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

First, let me make sure I understand the problem. We have this quadratic function f(x) = λx(1 - x), which is a parabola opening downward because the coefficient of x² is -λ, which is negative since λ is non-negative. The domain and codomain are both [0,1], so we're looking at the function's behavior on that interval.

The question is asking for λ values where there's at least one x in [0,1] such that applying f twice brings you 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, first, maybe I should recall that fixed points of f are solutions to f(x) = x. So solving λx(1 - x) = x. Let me write that equation:

λx(1 - x) = x

Subtract x from both sides:

λx(1 - x) - x = 0

Factor out x:

x[λ(1 - x) - 1] = 0

So, either x = 0 or λ(1 - x) - 1 = 0. Let's solve the second equation:

λ(1 - x) - 1 = 0 ⇒ λ(1 - x) = 1 ⇒ 1 - x = 1/λ ⇒ x = 1 - 1/λ

But since x must be in [0,1], we need 1 - 1/λ to be in [0,1]. Let's check when that happens.

First, 1 - 1/λ ≥ 0 ⇒ 1 ≥ 1/λ ⇒ λ ≥ 1. Also, 1 - 1/λ ≤ 1 ⇒ -1/λ ≤ 0, which is always true since λ is non-negative. So, for λ ≥ 1, there is a fixed point at x = 1 - 1/λ. For λ < 1, the only fixed point is x = 0.

Wait, but when λ = 1, x = 1 - 1/1 = 0. So, for λ = 1, the fixed points are x = 0 and x = 1 - 1/1 = 0? Wait, no. Wait, solving λ(1 - x) - 1 = 0 when λ = 1 gives 1*(1 - x) - 1 = 0 ⇒ 1 - x - 1 = -x = 0 ⇒ x = 0. So, for λ = 1, the fixed points are x = 0 (from the first factor) and x = 0 again? Wait, that seems like only one fixed point? Wait, but when λ = 1, the equation f(x) = x becomes x = 1*x*(1 - x) ⇒ x = x - x² ⇒ 0 = -x² ⇒ x = 0. So, only x = 0 is a fixed point. So, for λ = 1, the only fixed point is x = 0.

But when λ > 1, then x = 1 - 1/λ is another fixed point. For example, if λ = 2, then x = 1 - 1/2 = 1/2. Let me check: f(1/2) = 2*(1/2)*(1 - 1/2) = 2*(1/2)*(1/2) = 2*(1/4) = 1/2. So yes, that's a fixed point.

So, in summary, fixed points of f(x) are:

- For λ < 1: only x = 0.

- For λ = 1: only x = 0.

- For λ > 1: x = 0 and x = 1 - 1/λ.

But the problem is not about fixed points. It's about points where f(f(x)) = x but f(x) ≠ x. So, these are points that are periodic with period 2.

So, to find such points, we need to solve f(f(x)) = x, and then exclude the fixed points.

So, first, let me find all solutions to f(f(x)) = x, then subtract the fixed points.

So, let me compute f(f(x)).

First, f(x) = λx(1 - x). Then f(f(x)) = λ * [f(x)] * [1 - f(x)] = λ * [λx(1 - x)] * [1 - λx(1 - x)].

So, f(f(x)) = λ²x(1 - x)[1 - λx(1 - x)].

So, setting f(f(x)) = x:

λ²x(1 - x)[1 - λx(1 - x)] = x.

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

λ²(1 - x)[1 - λx(1 - x)] = 1.

But if x = 0, then f(0) = 0, so x = 0 is a fixed point. So, we need to consider x ≠ 0.

So, the equation becomes:

λ²(1 - x)[1 - λx(1 - x)] = 1.

But this seems complicated. Maybe expanding it would help? Let me try.

First, let me denote y = x(1 - x). Then f(x) = λy, and f(f(x)) = λ * (λy) * (1 - λy) = λ² y (1 - λy). So, f(f(x)) = λ² y (1 - λy).

But we want f(f(x)) = x, so:

λ² y (1 - λy) = x.

But y = x(1 - x), so substituting:

λ² x(1 - x) [1 - λx(1 - x)] = x.

Which is the same equation as before.

So, again, if x ≠ 0, divide both sides by x:

λ² (1 - x)[1 - λx(1 - x)] = 1.

Let me expand the left-hand side:

First, compute 1 - λx(1 - x):

1 - λx + λx².

So, the equation becomes:

λ² (1 - x)(1 - λx + λx²) = 1.

Let me expand (1 - x)(1 - λx + λx²):

Multiply term by term:

First, 1*(1 - λx + λx²) = 1 - λx + λx².

Then, -x*(1 - λx + λx²) = -x + λx² - λx³.

Adding these together:

1 - λx + λx² - x + λx² - λx³.

Combine like terms:

Constant term: 1.

x terms: -λx - x = -x(λ + 1).

x² terms: λx² + λx² = 2λx².

x³ term: -λx³.

So overall:

1 - (λ + 1)x + 2λx² - λx³.

Therefore, the equation is:

λ² [1 - (λ + 1)x + 2λx² - λx³] = 1.

Expanding this:

λ² - λ²(λ + 1)x + 2λ³x² - λ³x³ = 1.

Bring the 1 to the left:

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

So, the equation is:

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

Let me write it in standard polynomial form:

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

Alternatively, multiplying both sides by -1:

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

So:

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

Hmm, this seems complicated. Maybe there's a better way. Let me think.

Alternatively, since we know that fixed points satisfy f(x) = x, which are solutions to the equation f(f(x)) = x. Therefore, the solutions to f(f(x)) = x include all fixed points and the period-2 points. Therefore, to find the period-2 points, we can factor out the fixed points from the equation f(f(x)) = x.

So, if we can factor the equation f(f(x)) = x, then the roots would be the fixed points and the period-2 points. Therefore, if we can factor out the fixed points, then the remaining roots would be the period-2 points.

So, let me try that approach.

First, we know that the fixed points are x = 0 and x = 1 - 1/λ (for λ ≥ 1). So, if we can factor these out from the equation f(f(x)) = x, then the remaining roots would be the period-2 points.

But since we have the equation f(f(x)) = x, which is a cubic equation? Wait, no. Wait, f(f(x)) is a composition of quadratics, so f(f(x)) is a quartic function. Wait, let me check:

Original function f(x) is quadratic, so f(f(x)) is quadratic composed with quadratic, which is quartic (degree 4). Therefore, the equation f(f(x)) = x is a quartic equation, which can have up to 4 roots. However, since we already know that x = 0 is a root (since f(0) = 0, so f(f(0)) = 0 = 0), and x = 1 - 1/λ (for λ ≥ 1) is also a root. Therefore, the quartic equation f(f(x)) = x can be factored as (x)(x - (1 - 1/λ)) * quadratic = 0, assuming λ ≥ 1.

But for λ < 1, the only fixed point is x = 0, so the quartic equation would have x = 0 as a root, and maybe other roots.

But perhaps I need to consider both cases. However, since the problem is asking for λ in [0,4], maybe I can proceed by considering the quartic equation and factoring out the fixed points.

Alternatively, maybe I can write f(f(x)) - x = 0 and factor it. Let me try that.

Let me compute f(f(x)) - x:

f(f(x)) - x = λ²x(1 - x)[1 - λx(1 - x)] - x.

Factor out x:

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

So, the solutions are x = 0 and solutions to λ²(1 - x)[1 - λx(1 - x)] - 1 = 0.

So, the equation we need to solve for period-2 points is:

λ²(1 - x)[1 - λx(1 - x)] - 1 = 0.

Which is the same as before.

Alternatively, maybe I can write this equation as:

Let me denote y = x(1 - x). Then, the equation becomes:

λ²(1 - x)(1 - λy) - 1 = 0.

But since y = x(1 - x), maybe not helpful.

Alternatively, perhaps I can consider that for period-2 points, we have f(x) ≠ x, but f(f(x)) = x. So, if we let y = f(x), then y ≠ x, but f(y) = x. Therefore, we have the system:

y = λx(1 - x)

x = λy(1 - y)

So, substituting the first equation into the second:

x = λ[λx(1 - x)] [1 - λx(1 - x)].

Which is the same as:

x = λ²x(1 - x)[1 - λx(1 - x)].

Which is the same equation as before.

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

1 = λ²(1 - x)[1 - λx(1 - x)].

Which is the same equation as before.

So, perhaps solving this equation for x in [0,1], and then ensuring that x ≠ f(x).

Alternatively, since we have the system:

y = λx(1 - x)

x = λy(1 - y)

So, substituting y from the first equation into the second:

x = λ[λx(1 - x)][1 - λx(1 - x)].

Which is the same as before.

Alternatively, maybe we can consider solving the system:

y = λx(1 - x)

x = λy(1 - y)

So, substituting y from the first equation into the second:

x = λ[λx(1 - x)][1 - λx(1 - x)].

Which is the same equation.

Alternatively, maybe we can consider subtracting the two equations? Not sure.

Alternatively, let me consider that if we have y = λx(1 - x) and x = λy(1 - y), then we can write:

y = λx(1 - x)

x = λy(1 - y)

So, multiplying both equations:

xy = λ²x(1 - x)y(1 - y)

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

1 = λ²(1 - x)(1 - y)

So, 1 = λ²(1 - x)(1 - y)

But from the first equation, y = λx(1 - x). Let me substitute that into the above equation:

1 = λ²(1 - x)[1 - λx(1 - x)]

Which is the same equation as before.

So, again, we get back to the same equation.

Therefore, perhaps I need to solve this equation for x in [0,1], and then check if x ≠ f(x).

Alternatively, maybe I can consider that for period-2 points, the solutions to the equation f(f(x)) = x are the fixed points and the period-2 points. Therefore, if I can find the number of solutions to f(f(x)) = x, and subtract the fixed points, then the remaining solutions would be the period-2 points.

But since the problem is asking for the existence of at least one such x, maybe I can find for which λ there exists such x.

Alternatively, maybe I can consider the equation f(f(x)) = x and find when it has solutions other than the fixed points.

Alternatively, maybe I can think graphically. The function f(x) is a parabola opening downward, with maximum at x = 1/2, f(1/2) = λ*(1/2)*(1 - 1/2) = λ*(1/2)*(1/2) = λ/4.

So, the maximum value of f(x) is λ/4. Since the domain is [0,1], and the codomain is [0,1], we need that λ/4 ≤ 1 ⇒ λ ≤ 4, which is given.

So, for λ in [0,4], the maximum of f(x) is between 0 and 1.

Now, the equation f(f(x)) = x. Let me think about the graph of f(f(x)) and the line y = x. The intersections of these two graphs will give the solutions.

But since f(f(x)) is a composition of two functions, it's a quartic function.

But maybe instead of dealing with the quartic equation, I can think about the system:

y = f(x)

x = f(y)

Which is equivalent to the period-2 points.

So, solving this system:

y = λx(1 - x)

x = λy(1 - y)

So, substituting the first into the second:

x = λ[λx(1 - x)][1 - λx(1 - x)]

Which is the same equation as before.

Alternatively, maybe I can consider that if we have x ≠ y, then we can write:

From the two equations:

y = λx(1 - x)

x = λy(1 - y)

Let me subtract the two equations:

y - x = λx(1 - x) - λy(1 - y)

But not sure if helpful.

Alternatively, let me consider dividing the two equations.

From y = λx(1 - x) and x = λy(1 - y), dividing the first by the second:

y / x = [λx(1 - x)] / [λy(1 - y)] ⇒ y / x = [x(1 - x)] / [y(1 - y)]

Cross multiplying:

y²(1 - y) = x²(1 - x)

So, y²(1 - y) = x²(1 - x)

But since y = λx(1 - x), substitute that in:

[λ²x²(1 - x)²][1 - λx(1 - x)] = x²(1 - x)

Assuming x ≠ 0 and 1 - x ≠ 0 (i.e., x ≠ 0,1), we can divide both sides by x²(1 - x):

λ²(1 - x)[1 - λx(1 - x)] = 1

Which is the same equation as before.

