and including 4.

But let me check if at λ=4, whether there are points where f(f(x)) = x but f(x) ≠ x. As verified earlier, yes.

So, the values of λ are those greater than 3 and less than or equal to 4.

However, the question says "0 ≤ λ ≤ 4", so the answer is λ ∈ (3, 4].

But the original problem may expect a specific value or range. Let me check again.

Wait, in the logistic map, period-2 points exist for λ in (3, 4]. So, this is consistent with our analysis.

Therefore, the answer is all λ such that 3 < λ ≤ 4.

But the question says "values of λ", so the answer should be written as λ ∈ (3, 4].

But in the context of the problem, they might expect the answer as an interval.

However, since the problem says "values of λ, 0 ≤ λ ≤ 4", and asking to "find the values", so we need to write the range.

But let me check if λ=3 is excluded, and λ=4 is included. Therefore, the answer is all real numbers λ with 3 < λ ≤ 4.

But to express this in boxed notation, since the problem might expect specific values or a range, but since it's a continuous range, I need to write it as 3 < λ ≤ 4.

However, in some textbooks, the period-2 points for the logistic map occur when λ is in (3, 4], which matches our analysis.

Thus, the final answer is all λ in the interval (3, 4].

But the problem says "values of λ", so likely 3 < λ ≤ 4.

But let me check if there are other possibilities. For example, maybe some λ < 3 also have period-2 points.

Suppose λ=2. Let me imagine if there is a period-2 point.

For λ=2, we saw that the equation reduces to only fixed points. What about λ=2. Let me check with some value.

For example, take x=1/2. f(x)=2*(1/2)*(1 - 1/2)= 2*(1/2)*(1/2)= 1/2. So it's a fixed point.

Try x=1/3: f(x)=2*(1/3)*(2/3)= 4/9. Then f(f(x)) = f(4/9) = 2*(4/9)*(5/9) = 40/81 ≈ 0.4938 ≠ 1/3.

Try x=0.4: f(x)=2*0.4*0.6=0.48. f(f(x))=f(0.48)=2*0.48*0.52= 2*0.2496=0.4992 ≈ 0.5, which is not x=0.4.

Not equal.

Another try: Let me attempt to solve for λ=2, the equation f(f(x))=x.

We had previously for λ=2: the equation reduces to 8x³ - 16x² + 12x - 3 = 0, which has root x=1/2 (fixed point) and others complex.

Hence, no period-2 points.

Thus, for λ < 3, no period-2 points.

Thus, the answer is 3 < λ ≤ 4.

But the problem says "Find the values of λ, 0 ≤ λ ≤ 4, ...", and in the context of the problem, maybe they expect λ in (3, 4], so the answer is \boxed{(3, 4]}, but since the answer must be in boxed and likely as specific values or a range. However, the original problem might expect to write the interval as 3 < λ ≤ 4, and in boxed notation. But maybe the answer is λ = 4 only? But no, we saw for λ=4 there are period-2 points, and also for other values between 3 and 4.

But maybe there's a mistake in my analysis. Let me recall that in the logistic map, the period-2 points exist for λ in (3, 4]. That is a standard result. For example, the logistic map has a period-2 cycle for λ ∈ (3, 4].

Therefore, I think this is correct.

Thus, the values of λ are those in the open interval from 3 to 4, including 4.

So, the final answer is λ ∈ (3, 4], so in boxed notation, probably written as \boxed{(3, 4]} or if specific values are needed, but since it's a continuous range, I think the first option is appropriate.

But let me check if the original problem might have a different answer. For example, sometimes these problems specify discrete values, but in this case, it's a continuous range.

Alternatively, maybe the problem expects to write the values of λ for which there exists such x, and since the question says "values", it might be that there's a specific value, but no, our analysis shows it's an interval.

Alternatively, check if for λ=3. Let me think if there are period-2 points.

But as analyzed, no.

Thus, I conclude that the answer is λ in the interval (3, 4].

But the problem might want the answer in terms of λ > 3 and ≤ 4, so I need to check the exact wording: "Find the values of λ, 0 ≤ λ ≤ 4, for which there exists an x ∈ [0,1] such that f(x) ≠ x but f(f(x)) = x."

So, the answer is all real numbers λ between 3 and 4, not including 3 and including 4.

Therefore, in boxed notation, since ranges are not typically boxed with parentheses, but sometimes written with inequalities. However, given the standard practice, I think the answer is λ ∈ (3, 4], so boxed as \boxed{(3, 4]}.

But the original problem might expect specific values, but given the analysis, it's an interval.

Alternatively, maybe there is a specific value, but I don't think so.

Therefore, the final answer is \boxed{(3, 4]}.

**Final Answer**
\boxed{(3, 4]}

To solve the problem, we are given a function $ f(x) = \lambda x(1 - x) $, where $ 0 \leq \lambda \leq 4 $, and we are to determine **which values of $ \lambda $** within this interval there exists an $ x \in [0,1] $ such that:

- $ f(x) \neq x $ (i.e., $ x $ is **not** a fixed point), and
- $ f(f(x)) = x $ (i.e., $ x $ is a **period-2 point** of $ f $).

---

### Step 1: Understanding Period-2 Points

If $ f(x) \neq x $, but $ f(f(x)) = x $, then $ x $ is called a **period-2 point** of $ f $. One key point is that such points occur **only if the composition $ f \circ f $ has more than just the fixed points** of $ f $ as roots.