So, again, we get back to the same equation.

Therefore, perhaps I need to solve this equation for x in [0,1], and then check if there are solutions where x ≠ f(x).

Alternatively, maybe I can consider that for period-2 points, the solutions to the equation f(f(x)) = x are the fixed points and the period-2 points. Therefore, if the equation f(f(x)) = x has more than two solutions (the fixed points), then there are period-2 points.

But how many solutions does f(f(x)) = x have?

Well, since f(f(x)) is a quartic function, it can have up to 4 real roots. However, depending on λ, the number of roots can vary.

But since we already know that x = 0 is a root, and for λ ≥ 1, x = 1 - 1/λ is another root. So, if there are other roots, they would be the period-2 points.

But maybe I can consider the case when there are period-2 points.

Alternatively, maybe I can think about the function g(x) = f(f(x)) - x. We need to find λ such that g(x) = 0 has solutions other than the fixed points.

Alternatively, maybe I can consider specific values of λ and see when such points exist.

But since the problem is asking for all λ in [0,4], maybe I need to find the range of λ where such points exist.

Alternatively, maybe I can consider the equation we derived earlier:

λ²(1 - x)[1 - λx(1 - x)] = 1.

Let me denote z = x(1 - x). Then, since x ∈ [0,1], z ∈ [0, 1/4], because the maximum of x(1 - x) is at x = 1/2, which is 1/4.

So, z ∈ [0, 1/4].

Then, the equation becomes:

λ²(1 - x)(1 - λz) = 1.

But z = x(1 - x), so 1 - x = (1 - x). Hmm, not sure.

Alternatively, maybe express 1 - x in terms of z?

Since z = x(1 - x), we can write 1 - x = (z)/x, but that might not help.

Alternatively, maybe express 1 - x as something else.

Alternatively, maybe think of this as a quadratic equation in terms of x.

Let me try expanding the equation again:

λ²(1 - x)[1 - λx(1 - x)] = 1.

Let me expand the term inside:

First, expand 1 - λx(1 - x):

1 - λx + λx².

So, the equation becomes:

λ²(1 - x)(1 - λx + λx²) = 1.

Earlier, we expanded this to:

λ²[1 - (λ + 1)x + 2λx² - λx³] = 1.

Which led to:

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

Alternatively, maybe I can write this as:

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

Wait, let me check:

Original expansion was:

λ²[1 - (λ + 1)x + 2λx² - λx³] = 1.

Which is:

λ² - λ²(λ + 1)x + 2λ³x² - λ³x³ = 1.

Then moving 1 to the left:

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

Which is:

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

Alternatively, multiplying by -1:

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

So:

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

Wait, because λ²(λ + 1)x = λ³x + λ²x.

So, grouping terms:

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

Hmm, maybe factor terms:

Let me factor λ³x³ - 2λ³x² + λ³x:

λ³x(x² - 2x + 1) = λ³x(x - 1)^2.

Then, the remaining terms are λ²x - λ² + 1.

So, the equation becomes:

λ³x(x - 1)^2 + λ²x - λ² + 1 = 0.

So:

λ³x(x - 1)^2 + λ²(x - 1) + 1 = 0.

Wait, because λ²x - λ² = λ²(x - 1).

So, the equation is:

λ³x(x - 1)^2 + λ²(x - 1) + 1 = 0.

Hmm, interesting. Let me check:

Yes, because:

Original equation after grouping:

λ³x(x - 1)^2 + λ²(x - 1) + 1 = 0.

So, this seems like a more manageable form.

So, let me write that:

λ³x(x - 1)^2 + λ²(x - 1) + 1 = 0.

Let me denote t = x - 1. Then, x = t + 1. But maybe not helpful. Alternatively, let me consider substituting u = x - 1, but maybe not.

Alternatively, note that x ∈ [0,1], so (x - 1) is between -1 and 0.

Alternatively, maybe factor out (x - 1):

But the equation is:

λ³x(x - 1)^2 + λ²(x - 1) + 1 = 0.

Let me factor out (x - 1) from the first two terms:

(x - 1)[λ³x(x - 1) + λ²] + 1 = 0.

But not sure.

Alternatively, maybe consider that for x ∈ [0,1], (x - 1) is negative or zero.

Alternatively, maybe try specific values of x.

Alternatively, think about the equation:

λ³x(x - 1)^2 + λ²(x - 1) + 1 = 0.

Let me rearrange:

λ³x(x - 1)^2 = - [λ²(x - 1) + 1].

Note that the left-hand side is λ³x(x - 1)^2. Since x ∈ [0,1], (x - 1) is negative or zero, so (x - 1)^2 is non-negative. Also, x is non-negative. Therefore, the left-hand side is non-negative (since λ³ is non-negative, x is non-negative, and (x - 1)^2 is non-negative).

The right-hand side is - [λ²(x - 1) + 1]. Let me compute the expression inside the brackets:

λ²(x - 1) + 1 = λ²x - λ² + 1.

So, the right-hand side is - [λ²x - λ² + 1] = -λ²x + λ² - 1.

Therefore, the equation becomes:

λ³x(x - 1)^2 = -λ²x + λ² - 1.

So:

λ³x(x - 1)^2 + λ²x - λ² + 1 = 0.

Wait, that's the same as before.

But since the left-hand side is non-negative, and the right-hand side is - [λ²(x - 1) + 1], which is equal to - [λ²x - λ² + 1].

But since the left-hand side is non-negative, the right-hand side must also be non-negative. Therefore:

- [λ²x - λ² + 1] ≥ 0 ⇒ λ²x - λ² + 1 ≤ 0 ⇒ λ²(x - 1) + 1 ≤ 0 ⇒ λ²(x - 1) ≤ -1 ⇒ x - 1 ≤ -1/λ² ⇒ x ≤ 1 - 1/λ².

But since x ∈ [0,1], this gives a condition on x.

So, for the equation to have solutions, we need x ≤ 1 - 1/λ².

But since x ∈ [0,1], this requires that 1 - 1/λ² ≥ 0 ⇒ λ² ≥ 1 ⇒ λ ≥ 1.

Therefore, for λ < 1, there are no solutions? Wait, but if λ < 1, then 1 - 1/λ² is negative, so x ≤ negative number, but x ∈ [0,1], so no solutions. Therefore, for λ < 1, there are no solutions to the equation?

But wait, if λ < 1, then the equation f(f(x)) = x would have solutions? For example, take λ = 0. Then f(x) = 0, so f(f(x)) = 0 = x. So, for λ = 0, every x in [0,1] satisfies f(f(x)) = x. But in this case, f(x) = 0 for all x, so f(x) = x only when x = 0. Therefore, for λ = 0, all x satisfy f(f(x)) = x, but only x = 0 satisfies f(x) = x. Therefore, for λ = 0, there are points (all x ≠ 0) where f(f(x)) = x but f(x) ≠ x.

But according to the previous reasoning, for λ < 1, there are no solutions? But that contradicts with λ = 0.

So, there must be a mistake in my reasoning.

Let me check again.

Earlier, I said that the left-hand side is non-negative, and the right-hand side is - [λ²(x - 1) + 1]. Therefore, for the equation to hold, the right-hand side must be non-negative.

But for λ = 0, the equation becomes:

0³x(x - 1)^2 + 0²(x - 1) + 1 = 0 ⇒ 0 + 0 + 1 = 0 ⇒ 1 = 0, which is false.

Wait, but earlier, for λ = 0, f(f(x)) = 0 = x, so x must be 0. But according to the equation we derived, which was after dividing by x (assuming x ≠ 0), we have:

λ²(1 - x)[1 - λx(1 - x)] = 1.

For λ = 0, this becomes 0 * ... = 1, which is 0 = 1, which is false. Therefore, the equation λ²(1 - x)[1 - λx(1 - x)] = 1 has no solutions for λ = 0, except x = 0.

But when λ = 0, f(f(x)) = 0 for all x, so f(f(x)) = x implies 0 = x, so x = 0. Therefore, the only solution is x = 0, which is a fixed point. Therefore, there are no points where f(f(x)) = x but f(x) ≠ x.

But earlier, I thought that for λ = 0, all x satisfy f(f(x)) = x, but actually, f(f(x)) = 0 for all x, so f(f(x)) = x implies x = 0. Therefore, only x = 0 is a solution. Therefore, for λ = 0, there are no such points.

So, my previous thought was wrong. Therefore, for λ = 0, there are no such points.

Therefore, the previous reasoning that for λ < 1, there are no solutions is correct? But then, for λ = 1, let's check.

For λ = 1, the equation becomes:

1²(1 - x)[1 - 1*x(1 - x)] = 1 ⇒ (1 - x)[1 - x + x²] = 1.

Let me compute this:

(1 - x)(1 - x + x²) = ?

Multiply out:

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

Set equal to 1:

1 - 2x + 2x² - x³ = 1 ⇒ -2x + 2x² - x³ = 0 ⇒ x(-2 + 2x - x²) = 0.

Therefore, solutions are x = 0 or -2 + 2x - x² = 0 ⇒ x² - 2x + 2 = 0. The discriminant is 4 - 8 = -4 < 0, so no real solutions. Therefore, only x = 0 is a solution. Therefore, for λ = 1, the only solution is x = 0, which is a fixed point. Therefore, no period-2 points.

So, for λ = 1, there are no such points.

Therefore, for λ < 1, the equation has only x = 0 as a solution, which is a fixed point. Therefore, no period-2 points.

For λ ≥ 1, we have the equation:

λ³x(x - 1)^2 + λ²(x - 1) + 1 = 0.

But earlier, we saw that for the equation to have solutions, we need x ≤ 1 - 1/λ².

But since x ∈ [0,1], and 1 - 1/λ² ≥ 0 when λ ≥ 1.

So, for λ ≥ 1, we can have solutions.

Therefore, the problem reduces to finding λ ≥ 1 such that there exists x ∈ [0,1] with x ≤ 1 - 1/λ² and satisfying the equation.

But how do we find such λ?

Alternatively, maybe we can consider that for λ ≥ 1, the equation f(f(x)) = x has solutions other than the fixed points.

But how can we determine for which λ this is true?

Alternatively, maybe we can consider the case when there are period-2 points.

Alternatively, think about the function g(x) = f(f(x)) - x. We need to find λ such that g(x) = 0 has solutions other than the fixed points.

Alternatively, maybe consider specific values of λ.

For example, take λ = 2.

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

Let me check if there are period-2 points.

So, solving f(f(x)) = x.

Let me compute f(f(x)) for λ = 2:

f(x) = 2x(1 - x).

Then f(f(x)) = 2 * [2x(1 - x)] * [1 - 2x(1 - x)].

Let me compute this:

First, compute 2x(1 - x) = 2x - 2x².

Then, 1 - 2x(1 - x) = 1 - 2x + 2x².

Therefore, f(f(x)) = 2*(2x - 2x²)*(1 - 2x + 2x²).

Let me compute this:

First, multiply (2x - 2x²)(1 - 2x + 2x²):

Let me expand term by term:

First, 2x*(1 - 2x + 2x²) = 2x - 4x² + 4x³.

Then, -2x²*(1 - 2x + 2x²) = -2x² + 4x³ - 4x⁴.

Adding these together:

2x - 4x² + 4x³ - 2x² + 4x³ - 4x⁴ = 2x - 6x² + 8x³ - 4x⁴.

Multiply by 2:

f(f(x)) = 4x - 12x² + 16x³ - 8x⁴.

Set equal to x:

4x - 12x² + 16x³ - 8x⁴ = x ⇒ 4x - 12x² + 16x³ - 8x⁴ - x = 0 ⇒ 3x - 12x² + 16x³ - 8x⁴ = 0 ⇒ -8x⁴ + 16x³ - 12x² + 3x = 0 ⇒ x(-8x³ + 16x² - 12x + 3) = 0.

So, solutions are x = 0 and solutions to -8x³ + 16x² - 12x + 3 = 0.

Let me write this as 8x³ - 16x² + 12x - 3 = 0.

Let me try to find rational roots using Rational Root Theorem. Possible rational roots are ±1, ±3, ±1/2, ±3/2, ±1/4, ±3/4, ±1/8, ±3/8.

Testing x = 1: 8 - 16 + 12 - 3 = 1 ≠ 0.

x = 3/2: Not in [0,1], so skip.

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

Yes! x = 1/2 is a root.

Therefore, we can factor (x - 1/2) from the cubic.

Let me perform polynomial division or use synthetic division.

Divide 8x³ - 16x² + 12x - 3 by (x - 1/2).

Using synthetic division:

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.

So, the cubic factors as (x - 1/2)(8x² - 12x + 6).

Therefore, the equation becomes x(x - 1/2)(8x² - 12x + 6) = 0.

So, solutions are x = 0, x = 1/2, and solutions to 8x² - 12x + 6 = 0.

Solving 8x² - 12x + 6 = 0:

Discriminant D = 144 - 192 = -48 < 0.

Therefore, no real roots.

Therefore, for λ = 2, the only real solutions are x = 0 and x = 1/2, which are fixed points. Therefore, there are no period-2 points for λ = 2.

Wait, but that contradicts my initial thought. Maybe λ = 2 is not in the desired range?

Wait, but maybe I made a mistake here. Let me check with specific values.

Take x = 1/2. Then f(x) = 2*(1/2)*(1 - 1/2) = 2*(1/2)*(1/2) = 1/2. So, fixed point.

What about another x? Let me try x = 1/4.

f(1/4) = 2*(1/4)*(1 - 1/4) = 2*(1/4)*(3/4) = 2*(3/16) = 3/8.

Then f(3/8) = 2*(3/8)*(1 - 3/8) = 2*(3/8)*(5/8) = 30/64 = 15/32 ≈ 0.46875.

Which is not equal to 1/4. So, f(f(1/4)) ≈ 0.46875 ≠ 1/4.

What about x = 3/4?

f(3/4) = 2*(3/4)*(1 - 3/4) = 2*(3/4)*(1/4) = 6/16 = 3/8.

Then f(3/8) = 15/32 as before. Not equal to 3/4.

So, seems like for λ = 2, there are no period-2 points.

But then, when does a period-2 point exist?

Let me try λ = 3.

For λ = 3, fixed points are x = 0 and x = 1 - 1/3 = 2/3.

Let me try solving f(f(x)) = x.

First, f(x) = 3x(1 - x).

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

Let me compute this:

First, compute 3x(1 - x) = 3x - 3x².

Then, 1 - 3x(1 - x) = 1 - 3x + 3x².

Therefore, f(f(x)) = 3*(3x - 3x²)*(1 - 3x + 3x²).

Let me compute (3x - 3x²)(1 - 3x + 3x²):

First, expand term by term:

3x*(1 - 3x + 3x²) = 3x - 9x² + 9x³.

-3x²*(1 - 3x + 3x²) = -3x² + 9x³ - 9x⁴.

Adding together:

3x - 9x² + 9x³ - 3x² + 9x³ - 9x⁴ = 3x - 12x² + 18x³ - 9x⁴.

Multiply by 3:

f(f(x)) = 9x - 36x² + 54x³ - 27x⁴.

Set equal to x:

9x - 36x² + 54x³ - 27x⁴ = x ⇒ 9x - 36x² + 54x³ - 27x⁴ - x = 0 ⇒ 8x - 36x² + 54x³ - 27x⁴ = 0 ⇒ -27x⁴ + 54x³ - 36x² + 8x = 0 ⇒ x(-27x³ + 54x² - 36x + 8) = 0.

So, solutions are x = 0 and solutions to -27x³ + 54x² - 36x + 8 = 0.

Let me write this as 27x³ - 54x² + 36x - 8 = 0.

Try rational roots. Possible roots: ±1, ±2, ±4, ±8, ±1/3, etc.

Testing x = 2/3:

27*(8/27) - 54*(4/9) + 36*(2/3) - 8 = 8 - 24 + 24 - 8 = 0.

Yes! x = 2/3 is a root.

Therefore, factor out (x - 2/3).

Using synthetic division:

Divide 27x³ - 54x² + 36x - 8 by (x - 2/3).

Using synthetic division with root 2/3:

Coefficients: 27 | -54 | 36 | -8

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.

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

Therefore, the equation becomes x(x - 2/3)(27x² - 36x + 12) = 0.

Solving 27x² - 36x + 12 = 0:

Divide by 3: 9x² - 12x + 4 = 0.

Discriminant D = 144 - 144 = 0. So, repeated root: x = [12 ± 0]/18 = 12/18 = 2/3.

Therefore, the equation factors as x(x - 2/3)^3 = 0.

Therefore, the only real solutions are x = 0 and x = 2/3. Which are the fixed points. Therefore, again, no period-2 points for λ = 3.

Hmm, so for λ = 2 and λ = 3, there are no period-2 points.

Wait, but maybe for some λ between 1 and 4, there are period-2 points?

Let me try λ = 4.

For λ = 4, fixed points are x = 0 and x = 1 - 1/4 = 3/4.

Let me compute f(f(x)) = x.

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

Then f(f(x)) = 4 * [4x(1 - x)] * [1 - 4x(1 - x)].

Compute 4x(1 - x) = 4x - 4x².

Then 1 - 4x(1 - x) = 1 - 4x + 4x².

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

Let me compute (4x - 4x²)(1 - 4x + 4x²):

First, expand term by term:

4x*(1 - 4x + 4x²) = 4x - 16x² + 16x³.

-4x²*(1 - 4x + 4x²) = -4x² + 16x³ - 16x⁴.

Adding together:

4x - 16x² + 16x³ - 4x² + 16x³ - 16x⁴ = 4x - 20x² + 32x³ - 16x⁴.

Multiply by 4:

f(f(x)) = 16x - 80x² + 128x³ - 64x⁴.

Set equal to x:

16x - 80x² + 128x³ - 64x⁴ = x ⇒ 16x - 80x² + 128x³ - 64x⁴ - x = 0 ⇒ 15x - 80x² + 128x³ - 64x⁴ = 0 ⇒ -64x⁴ + 128x³ - 80x² + 15x = 0 ⇒ x(-64x³ + 128x² - 80x + 15) = 0.

So, solutions are x = 0 and solutions to -64x³ + 128x² - 80x + 15 = 0.

Let me write this as 64x³ - 128x² + 80x - 15 = 0.

Try rational roots. Possible roots: ±1, ±3, ±5, ±15, ±1/2, etc.

Testing x = 3/4:

64*(27/64) - 128*(9/16) + 80*(3/4) - 15 = 27 - 72 + 60 - 15 = 0.

Yes! x = 3/4 is a root.

Therefore, factor out (x - 3/4).

Using synthetic division:

Divide 64x³ - 128x² + 80x - 15 by (x - 3/4).

Using synthetic division with root 3/4:

Coefficients: 64 | -128 | 80 | -15

Bring down 64.

Multiply by 3/4: 48.

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

Multiply by 3/4: -60.

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

Multiply by 3/4: 15.

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

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

Therefore, the equation becomes x(x - 3/4)(64x² - 80x + 20) = 0.

Solving 64x² - 80x + 20 = 0:

Divide by 4: 16x² - 20x + 5 = 0.

Discriminant D = 400 - 320 = 80.

Solutions: [20 ± √80]/32 = [20 ± 4√5]/32 = [5 ± √5]/8 ≈ [5 ± 2.236]/8.

So, approximately (5 + 2.236)/8 ≈ 7.236/8 ≈ 0.9045 and (5 - 2.236)/8 ≈ 2.764/8 ≈ 0.3455.

So, these are two real roots.

Therefore, for λ = 4, the solutions to f(f(x)) = x are x = 0, x = 3/4, and x ≈ 0.9045 and x ≈ 0.3455.

Now, check if these are fixed points or period-2 points.

First, x = 0 and x = 3/4 are fixed points.

Check x ≈ 0.9045: compute f(x) and see if it's equal to x or not.

Let me compute f(0.9045) = 4 * 0.9045 * (1 - 0.9045) ≈ 4 * 0.9045 * 0.0955 ≈ 4 * 0.0863 ≈ 0.345.

Which is approximately the other root, 0.3455. Then f(0.3455) ≈ 4 * 0.3455 * (1 - 0.3455) ≈ 4 * 0.3455 * 0.6545 ≈ 4 * 0.226 ≈ 0.904.

Therefore, these two points form a 2-cycle: f(0.9045) ≈ 0.3455 and f(0.3455) ≈ 0.9045. Therefore, they are period-2 points.

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

So, for λ = 4, there are such points.

Therefore, the question is, for which λ between 1 and 4 does this happen?

Earlier, for λ = 2 and λ = 3, there were no period-2 points, but for λ = 4, there are.

Therefore, there must be a critical value of λ where period-2 points start to exist.

This suggests that the period-2 points exist for some λ ≥ some value.

To find this critical value, maybe we need to find when the equation f(f(x)) = x has solutions other than the fixed points.

Alternatively, since for λ = 4, there are period-2 points, and for λ = 2 and 3, there are not, maybe there is a range of λ where period-2 points exist.

Alternatively, maybe the period-2 points exist when the function f has a certain property, like when the function is chaotic or something.

Alternatively, maybe we can find the values of λ for which the equation f(f(x)) = x has solutions other than the fixed points.

Let me recall that for the logistic map f(x) = λx(1 - x), the period-doubling bifurcations occur at specific λ values.

But I might not remember the exact values, but maybe I can derive them.

Alternatively, since we have the equation for period-2 points, which is:

λ³x(x - 1)^2 + λ²(x - 1) + 1 = 0.

But earlier, we had:

For λ ≥ 1, the equation is:

λ³x(x - 1)^2 + λ²(x - 1) + 1 = 0.

Let me denote t = x - 1, so x = t + 1. But since x ∈ [0,1], t ∈ [-1, 0].

But maybe not helpful.

Alternatively, let me consider that for period-2 points, we have x ≠ f(x), and f(f(x)) = x.

Let me consider the system:

y = λx(1 - x)

x = λy(1 - y)

Which can be rewritten as:

y = λx(1 - x)

x = λy(1 - y)

Let me subtract the two equations:

y - x = λx(1 - x) - λy(1 - y)

But not sure.

Alternatively, let me consider that if we have a period-2 cycle, then the points x and y satisfy:

y = λx(1 - x)

x = λy(1 - y)

So, substituting y from the first equation into the second:

x = λ[λx(1 - x)][1 - λx(1 - x)]

Which is the same equation as before.

Let me denote z = x(1 - x). Then, y = λz, and the second equation becomes:

x = λ(λz)(1 - λz) = λ²z(1 - λz).

But z = x(1 - x), so:

x = λ²x(1 - x)(1 - λx(1 - x)).

Assuming x ≠ 0, divide both sides by x:

1 = λ²(1 - x)(1 - λx(1 - x)).

Which is the same equation as before.

Let me denote u = x(1 - x). Then, the equation becomes:

1 = λ²(1 - x)(1 - λu).

But since u = x(1 - x), and 1 - x = (1 - x).

Alternatively, maybe express 1 - x in terms of u.

Note that u = x(1 - x) ⇒ 1 - x = u/x, assuming x ≠ 0.