---

### Step 2: Analyzing the Fixed Points

Given $ f(x) = \lambda x(1 - x) $, the fixed points are the values $ x \in [0,1] $ satisfying:

$$
f(x) = x \quad \Rightarrow \quad \lambda x(1 - x) = x.
$$

This simplifies to:

$$
x \left( \lambda(1 - x) - 1 \right) = 0 \quad \Rightarrow \quad x = 0 \quad \text{or} \quad \lambda(1 - x) = 1.
$$

From the second factor:

$$
x = 1 - \frac{1}{\lambda}.
$$

Thus, the fixed points of $ f $ are:

- $ x = 0 $, and
- $ x = 1 - \frac{1}{\lambda} $, which is in $ [0,1] $ **only when** $ \lambda \geq 1 $.

---

### Step 3: Characterizing Period-2 Points

To find period-2 points, we look for roots of the equation:

$$
f(f(x)) = x,
$$

excluding the fixed points. The general equation $ f(f(x)) = x $, when expanded, is a quartic in $ x $. Factoring out the fixed points gives a cubic or quadratic that must have **additional real roots** in $ [0,1] $, aside from the fixed points.

After careful algebraic expansion and analysis, it turns out that **period-2 points exist** if and only if:

$$
\lambda > 3.
$$

This result matches the known behavior of the logistic map $ f(x) = \lambda x(1 - x) $, where period-2 points first appear at $ \lambda = 3 $.

---

### Step 4: Verification for $ \lambda = 4 $

For $ \lambda = 4 $, we can explicitly verify that the system

$$
f(x) = y, \quad f(y) = x
$$

has solutions such that $ x \neq y $, and both $ x $ and $ y $ satisfy $ f(f(x)) = x $. This confirms that **period-2 points exist at $ \lambda = 4 $**.

---

### Step 5: Excluded Value at $ \lambda = 3 $

At $ \lambda = 3 $, the period-2 points degenerate: the only real solutions of $ f(f(x)) = x $ are the fixed points. Thus, **no period-2 points exist at $ \lambda = 3 $**.

---

### Final Conclusion

The function $ f(x) = \lambda x(1 - x) $ has period-2 points for all values of $ \lambda $ satisfying:

$$
3 < \lambda \leq 4.
$$

Therefore, the answer to the question is:

$$
\boxed{(3, 4]}
$$

Solution:
We have that
\[f(f(x)) = f(\lambda x(1 - x)) = \lambda \cdot \lambda x(1 - x) (1 - \lambda x(1 - x)),\]so we want to solve $\lambda \cdot \lambda x(1 - x) (1 - \lambda x(1 - x)) = x.$

Note that if $f(x) = x,$ then $f(f(x)) = f(x) = x,$ so any roots of $\lambda x(1 - x) = x$ will also be roots of $\lambda \cdot \lambda x(1 - x) (1 - \lambda x(1 - x)) = x.$  Thus, we should expect $\lambda x(1 - x) - x$ to be a factor of $\lambda \cdot \lambda x(1 - x) (1 - \lambda x(1 - x)) - x.$  Indeed,
\[\lambda \cdot \lambda x(1 - x) (1 - \lambda x(1 - x)) - x = (\lambda x(1 - x) - x)(\lambda^2 x^2 - (\lambda^2 + \lambda) x + \lambda + 1).\]The discriminant of $\lambda^2 x^2 - (\lambda^2 + \lambda) x + \lambda + 1$ is
\[(\lambda^2 + \lambda)^2 - 4 \lambda^2 (\lambda + 1) = \lambda^4 - 2 \lambda^3 - 3 \lambda^2 = \lambda^2 (\lambda + 1)(\lambda - 3).\]This is nonnegative when $\lambda = 0$ or $3 \le \lambda \le 4.$

If $\lambda = 0,$ then $f(x) = 0$ for all $x \in [0,1].$

If $\lambda = 3,$ then the equation $f(f(x)) = x$ becomes
\[(3x(1 - x) - x)(9x^2 - 12x + 4) = 0.\]The roots of $9x^2 - 12x + 4 = 0$ are both $\frac{2}{3},$ which satisfy $f(x) = x.$

On the other hand, for $\lambda > 3,$ the roots of $\lambda x(1 - x) = x$ are $x = 0$ and $x = \frac{\lambda - 1}{\lambda}.$  Clearly $x = 0$ is not a root of $\lambda^2 x^2 - (\lambda^2 + \lambda) x + \lambda + 1 = 0.$  Also, if $x = \frac{\lambda - 1}{\lambda},$ then
\[\lambda^2 x^2 - (\lambda^2 + \lambda) x + \lambda + 1 = \lambda^2 \left( \frac{\lambda - 1}{\lambda} \right)^2 - (\lambda^2 + \lambda) \cdot \frac{\lambda - 1}{\lambda} + \lambda + 1 = 3 - \lambda \neq 0.\]Furthermore, the product of the roots is $\frac{\lambda + 1}{\lambda^2},$ which is positive, so either both roots are positive or both roots are negative.  Since the sum of the roots is $\frac{\lambda^2 + \lambda}{\lambda^2} > 0,$ both roots are positive.  Also,
\[\frac{\lambda^2 + \lambda}{\lambda} = 1 + \frac{1}{\lambda} < \frac{4}{3},\]so at least one root must be less than 1.

Therefore, the set of $\lambda$ that satisfy the given condition is $\lambda \in \boxed{(3,4]}.$