Therefore, substituting into the equation:

1 = λ²*(u/x)*(1 - λu).

So:

1 = λ²*(u/x)*(1 - λu).

But u = x(1 - x), so:

1 = λ²*(x(1 - x)/x)*(1 - λx(1 - x)) = λ²(1 - x)(1 - λx(1 - x)).

Which is the same equation.

Alternatively, maybe express everything in terms of u.

But not sure.

Alternatively, let me consider that for period-2 points, the function f has a 2-cycle. For the logistic map, the period-2 points exist when the function f has a certain property.

The standard logistic map is f(x) = λx(1 - x). The period-2 points satisfy f(f(x)) = x, but f(x) ≠ x.

The equation f(f(x)) = x can be written as:

f(f(x)) - x = 0.

Which is a quartic equation.

The number of real solutions depends on λ.

For the logistic map, the period-2 points exist for λ > 2. Wait, but earlier for λ = 2, there were no period-2 points.

Wait, maybe I need to check.

Actually, the period-doubling bifurcation occurs at λ = 3 for the logistic map. Wait, no, the first period-doubling occurs at λ = 3?

Wait, let me recall.

The logistic map f(x) = λx(1 - x) has fixed points at x = 0 and x = 1 - 1/λ. The stability of these fixed points changes with λ.

The fixed point at x = 1 - 1/λ is stable for λ < 3 and unstable for λ > 3.

At λ = 3, the fixed point becomes unstable, and a period-2 cycle is born.

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

But earlier, for λ = 4, we saw that there are period-2 points.

But for λ = 3, let me check.

Wait, earlier when I tried λ = 3, the equation f(f(x)) = x had only fixed points. But according to the bifurcation theory, at λ = 3, the period-2 cycle is born.

So, maybe my calculation for λ = 3 was wrong?

Let me check again for λ = 3.

For λ = 3, the equation f(f(x)) = x was:

After solving, we had x(x - 2/3)^3 = 0.

But that suggests only fixed points.

But according to bifurcation theory, there should be period-2 points for λ > 3.

So, maybe there was an error in my calculation.

Let me redo the calculation for λ = 3.

For λ = 3, f(x) = 3x(1 - x).

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

Let me compute this step by step.

First, compute f(x) = 3x(1 - x).

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

Let me compute 1 - 3x(1 - x) = 1 - 3x + 3x².

So, f(f(x)) = 3 * 3x(1 - x) * (1 - 3x + 3x²) = 9x(1 - x)(1 - 3x + 3x²).

Set equal to x:

9x(1 - x)(1 - 3x + 3x²) = x.

Assuming x ≠ 0, divide both sides by x:

9(1 - x)(1 - 3x + 3x²) = 1.

Expand the left-hand side:

First, expand (1 - x)(1 - 3x + 3x²):

Multiply term by term:

1*(1 - 3x + 3x²) = 1 - 3x + 3x².

- x*(1 - 3x + 3x²) = -x + 3x² - 3x³.

Adding together:

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

Multiply by 9:

9 - 36x + 54x² - 27x³ = 1.

Subtract 1:

8 - 36x + 54x² - 27x³ = 0 ⇒ -27x³ + 54x² - 36x + 8 = 0 ⇒ 27x³ - 54x² + 36x - 8 = 0.

Earlier, I factored this as (x - 2/3)(27x² - 36x + 12) = 0.

But let me check if this is correct.

Let me perform polynomial division of 27x³ - 54x² + 36x - 8 by (x - 2/3).

Using synthetic division with root 2/3:

Coefficients: 27 | -54 | 36 | -8

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.

So, the quotient is 27x² - 36x + 12.

Then, 27x² - 36x + 12 = 0 ⇒ divide by 3: 9x² - 12x + 4 = 0 ⇒ discriminant D = 144 - 144 = 0 ⇒ repeated root x = 12/(2*9) = 12/18 = 2/3.

Therefore, the equation factors as (x - 2/3)^3 * 27 = 0? Wait, no.

Wait, the original cubic was 27x³ - 54x² + 36x - 8 = 0, which factors as (x - 2/3)(27x² - 36x + 12) = 0. And the quadratic factors as 3(9x² - 12x + 4) = 3(3x - 2)^2.

Therefore, the equation is (x - 2/3) * 3(3x - 2)^2 = 0 ⇒ 3(x - 2/3)(3x - 2)^2 = 0.

Therefore, the solutions are x = 2/3 (with multiplicity 1) and x = 2/3 (with multiplicity 2 from the quadratic). Wait, no.

Wait, 3x - 2 = 0 ⇒ x = 2/3. So, the quadratic is (3x - 2)^2, so the equation is (x - 2/3) * 3*(3x - 2)^2 = 0 ⇒ (x - 2/3) * 3*(x - 2/3)^2 = 0 ⇒ 3(x - 2/3)^3 = 0.

Therefore, the only real solution is x = 2/3 with multiplicity 3.

Therefore, for λ = 3, the equation f(f(x)) = x has only x = 0 and x = 2/3 as solutions.

But according to bifurcation theory, at λ = 3, the period-2 cycle is born. So, why isn't there a period-2 cycle for λ = 3?

Wait, maybe the period-2 cycle is born at λ = 3, but it's a double root? Or maybe the period-2 points coincide with the fixed points?

Wait, for λ = 3, the fixed point at x = 2/3 is unstable, and the period-2 points might be at the same location?

Wait, no. If the period-2 cycle is born at λ = 3, then for λ > 3, there are period-2 points.

But for λ = 3, the period-2 points might be at the same location as the fixed point?

Wait, let me check.

If we have a period-2 cycle, then the points x and y satisfy f(x) = y and f(y) = x.

If x = y, then it's a fixed point. So, for λ = 3, the equation f(f(x)) = x has only the fixed points, but the period-2 points are tangent to the fixed points?

This suggests that at λ = 3, the period-2 cycle is born with a tangency, i.e., the period-2 points coincide with the fixed points.

Therefore, for λ > 3, the period-2 points exist.

But earlier, for λ = 4, we saw that there are period-2 points.

Therefore, the answer might be λ > 3.

But let me check for λ = 3.5.

Let me try λ = 3.5.

For λ = 3.5, fixed points are x = 0 and x = 1 - 1/3.5 = 1 - 2/7 = 5/7 ≈ 0.714.

Let me try solving f(f(x)) = x.

But this might be tedious. Alternatively, let me consider the equation we had earlier:

For λ ≥ 1, the equation is:

λ³x(x - 1)^2 + λ²(x - 1) + 1 = 0.

Let me denote this as:

λ³x(x - 1)^2 + λ²(x - 1) + 1 = 0.

Let me make a substitution. Let me set t = x - 1, so x = t + 1, where t ∈ [-1, 0].

Substituting into the equation:

λ³(t + 1)(t)^2 + λ²(t) + 1 = 0.

Expanding:

λ³(t + 1)t² + λ²t + 1 = 0.

Which is:

λ³t³ + λ³t² + λ²t + 1 = 0.

So:

λ³t³ + λ³t² + λ²t + 1 = 0.

This is a cubic equation in t.

Let me factor this equation.

Let me try to factor it.

Let me write it as:

λ³t³ + λ³t² + λ²t + 1 = 0.

Group terms:

(λ³t³ + λ³t²) + (λ²t + 1) = 0 ⇒ λ³t²(t + 1) + (λ²t + 1) = 0.

Not helpful.

Alternatively, try to factor as (at + b)(ct² + dt + e) = 0.

Assume it factors as (at + b)(ct² + dt + e) = 0.

Expanding: act³ + (ad + bc)t² + (ae + bd)t + be = 0.

Comparing coefficients:

act³ + (ad + bc)t² + (ae + bd)t + be = λ³t³ + λ³t² + λ²t + 1.

Therefore:

ac = λ³,

ad + bc = λ³,

ae + bd = λ²,

be = 1.

Assume that b and e are constants such that be = 1. Let me assume b = 1, e = 1.

Then:

ac = λ³,

ad + c = λ³,

a*1 + d*1 = λ²,

be = 1.

From the third equation: a + d = λ².

From the second equation: ad + c = λ³.

From the first equation: ac = λ³.

Let me solve for c from the first equation: c = λ³ / a.

Substitute into the second equation: ad + λ³ / a = λ³ ⇒ ad = λ³ - λ³ / a = λ³(1 - 1/a).

From the third equation: d = λ² - a.

Substitute into the above: a(λ² - a) = λ³(1 - 1/a).

Multiply both sides by a:

a²(λ² - a) = λ³(a - 1).

Expand left side: a²λ² - a³.

Right side: λ³a - λ³.

Bring all terms to left:

a²λ² - a³ - λ³a + λ³ = 0.

Factor:

- a³ + a²λ² - λ³a + λ³ = 0.

Factor by grouping:

(-a³ + a²λ²) + (-λ³a + λ³) = 0 ⇒ -a²(a - λ²) - λ³(a - 1) = 0.

Not sure.

Alternatively, let me try specific values.

Assume that a = λ. Then:

From ac = λ³ ⇒ c = λ³ / a = λ³ / λ = λ².

From third equation: d = λ² - a = λ² - λ.

From second equation: ad + c = λ³ ⇒ λ*(λ² - λ) + λ² = λ³ - λ² + λ² = λ³, which matches.

Therefore, this works!

Therefore, the cubic factors as (at + b)(ct² + dt + e) = (λt + 1)(λ²t² + (λ² - λ)t + 1).

Let me check:

(λt + 1)(λ²t² + (λ² - λ)t + 1) = λt * λ²t² + λt * (λ² - λ)t + λt * 1 + 1 * λ²t² + 1 * (λ² - λ)t + 1 * 1.

Compute term by term:

λ³t³ + λ(λ² - λ)t² + λt + λ²t² + (λ² - λ)t + 1.

Simplify:

λ³t³ + [λ³ - λ² + λ²]t² + [λ + λ² - λ]t + 1 = λ³t³ + λ³t² + λ²t + 1.

Which matches the original equation.

Therefore, the equation factors as:

(λt + 1)(λ²t² + (λ² - λ)t + 1) = 0.

Therefore, the solutions are:

1. λt + 1 = 0 ⇒ t = -1/λ.

2. λ²t² + (λ² - λ)t + 1 = 0.

So, for the first factor, t = -1/λ. Recall that t = x - 1, so x = t + 1 = 1 - 1/λ. Which is the fixed point we already knew.

For the second factor, solve λ²t² + (λ² - λ)t + 1 = 0.

Let me compute the discriminant D:

D = (λ² - λ)^2 - 4 * λ² * 1 = λ^4 - 2λ³ + λ² - 4λ² = λ^4 - 2λ³ - 3λ².

Factor:

D = λ²(λ² - 2λ - 3) = λ²(λ - 3)(λ + 1).

Since λ ≥ 1, and λ + 1 > 0, the discriminant is non-negative when λ - 3 ≥ 0 ⇒ λ ≥ 3.

Therefore, for λ ≥ 3, the quadratic equation has real roots.

Therefore, for λ ≥ 3, the equation f(f(x)) = x has solutions from the second factor.

Therefore, the solutions for t are:

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

Simplify:

√(λ²(λ - 3)(λ + 1)) = λ√[(λ - 3)(λ + 1)].

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

Therefore, the solutions for t are:

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

Recall that t = x - 1, so x = t + 1 = 1 + t.

Therefore, the solutions for x are:

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

Simplify:

x = [λ + 1 ± √((λ - 3)(λ + 1))]/(2λ).

Let me simplify the expression under the square root:

(λ - 3)(λ + 1) = λ² - 2λ - 3.

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

Now, since x must be in [0,1], we need to check if these solutions are in [0,1].

Let me analyze this expression.

First, for λ ≥ 3, the discriminant is non-negative.

Let me consider the two roots:

x₁ = [λ + 1 + √(λ² - 2λ - 3)]/(2λ)

x₂ = [λ + 1 - √(λ² - 2λ - 3)]/(2λ)

Let me check if these are in [0,1].

First, note that for λ ≥ 3, the term under the square root is λ² - 2λ - 3. Let me compute it for λ = 3: 9 - 6 - 3 = 0. For λ > 3, it's positive.

Let me check x₁ and x₂ for λ = 3:

At λ = 3, √(9 - 6 - 3) = √0 = 0.

So, x₁ = [3 + 1 + 0]/(6) = 4/6 = 2/3, which is the fixed point.

x₂ = [3 + 1 - 0]/6 = 4/6 = 2/3. So, both roots coincide with the fixed point.

For λ > 3, let's take λ = 4:

x₁ = [4 + 1 + √(16 - 8 - 3)]/(8) = [5 + √5]/8 ≈ [5 + 2.236]/8 ≈ 7.236/8 ≈ 0.9045, which is in [0,1].

x₂ = [5 - √5]/8 ≈ [5 - 2.236]/8 ≈ 2.764/8 ≈ 0.3455, which is also in [0,1].

Therefore, for λ > 3, we have two distinct solutions x₁ and x₂ in [0,1], which are period-2 points.

For λ = 3, they coincide with the fixed point.

Therefore, the period-2 points exist for λ > 3.

Therefore, the values of λ for which there exists an x ∈ [0,1] such that f(x) ≠ x but f(f(x)) = x are λ ∈ (3, 4].

But wait, the problem says 0 ≤ λ ≤ 4.

But we need to check if for λ = 3, there are such points. At λ = 3, the solutions x₁ and x₂ coincide with the fixed point x = 2/3. Therefore, for λ = 3, the only solutions are the fixed points, so there are no period-2 points.

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

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

For λ = 3.1, the discriminant is (3.1)^2 - 2*3.1 - 3 = 9.61 - 6.2 - 3 = 0.41 > 0.

Therefore, there are two distinct solutions x₁ and x₂, which are different from the fixed points.

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

Therefore, the answer is λ ∈ (3, 4].

But the problem says "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."

Therefore, the answer is λ ∈ (3, 4].

But let me check for λ = 3.

At λ = 3, the period-2 points coincide with the fixed point, so there are no such x.

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

But the problem says "values of λ, 0 ≤ λ ≤ 4", so including 4.

But let me check if for λ = 4, there are such points. As we saw earlier, yes, there are two period-2 points.

Therefore, the answer is λ ∈ (3, 4].

But the problem might expect a closed interval or something else.

Wait, but let me check for λ = 3.

At λ = 3, the equation f(f(x)) = x has only fixed points. Therefore, no period-2 points.

Therefore, the values of λ are those strictly greater than 3 and up to 4.

But the problem says "values of λ, 0 ≤ λ ≤ 4", so the answer is λ ∈ (3, 4].

But the question says "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."

Therefore, the answer is λ ∈ (3, 4].

But let me check if there are any other λ values where period-2 points exist.

For example, for λ between 1 and 3, we saw that for λ = 2 and λ = 3, there are no period-2 points.

But what about for some λ between 1 and 3?

Suppose λ = 2.5.

Let me try solving the equation for λ = 2.5.

But this might be time-consuming.

Alternatively, recall that the discriminant of the quadratic factor is D = λ²(λ - 3)(λ + 1). For λ < 3, D is negative, so no real solutions. Therefore, for λ < 3, the only real solutions to f(f(x)) = x are the fixed points.

Therefore, for λ < 3, there are no period-2 points.

Therefore, the answer is λ ∈ (3, 4].

But the problem says "values of λ, 0 ≤ λ ≤ 4", so including 4.

But the original question says "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."

Therefore, the answer is λ ∈ (3, 4].

But the problem might expect the answer in terms of intervals.

But let me check if for λ = 3, there are any such points.

At λ = 3, the equation f(f(x)) = x has only fixed points. Therefore, no such x.

Therefore, the answer is λ ∈ (3, 4].

But the problem might want it in boxed notation.

But let me check if there's a mistake.

Wait, the original equation for the period-2 points is derived from the quadratic factor, which has real solutions only when λ ≥ 3. For λ = 3, the solutions coincide with the fixed point. For λ > 3, they are distinct.

Therefore, the answer is λ ∈ (3, 4].

But the problem says "values of λ, 0 ≤ λ ≤ 4", so including 4.

But the question is from a math competition or similar, and often such problems expect answers in specific forms.

But let me check if there's a possibility that for λ = 3, there are period-2 points.

At λ = 3, the fixed point x = 2/3 is unstable. The period-2 points might be at the same location?

But according to our calculation, the period-2 points coincide with the fixed point.

Alternatively, maybe there are other solutions?

Wait, for λ = 3, the equation f(f(x)) = x has solutions x = 0, x = 2/3 (with multiplicity 3).

But x = 0 is a fixed point.

Therefore, there are no other solutions.

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

Therefore, the answer is λ ∈ (3, 4].

But the problem says "values of λ, 0 ≤ λ ≤ 4", so the answer is all λ in (3, 4].

But the problem might expect the answer as λ > 3, but since λ is in [0,4], it's λ ∈ (3, 4].

But let me check if for λ = 3, there are any points where f(f(x)) = x but f(x) ≠ x.

At λ = 3, suppose there is such a point. Then, it would have to satisfy f(f(x)) = x and f(x) ≠ x.

But according to our previous calculation, the only solutions are x = 0 and x = 2/3.

At x = 0, f(x) = 0, so it's a fixed point.

At x = 2/3, f(x) = 3*(2/3)*(1 - 2/3) = 3*(2/3)*(1/3) = 2/3. So, also a fixed point.

Therefore, no such points.

Therefore, the answer is λ ∈ (3, 4].

But the problem says "values of λ, 0 ≤ λ ≤ 4", so the answer is λ in (3, 4].

But in some contexts, the interval might be closed, but since at λ = 3 there are no such points, it's open at 3.

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

But the problem might expect it written as 3 < λ ≤ 4.

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

So, boxed answer would be \boxed{(3, 4]}.

But let me check if the problem expects a specific format.

The original problem says "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."

Since the answer is an interval, and in many math competitions, they might expect it written as 3 < λ ≤ 4, so boxed as \boxed{(3, 4]}.

But sometimes, depending on the convention, they might write it as 3 < λ ≤ 4, but since the user asked for the answer in boxed notation, likely \boxed{(3, 4]}.

But let me check if there's a possibility that λ = 3 is included.

But as discussed, at λ = 3, there are no such points.

Therefore, the answer is λ ∈ (3, 4].

**Final Answer**
The values of $\lambda$ are $\boxed{(3, 4]}$.

To solve the problem, we are given a function:

$$
f(x) = \lambda x(1 - x), \quad \text{with } 0 \leq \lambda \leq 4
$$

and we are asked to find the values of $\lambda$ for which there exists an $x \in [0,1]$ such that:

$$
f(x) \neq x \quad \text{but} \quad f(f(x)) = x
$$

---

### Step 1: Understanding the Condition

We are looking for **period-2 points** of the function $f$, i.e., points $x$ such that:

- $f(x) \neq x$ (not fixed points),
- $f(f(x)) = x$ (period-2 behavior).

This means that $x$ is not a fixed point, but it is part of a 2-cycle: $x \to f(x) \to x$.

---

### Step 2: Fixed Points of $f(x)$

To find fixed points, solve:

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

This simplifies to:

$$
x(\lambda(1 - x) - 1) = 0
$$

So the fixed points are:

- $x = 0$
- $x = 1 - \frac{1}{\lambda}$, provided $\lambda \geq 1$

---

### Step 3: Period-2 Points

To find period-2 points, we solve:

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

This is a quartic equation in $x$, and it includes all fixed points as solutions. To find **non-fixed** solutions, we subtract the fixed points and look for other real roots.

After algebraic manipulation and factoring, we find that the equation $f(f(x)) = x$ can be factored as:

$$
f(f(x)) - x = (x)(x - (1 - \frac{1}{\lambda})) \cdot \text{(quadratic)}
$$

The quadratic factor has real roots **only when** the discriminant is non-negative. This leads to the condition:

$$
\lambda^2(\lambda - 3)(\lambda + 1) \geq 0
$$

Since $\lambda \in [0,4]$, and $\lambda + 1 > 0$, the inequality reduces to:

$$
\lambda - 3 \geq 0 \Rightarrow \lambda \geq 3
$$

---

### Step 4: Behavior at $\lambda = 3$

At $\lambda = 3$, the discriminant is zero, and the quadratic factor has a **double root**. This means the period-2 points **coincide** with the fixed point $x = 1 - \frac{1}{3} = \frac{2}{3}$. Thus, **no new period-2 points** are introduced at $\lambda = 3$.

---

### Step 5: Behavior for $\lambda > 3$

For $\lambda > 3$, the discriminant is positive, and the quadratic factor has **two distinct real roots**. These roots correspond to **period-2 points** that are **not fixed points**, satisfying the condition $f(x) \neq x$ but $f(f(x)) = x$.

---

### Final Conclusion

The values of $\lambda$ for which there exists an $x \in [0,1]$ such that $f(x) \neq x$ but $f(f(x)) = x$ are:

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

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

First, let me make sure I understand the problem. We have this quadratic function f(x) = λx(1 - x), which is a parabola opening downward because the coefficient of x² is -λ, which is negative since λ is non-negative. The domain and codomain are both [0,1], so we're looking at the function's behavior on that interval.

The question is asking for λ values where there's at least one x in [0,1] such that applying f twice brings you 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, first, maybe I should recall that fixed points of f are solutions to f(x) = x. So solving λx(1 - x) = x. Let me write that equation:

λx(1 - x) = x

Subtract x from both sides:

λx(1 - x) - x = 0

Factor out x:

x[λ(1 - x) - 1] = 0

So, either x = 0 or λ(1 - x) - 1 = 0. Let's solve the second equation:

λ(1 - x) - 1 = 0 ⇒ λ(1 - x) = 1 ⇒ 1 - x = 1/λ ⇒ x = 1 - 1/λ

But since x must be in [0,1], we need 1 - 1/λ to be in [0,1]. Let's check when that happens.

First, 1 - 1/λ ≥ 0 ⇒ 1 ≥ 1/λ ⇒ λ ≥ 1. Also, 1 - 1/λ ≤ 1 ⇒ -1/λ ≤ 0, which is always true since λ is non-negative. So, for λ ≥ 1, there is a fixed point at x = 1 - 1/λ. For λ < 1, the only fixed point is x = 0.

Wait, but when λ = 1, x = 1 - 1/1 = 0. So, for λ = 1, the fixed points are x = 0 and x = 1 - 1/1 = 0? Wait, no. Wait, solving λ(1 - x) - 1 = 0 when λ = 1 gives 1*(1 - x) - 1 = 0 ⇒ 1 - x - 1 = -x = 0 ⇒ x = 0. So, for λ = 1, the fixed points are x = 0 (from the first factor) and x = 0 again? Wait, that seems like only one fixed point? Wait, but when λ = 1, the equation f(x) = x becomes x = 1*x*(1 - x) ⇒ x = x - x² ⇒ 0 = -x² ⇒ x = 0. So, only x = 0 is a fixed point. So, for λ = 1, the only fixed point is x = 0.

But when λ > 1, then x = 1 - 1/λ is another fixed point. For example, if λ = 2, then x = 1 - 1/2 = 1/2. Let me check: f(1/2) = 2*(1/2)*(1 - 1/2) = 2*(1/2)*(1/2) = 2*(1/4) = 1/2. So yes, that's a fixed point.

So, in summary, fixed points of f(x) are:

- For λ < 1: only x = 0.

- For λ = 1: only x = 0.

- For λ > 1: x = 0 and x = 1 - 1/λ.

But the problem is not about fixed points. It's about points where f(f(x)) = x but f(x) ≠ x. So, these are points that are periodic with period 2.

So, to find such points, we need to solve f(f(x)) = x, and then exclude the fixed points.

So, first, let me find all solutions to f(f(x)) = x, then subtract the fixed points.

So, let me compute f(f(x)).

First, f(x) = λx(1 - x). Then f(f(x)) = λ * [f(x)] * [1 - f(x)] = λ * [λx(1 - x)] * [1 - λx(1 - x)].

So, f(f(x)) = λ²x(1 - x)[1 - λx(1 - x)].

So, setting f(f(x)) = x:

λ²x(1 - x)[1 - λx(1 - x)] = x.

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

λ²(1 - x)[1 - λx(1 - x)] = 1.

But if x = 0, then f(0) = 0, so x = 0 is a fixed point. So, we need to consider x ≠ 0.

So, the equation becomes:

λ²(1 - x)[1 - λx(1 - x)] = 1.

But this seems complicated. Maybe expanding it would help? Let me try.

First, let me denote y = x(1 - x). Then f(x) = λy, and f(f(x)) = λ * (λy) * (1 - λy) = λ² y (1 - λy). So, f(f(x)) = λ² y (1 - λy).

But we want f(f(x)) = x, so:

λ² y (1 - λy) = x.

But y = x(1 - x), so substituting:

λ² x(1 - x) [1 - λx(1 - x)] = x.

Which is the same equation as before.

So, again, if x ≠ 0, divide both sides by x:

λ² (1 - x)[1 - λx(1 - x)] = 1.

Let me expand the left-hand side:

First, compute 1 - λx(1 - x):

1 - λx + λx².

So, the equation becomes:

λ² (1 - x)(1 - λx + λx²) = 1.

Let me expand (1 - x)(1 - λx + λx²):

Multiply term by term:

First, 1*(1 - λx + λx²) = 1 - λx + λx².

Then, -x*(1 - λx + λx²) = -x + λx² - λx³.

Adding these together:

1 - λx + λx² - x + λx² - λx³.

Combine like terms:

Constant term: 1.

x terms: -λx - x = -x(λ + 1).

x² terms: λx² + λx² = 2λx².

x³ term: -λx³.

So overall:

1 - (λ + 1)x + 2λx² - λx³.

Therefore, the equation is:

λ² [1 - (λ + 1)x + 2λx² - λx³] = 1.

Expanding this:

λ² - λ²(λ + 1)x + 2λ³x² - λ³x³ = 1.

Bring the 1 to the left:

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

So, the equation is:

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

Let me write it in standard polynomial form:

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

Alternatively, multiplying both sides by -1:

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

So:

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

Hmm, this seems complicated. Maybe there's a better way. Let me think.

Alternatively, since we know that fixed points satisfy f(x) = x, which are solutions to the equation f(f(x)) = x. Therefore, the solutions to f(f(x)) = x include all fixed points and the period-2 points. Therefore, to find the period-2 points, we can factor out the fixed points from the equation f(f(x)) = x.

So, if we can factor the equation f(f(x)) = x, then the roots would be the fixed points and the period-2 points. Therefore, if we can factor out the fixed points, then the remaining roots would be the period-2 points.

So, let me try that approach.

First, we know that the fixed points are x = 0 and x = 1 - 1/λ (for λ ≥ 1). So, if we can factor these out from the equation f(f(x)) = x, then the remaining roots would be the period-2 points.

But since we have the equation f(f(x)) = x, which is a cubic equation? Wait, no. Wait, f(f(x)) is a composition of quadratics, so f(f(x)) is a quartic function. Wait, let me check:

Original function f(x) is quadratic, so f(f(x)) is quadratic composed with quadratic, which is quartic (degree 4). Therefore, the equation f(f(x)) = x is a quartic equation, which can have up to 4 roots. However, since we already know that x = 0 is a root (since f(0) = 0, so f(f(0)) = 0 = 0), and x = 1 - 1/λ (for λ ≥ 1) is also a root. Therefore, the quartic equation f(f(x)) = x can be factored as (x)(x - (1 - 1/λ)) * quadratic = 0, assuming λ ≥ 1.

But for λ < 1, the only fixed point is x = 0, so the quartic equation would have x = 0 as a root, and maybe other roots.

But perhaps I need to consider both cases. However, since the problem is asking for λ in [0,4], maybe I can proceed by considering the quartic equation and factoring out the fixed points.

Alternatively, maybe I can write f(f(x)) - x = 0 and factor it. Let me try that.

Let me compute f(f(x)) - x:

f(f(x)) - x = λ²x(1 - x)[1 - λx(1 - x)] - x.

Factor out x:

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

So, the solutions are x = 0 and solutions to λ²(1 - x)[1 - λx(1 - x)] - 1 = 0.

So, the equation we need to solve for period-2 points is:

λ²(1 - x)[1 - λx(1 - x)] - 1 = 0.

Which is the same as before.

Alternatively, maybe I can write this equation as:

Let me denote y = x(1 - x). Then, the equation becomes:

λ²(1 - x)(1 - λy) - 1 = 0.

But since y = x(1 - x), maybe not helpful.

Alternatively, perhaps I can consider that for period-2 points, we have f(x) ≠ x, but f(f(x)) = x. So, if we let y = f(x), then y ≠ x, but f(y) = x. Therefore, we have the system:

y = λx(1 - x)

x = λy(1 - y)

So, substituting the first equation into the second:

x = λ[λx(1 - x)] [1 - λx(1 - x)].

Which is the same as:

x = λ²x(1 - x)[1 - λx(1 - x)].

Which is the same equation as before.

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

1 = λ²(1 - x)[1 - λx(1 - x)].

Which is the same equation as before.

So, perhaps solving this equation for x in [0,1], and then ensuring that x ≠ f(x).

Alternatively, since we have the system:

y = λx(1 - x)

x = λy(1 - y)

So, substituting y from the first equation into the second:

x = λ[λx(1 - x)][1 - λx(1 - x)].

Which is the same as before.

Alternatively, maybe we can consider solving the system:

y = λx(1 - x)

x = λy(1 - y)

So, substituting y from the first equation into the second:

x = λ[λx(1 - x)][1 - λx(1 - x)].

Which is the same equation.

Alternatively, maybe we can consider subtracting the two equations? Not sure.

Alternatively, let me consider that if we have y = λx(1 - x) and x = λy(1 - y), then we can write:

y = λx(1 - x)

x = λy(1 - y)

So, multiplying both equations:

xy = λ²x(1 - x)y(1 - y)

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

1 = λ²(1 - x)(1 - y)

So, 1 = λ²(1 - x)(1 - y)

But from the first equation, y = λx(1 - x). Let me substitute that into the above equation:

1 = λ²(1 - x)[1 - λx(1 - x)]

Which is the same equation as before.

So, again, we get back to the same equation.

Therefore, perhaps I need to solve this equation for x in [0,1], and then check if x ≠ f(x).

Alternatively, maybe I can consider that for period-2 points, the solutions to the equation f(f(x)) = x are the fixed points and the period-2 points. Therefore, if I can find the number of solutions to f(f(x)) = x, and subtract the fixed points, then the remaining solutions would be the period-2 points.

But since the problem is asking for the existence of at least one such x, maybe I can find for which λ there exists such x.

Alternatively, maybe I can consider the equation f(f(x)) = x and find when it has solutions other than the fixed points.

Alternatively, maybe I can think graphically. The function f(x) is a parabola opening downward, with maximum at x = 1/2, f(1/2) = λ*(1/2)*(1 - 1/2) = λ*(1/2)*(1/2) = λ/4.

So, the maximum value of f(x) is λ/4. Since the domain is [0,1], and the codomain is [0,1], we need that λ/4 ≤ 1 ⇒ λ ≤ 4, which is given.

So, for λ in [0,4], the maximum of f(x) is between 0 and 1.

Now, the equation f(f(x)) = x. Let me think about the graph of f(f(x)) and the line y = x. The intersections of these two graphs will give the solutions.

But since f(f(x)) is a composition of two functions, it's a quartic function.

But maybe instead of dealing with the quartic equation, I can think about the system:

y = f(x)

x = f(y)

Which is equivalent to the period-2 points.

So, solving this system:

y = λx(1 - x)

x = λy(1 - y)

So, substituting the first into the second:

x = λ[λx(1 - x)][1 - λx(1 - x)]

Which is the same equation as before.

Alternatively, maybe I can consider that if we have x ≠ y, then we can write:

From the two equations:

y = λx(1 - x)

x = λy(1 - y)

Let me subtract the two equations:

y - x = λx(1 - x) - λy(1 - y)

But not sure if helpful.

Alternatively, let me consider dividing the two equations.

From y = λx(1 - x) and x = λy(1 - y), dividing the first by the second:

y / x = [λx(1 - x)] / [λy(1 - y)] ⇒ y / x = [x(1 - x)] / [y(1 - y)]

Cross multiplying:

y²(1 - y) = x²(1 - x)

So, y²(1 - y) = x²(1 - x)

But since y = λx(1 - x), substitute that in:

[λ²x²(1 - x)²][1 - λx(1 - x)] = x²(1 - x)

Assuming x ≠ 0 and 1 - x ≠ 0 (i.e., x ≠ 0,1), we can divide both sides by x²(1 - x):

λ²(1 - x)[1 - λx(1 - x)] = 1

Which is the same equation as before.

So, again, we get back to the same equation.

Therefore, perhaps I need to solve this equation for x in [0,1], and then check if there are solutions where x ≠ f(x).

Alternatively, maybe I can consider that for period-2 points, the solutions to the equation f(f(x)) = x are the fixed points and the period-2 points. Therefore, if the equation f(f(x)) = x has more than two solutions (the fixed points), then there are period-2 points.

But how many solutions does f(f(x)) = x have?

Well, since f(f(x)) is a quartic function, it can have up to 4 real roots. However, depending on λ, the number of roots can vary.

But since we already know that x = 0 is a root, and for λ ≥ 1, x = 1 - 1/λ is another root. So, if there are other roots, they would be the period-2 points.

But maybe I can consider the case when there are period-2 points.

Alternatively, maybe I can think about the function g(x) = f(f(x)) - x. We need to find λ such that g(x) = 0 has solutions other than the fixed points.

Alternatively, maybe I can consider specific values of λ and see when such points exist.

But since the problem is asking for all λ in [0,4], maybe I need to find the range of λ where such points exist.

Alternatively, maybe I can consider the equation we derived earlier:

λ²(1 - x)[1 - λx(1 - x)] = 1.

Let me denote z = x(1 - x). Then, since x ∈ [0,1], z ∈ [0, 1/4], because the maximum of x(1 - x) is at x = 1/2, which is 1/4.

So, z ∈ [0, 1/4].

Then, the equation becomes:

λ²(1 - x)(1 - λz) = 1.

But z = x(1 - x), so 1 - x = (1 - x). Hmm, not sure.

Alternatively, maybe express 1 - x in terms of z?

Since z = x(1 - x), we can write 1 - x = (z)/x, but that might not help.

Alternatively, maybe express 1 - x as something else.

Alternatively, maybe think of this as a quadratic equation in terms of x.

Let me try expanding the equation again:

λ²(1 - x)[1 - λx(1 - x)] = 1.

Let me expand the term inside:

First, expand 1 - λx(1 - x):

1 - λx + λx².

So, the equation becomes:

λ²(1 - x)(1 - λx + λx²) = 1.

Earlier, we expanded this to:

λ²[1 - (λ + 1)x + 2λx² - λx³] = 1.

Which led to:

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

Alternatively, maybe I can write this as:

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

Wait, let me check:

Original expansion was:

λ²[1 - (λ + 1)x + 2λx² - λx³] = 1.

Which is:

λ² - λ²(λ + 1)x + 2λ³x² - λ³x³ = 1.

Then moving 1 to the left:

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

Which is:

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

Alternatively, multiplying by -1:

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

So:

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

Wait, because λ²(λ + 1)x = λ³x + λ²x.

So, grouping terms:

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

Hmm, maybe factor terms:

Let me factor λ³x³ - 2λ³x² + λ³x:

λ³x(x² - 2x + 1) = λ³x(x - 1)^2.

Then, the remaining terms are λ²x - λ² + 1.

So, the equation becomes:

λ³x(x - 1)^2 + λ²x - λ² + 1 = 0.

So:

λ³x(x - 1)^2 + λ²(x - 1) + 1 = 0.

Wait, because λ²x - λ² = λ²(x - 1).

So, the equation is:

λ³x(x - 1)^2 + λ²(x - 1) + 1 = 0.

Hmm, interesting. Let me check:

Yes, because:

Original equation after grouping:

λ³x(x - 1)^2 + λ²(x - 1) + 1 = 0.

So, this seems like a more manageable form.

So, let me write that:

λ³x(x - 1)^2 + λ²(x - 1) + 1 = 0.

Let me denote t = x - 1. Then, x = t + 1. But maybe not helpful. Alternatively, let me consider substituting u = x - 1, but maybe not.

Alternatively, note that x ∈ [0,1], so (x - 1) is between -1 and 0.

Alternatively, maybe factor out (x - 1):

But the equation is:

λ³x(x - 1)^2 + λ²(x - 1) + 1 = 0.

Let me factor out (x - 1) from the first two terms:

(x - 1)[λ³x(x - 1) + λ²] + 1 = 0.

But not sure.

Alternatively, maybe consider that for x ∈ [0,1], (x - 1) is negative or zero.

Alternatively, maybe try specific values of x.

Alternatively, think about the equation:

λ³x(x - 1)^2 + λ²(x - 1) + 1 = 0.

Let me rearrange:

λ³x(x - 1)^2 = - [λ²(x - 1) + 1].

Note that the left-hand side is λ³x(x - 1)^2. Since x ∈ [0,1], (x - 1) is negative or zero, so (x - 1)^2 is non-negative. Also, x is non-negative. Therefore, the left-hand side is non-negative (since λ³ is non-negative, x is non-negative, and (x - 1)^2 is non-negative).

The right-hand side is - [λ²(x - 1) + 1]. Let me compute the expression inside the brackets:

λ²(x - 1) + 1 = λ²x - λ² + 1.

So, the right-hand side is - [λ²x - λ² + 1] = -λ²x + λ² - 1.

Therefore, the equation becomes:

λ³x(x - 1)^2 = -λ²x + λ² - 1.

So:

λ³x(x - 1)^2 + λ²x - λ² + 1 = 0.

Wait, that's the same as before.

But since the left-hand side is non-negative, and the right-hand side is - [λ²(x - 1) + 1], which is equal to - [λ²x - λ² + 1].

But since the left-hand side is non-negative, the right-hand side must also be non-negative. Therefore:

- [λ²x - λ² + 1] ≥ 0 ⇒ λ²x - λ² + 1 ≤ 0 ⇒ λ²(x - 1) + 1 ≤ 0 ⇒ λ²(x - 1) ≤ -1 ⇒ x - 1 ≤ -1/λ² ⇒ x ≤ 1 - 1/λ².

But since x ∈ [0,1], this gives a condition on x.

So, for the equation to have solutions, we need x ≤ 1 - 1/λ².

But since x ∈ [0,1], this requires that 1 - 1/λ² ≥ 0 ⇒ λ² ≥ 1 ⇒ λ ≥ 1.

Therefore, for λ < 1, there are no solutions? Wait, but if λ < 1, then 1 - 1/λ² is negative, so x ≤ negative number, but x ∈ [0,1], so no solutions. Therefore, for λ < 1, there are no solutions to the equation?

But wait, if λ < 1, then the equation f(f(x)) = x would have solutions? For example, take λ = 0. Then f(x) = 0, so f(f(x)) = 0 = x. So, for λ = 0, every x in [0,1] satisfies f(f(x)) = x. But in this case, f(x) = 0 for all x, so f(x) = x only when x = 0. Therefore, for λ = 0, all x satisfy f(f(x)) = x, but only x = 0 satisfies f(x) = x. Therefore, for λ = 0, there are points (all x ≠ 0) where f(f(x)) = x but f(x) ≠ x.

But according to the previous reasoning, for λ < 1, there are no solutions? But that contradicts with λ = 0.

So, there must be a mistake in my reasoning.

Let me check again.

Earlier, I said that the left-hand side is non-negative, and the right-hand side is - [λ²(x - 1) + 1]. Therefore, for the equation to hold, the right-hand side must be non-negative.

But for λ = 0, the equation becomes:

0³x(x - 1)^2 + 0²(x - 1) + 1 = 0 ⇒ 0 + 0 + 1 = 0 ⇒ 1 = 0, which is false.

Wait, but earlier, for λ = 0, f(f(x)) = 0 = x, so x must be 0. But according to the equation we derived, which was after dividing by x (assuming x ≠ 0), we have:

λ²(1 - x)[1 - λx(1 - x)] = 1.

For λ = 0, this becomes 0 * ... = 1, which is 0 = 1, which is false. Therefore, the equation λ²(1 - x)[1 - λx(1 - x)] = 1 has no solutions for λ = 0, except x = 0.

But when λ = 0, f(f(x)) = 0 for all x, so f(f(x)) = x implies 0 = x, so x = 0. Therefore, the only solution is x = 0, which is a fixed point. Therefore, there are no points where f(f(x)) = x but f(x) ≠ x.

But earlier, I thought that for λ = 0, all x satisfy f(f(x)) = x, but actually, f(f(x)) = 0 for all x, so f(f(x)) = x implies x = 0. Therefore, only x = 0 is a solution. Therefore, for λ = 0, there are no such points.

So, my previous thought was wrong. Therefore, for λ = 0, there are no such points.

Therefore, the previous reasoning that for λ < 1, there are no solutions is correct? But then, for λ = 1, let's check.

For λ = 1, the equation becomes:

1²(1 - x)[1 - 1*x(1 - x)] = 1 ⇒ (1 - x)[1 - x + x²] = 1.

Let me compute this:

(1 - x)(1 - x + x²) = ?

Multiply out:

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

Set equal to 1:

1 - 2x + 2x² - x³ = 1 ⇒ -2x + 2x² - x³ = 0 ⇒ x(-2 + 2x - x²) = 0.

Therefore, solutions are x = 0 or -2 + 2x - x² = 0 ⇒ x² - 2x + 2 = 0. The discriminant is 4 - 8 = -4 < 0, so no real solutions. Therefore, only x = 0 is a solution. Therefore, for λ = 1, the only solution is x = 0, which is a fixed point. Therefore, no period-2 points.

So, for λ = 1, there are no such points.

Therefore, for λ < 1, the equation has only x = 0 as a solution, which is a fixed point. Therefore, no period-2 points.

For λ ≥ 1, we have the equation:

λ³x(x - 1)^2 + λ²(x - 1) + 1 = 0.

But earlier, we saw that for the equation to have solutions, we need x ≤ 1 - 1/λ².

But since x ∈ [0,1], and 1 - 1/λ² ≥ 0 when λ ≥ 1.

So, for λ ≥ 1, we can have solutions.

Therefore, the problem reduces to finding λ ≥ 1 such that there exists x ∈ [0,1] with x ≤ 1 - 1/λ² and satisfying the equation.

But how do we find such λ?

Alternatively, maybe we can consider that for λ ≥ 1, the equation f(f(x)) = x has solutions other than the fixed points.

But how can we determine for which λ this is true?

Alternatively, maybe we can consider the case when there are period-2 points.

Alternatively, think about the function g(x) = f(f(x)) - x. We need to find λ such that g(x) = 0 has solutions other than the fixed points.

Alternatively, maybe consider specific values of λ.

For example, take λ = 2.

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

Let me check if there are period-2 points.

So, solving f(f(x)) = x.

Let me compute f(f(x)) for λ = 2:

f(x) = 2x(1 - x).

Then f(f(x)) = 2 * [2x(1 - x)] * [1 - 2x(1 - x)].

Let me compute this:

First, compute 2x(1 - x) = 2x - 2x².

Then, 1 - 2x(1 - x) = 1 - 2x + 2x².

Therefore, f(f(x)) = 2*(2x - 2x²)*(1 - 2x + 2x²).

Let me compute this:

First, multiply (2x - 2x²)(1 - 2x + 2x²):

Let me expand term by term:

First, 2x*(1 - 2x + 2x²) = 2x - 4x² + 4x³.

Then, -2x²*(1 - 2x + 2x²) = -2x² + 4x³ - 4x⁴.

Adding these together:

2x - 4x² + 4x³ - 2x² + 4x³ - 4x⁴ = 2x - 6x² + 8x³ - 4x⁴.

Multiply by 2:

f(f(x)) = 4x - 12x² + 16x³ - 8x⁴.

Set equal to x:

4x - 12x² + 16x³ - 8x⁴ = x ⇒ 4x - 12x² + 16x³ - 8x⁴ - x = 0 ⇒ 3x - 12x² + 16x³ - 8x⁴ = 0 ⇒ -8x⁴ + 16x³ - 12x² + 3x = 0 ⇒ x(-8x³ + 16x² - 12x + 3) = 0.

So, solutions are x = 0 and solutions to -8x³ + 16x² - 12x + 3 = 0.

Let me write this as 8x³ - 16x² + 12x - 3 = 0.

Let me try to find rational roots using Rational Root Theorem. Possible rational roots are ±1, ±3, ±1/2, ±3/2, ±1/4, ±3/4, ±1/8, ±3/8.

Testing x = 1: 8 - 16 + 12 - 3 = 1 ≠ 0.

x = 3/2: Not in [0,1], so skip.

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

Yes! x = 1/2 is a root.

Therefore, we can factor (x - 1/2) from the cubic.

Let me perform polynomial division or use synthetic division.

Divide 8x³ - 16x² + 12x - 3 by (x - 1/2).

Using synthetic division:

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.

So, the cubic factors as (x - 1/2)(8x² - 12x + 6).

Therefore, the equation becomes x(x - 1/2)(8x² - 12x + 6) = 0.

So, solutions are x = 0, x = 1/2, and solutions to 8x² - 12x + 6 = 0.

Solving 8x² - 12x + 6 = 0:

Discriminant D = 144 - 192 = -48 < 0.

Therefore, no real roots.

Therefore, for λ = 2, the only real solutions are x = 0 and x = 1/2, which are fixed points. Therefore, there are no period-2 points for λ = 2.

Wait, but that contradicts my initial thought. Maybe λ = 2 is not in the desired range?

Wait, but maybe I made a mistake here. Let me check with specific values.

Take x = 1/2. Then f(x) = 2*(1/2)*(1 - 1/2) = 2*(1/2)*(1/2) = 1/2. So, fixed point.

What about another x? Let me try x = 1/4.

f(1/4) = 2*(1/4)*(1 - 1/4) = 2*(1/4)*(3/4) = 2*(3/16) = 3/8.

Then f(3/8) = 2*(3/8)*(1 - 3/8) = 2*(3/8)*(5/8) = 30/64 = 15/32 ≈ 0.46875.

Which is not equal to 1/4. So, f(f(1/4)) ≈ 0.46875 ≠ 1/4.

What about x = 3/4?

f(3/4) = 2*(3/4)*(1 - 3/4) = 2*(3/4)*(1/4) = 6/16 = 3/8.

Then f(3/8) = 15/32 as before. Not equal to 3/4.

So, seems like for λ = 2, there are no period-2 points.

But then, when does a period-2 point exist?

Let me try λ = 3.

For λ = 3, fixed points are x = 0 and x = 1 - 1/3 = 2/3.

Let me try solving f(f(x)) = x.

First, f(x) = 3x(1 - x).

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

Let me compute this:

First, compute 3x(1 - x) = 3x - 3x².

Then, 1 - 3x(1 - x) = 1 - 3x + 3x².

Therefore, f(f(x)) = 3*(3x - 3x²)*(1 - 3x + 3x²).

Let me compute (3x - 3x²)(1 - 3x + 3x²):

First, expand term by term:

3x*(1 - 3x + 3x²) = 3x - 9x² + 9x³.

-3x²*(1 - 3x + 3x²) = -3x² + 9x³ - 9x⁴.

Adding together:

3x - 9x² + 9x³ - 3x² + 9x³ - 9x⁴ = 3x - 12x² + 18x³ - 9x⁴.

Multiply by 3:

f(f(x)) = 9x - 36x² + 54x³ - 27x⁴.

Set equal to x:

9x - 36x² + 54x³ - 27x⁴ = x ⇒ 9x - 36x² + 54x³ - 27x⁴ - x = 0 ⇒ 8x - 36x² + 54x³ - 27x⁴ = 0 ⇒ -27x⁴ + 54x³ - 36x² + 8x = 0 ⇒ x(-27x³ + 54x² - 36x + 8) = 0.

So, solutions are x = 0 and solutions to -27x³ + 54x² - 36x + 8 = 0.

Let me write this as 27x³ - 54x² + 36x - 8 = 0.

Try rational roots. Possible roots: ±1, ±2, ±4, ±8, ±1/3, etc.

Testing x = 2/3:

27*(8/27) - 54*(4/9) + 36*(2/3) - 8 = 8 - 24 + 24 - 8 = 0.

Yes! x = 2/3 is a root.

Therefore, factor out (x - 2/3).

Using synthetic division:

Divide 27x³ - 54x² + 36x - 8 by (x - 2/3).

Using synthetic division with root 2/3:

Coefficients: 27 | -54 | 36 | -8

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.

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

Therefore, the equation becomes x(x - 2/3)(27x² - 36x + 12) = 0.

Solving 27x² - 36x + 12 = 0:

Divide by 3: 9x² - 12x + 4 = 0.

Discriminant D = 144 - 144 = 0. So, repeated root: x = [12 ± 0]/18 = 12/18 = 2/3.

Therefore, the equation factors as x(x - 2/3)^3 = 0.

Therefore, the only real solutions are x = 0 and x = 2/3. Which are the fixed points. Therefore, again, no period-2 points for λ = 3.

Hmm, so for λ = 2 and λ = 3, there are no period-2 points.

Wait, but maybe for some λ between 1 and 4, there are period-2 points?

Let me try λ = 4.

For λ = 4, fixed points are x = 0 and x = 1 - 1/4 = 3/4.

Let me compute f(f(x)) = x.

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

Then f(f(x)) = 4 * [4x(1 - x)] * [1 - 4x(1 - x)].

Compute 4x(1 - x) = 4x - 4x².

Then 1 - 4x(1 - x) = 1 - 4x + 4x².

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

Let me compute (4x - 4x²)(1 - 4x + 4x²):

First, expand term by term:

4x*(1 - 4x + 4x²) = 4x - 16x² + 16x³.

-4x²*(1 - 4x + 4x²) = -4x² + 16x³ - 16x⁴.

Adding together:

4x - 16x² + 16x³ - 4x² + 16x³ - 16x⁴ = 4x - 20x² + 32x³ - 16x⁴.

Multiply by 4:

f(f(x)) = 16x - 80x² + 128x³ - 64x⁴.

Set equal to x:

16x - 80x² + 128x³ - 64x⁴ = x ⇒ 16x - 80x² + 128x³ - 64x⁴ - x = 0 ⇒ 15x - 80x² + 128x³ - 64x⁴ = 0 ⇒ -64x⁴ + 128x³ - 80x² + 15x = 0 ⇒ x(-64x³ + 128x² - 80x + 15) = 0.

So, solutions are x = 0 and solutions to -64x³ + 128x² - 80x + 15 = 0.

Let me write this as 64x³ - 128x² + 80x - 15 = 0.

Try rational roots. Possible roots: ±1, ±3, ±5, ±15, ±1/2, etc.

Testing x = 3/4:

64*(27/64) - 128*(9/16) + 80*(3/4) - 15 = 27 - 72 + 60 - 15 = 0.

Yes! x = 3/4 is a root.

Therefore, factor out (x - 3/4).

Using synthetic division:

Divide 64x³ - 128x² + 80x - 15 by (x - 3/4).

Using synthetic division with root 3/4:

Coefficients: 64 | -128 | 80 | -15

Bring down 64.

Multiply by 3/4: 48.

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

Multiply by 3/4: -60.

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

Multiply by 3/4: 15.

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

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

Therefore, the equation becomes x(x - 3/4)(64x² - 80x + 20) = 0.

Solving 64x² - 80x + 20 = 0:

Divide by 4: 16x² - 20x + 5 = 0.

Discriminant D = 400 - 320 = 80.

Solutions: [20 ± √80]/32 = [20 ± 4√5]/32 = [5 ± √5]/8 ≈ [5 ± 2.236]/8.

So, approximately (5 + 2.236)/8 ≈ 7.236/8 ≈ 0.9045 and (5 - 2.236)/8 ≈ 2.764/8 ≈ 0.3455.

So, these are two real roots.

Therefore, for λ = 4, the solutions to f(f(x)) = x are x = 0, x = 3/4, and x ≈ 0.9045 and x ≈ 0.3455.

Now, check if these are fixed points or period-2 points.

First, x = 0 and x = 3/4 are fixed points.

Check x ≈ 0.9045: compute f(x) and see if it's equal to x or not.

Let me compute f(0.9045) = 4 * 0.9045 * (1 - 0.9045) ≈ 4 * 0.9045 * 0.0955 ≈ 4 * 0.0863 ≈ 0.345.

Which is approximately the other root, 0.3455. Then f(0.3455) ≈ 4 * 0.3455 * (1 - 0.3455) ≈ 4 * 0.3455 * 0.6545 ≈ 4 * 0.226 ≈ 0.904.

Therefore, these two points form a 2-cycle: f(0.9045) ≈ 0.3455 and f(0.3455) ≈ 0.9045. Therefore, they are period-2 points.

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

So, for λ = 4, there are such points.

Therefore, the question is, for which λ between 1 and 4 does this happen?

Earlier, for λ = 2 and λ = 3, there were no period-2 points, but for λ = 4, there are.

Therefore, there must be a critical value of λ where period-2 points start to exist.

This suggests that the period-2 points exist for some λ ≥ some value.

To find this critical value, maybe we need to find when the equation f(f(x)) = x has solutions other than the fixed points.

Alternatively, since for λ = 4, there are period-2 points, and for λ = 2 and 3, there are not, maybe there is a range of λ where period-2 points exist.

Alternatively, maybe the period-2 points exist when the function f has a certain property, like when the function is chaotic or something.

Alternatively, maybe we can find the values of λ for which the equation f(f(x)) = x has solutions other than the fixed points.

Let me recall that for the logistic map f(x) = λx(1 - x), the period-doubling bifurcations occur at specific λ values.

But I might not remember the exact values, but maybe I can derive them.

Alternatively, since we have the equation for period-2 points, which is:

λ³x(x - 1)^2 + λ²(x - 1) + 1 = 0.

But earlier, we had:

For λ ≥ 1, the equation is:

λ³x(x - 1)^2 + λ²(x - 1) + 1 = 0.

Let me denote t = x - 1, so x = t + 1. But since x ∈ [0,1], t ∈ [-1, 0].

But maybe not helpful.

Alternatively, let me consider that for period-2 points, we have x ≠ f(x), and f(f(x)) = x.

Let me consider the system:

y = λx(1 - x)

x = λy(1 - y)

Which can be rewritten as:

y = λx(1 - x)

x = λy(1 - y)

Let me subtract the two equations:

y - x = λx(1 - x) - λy(1 - y)

But not sure.

Alternatively, let me consider that if we have a period-2 cycle, then the points x and y satisfy:

y = λx(1 - x)

x = λy(1 - y)

So, substituting y from the first equation into the second:

x = λ[λx(1 - x)][1 - λx(1 - x)]

Which is the same equation as before.

Let me denote z = x(1 - x). Then, y = λz, and the second equation becomes:

x = λ(λz)(1 - λz) = λ²z(1 - λz).

But z = x(1 - x), so:

x = λ²x(1 - x)(1 - λx(1 - x)).

Assuming x ≠ 0, divide both sides by x:

1 = λ²(1 - x)(1 - λx(1 - x)).

Which is the same equation as before.

Let me denote u = x(1 - x). Then, the equation becomes:

1 = λ²(1 - x)(1 - λu).

But since u = x(1 - x), and 1 - x = (1 - x).

Alternatively, maybe express 1 - x in terms of u.

Note that u = x(1 - x) ⇒ 1 - x = u/x, assuming x ≠ 0.

Therefore, substituting into the equation:

1 = λ²*(u/x)*(1 - λu).

So:

1 = λ²*(u/x)*(1 - λu).

But u = x(1 - x), so:

1 = λ²*(x(1 - x)/x)*(1 - λx(1 - x)) = λ²(1 - x)(1 - λx(1 - x)).

Which is the same equation.

Alternatively, maybe express everything in terms of u.

But not sure.

Alternatively, let me consider that for period-2 points, the function f has a 2-cycle. For the logistic map, the period-2 points exist when the function f has a certain property.

The standard logistic map is f(x) = λx(1 - x). The period-2 points satisfy f(f(x)) = x, but f(x) ≠ x.

The equation f(f(x)) = x can be written as:

f(f(x)) - x = 0.

Which is a quartic equation.

The number of real solutions depends on λ.

For the logistic map, the period-2 points exist for λ > 2. Wait, but earlier for λ = 2, there were no period-2 points.

Wait, maybe I need to check.

Actually, the period-doubling bifurcation occurs at λ = 3 for the logistic map. Wait, no, the first period-doubling occurs at λ = 3?

Wait, let me recall.

The logistic map f(x) = λx(1 - x) has fixed points at x = 0 and x = 1 - 1/λ. The stability of these fixed points changes with λ.

The fixed point at x = 1 - 1/λ is stable for λ < 3 and unstable for λ > 3.

At λ = 3, the fixed point becomes unstable, and a period-2 cycle is born.

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

But earlier, for λ = 4, we saw that there are period-2 points.

But for λ = 3, let me check.

Wait, earlier when I tried λ = 3, the equation f(f(x)) = x had only fixed points. But according to the bifurcation theory, at λ = 3, the period-2 cycle is born.

So, maybe my calculation for λ = 3 was wrong?

Let me check again for λ = 3.

For λ = 3, the equation f(f(x)) = x was:

After solving, we had x(x - 2/3)^3 = 0.

But that suggests only fixed points.

But according to bifurcation theory, there should be period-2 points for λ > 3.

So, maybe there was an error in my calculation.

Let me redo the calculation for λ = 3.

For λ = 3, f(x) = 3x(1 - x).

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

Let me compute this step by step.

First, compute f(x) = 3x(1 - x).

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

Let me compute 1 - 3x(1 - x) = 1 - 3x + 3x².

So, f(f(x)) = 3 * 3x(1 - x) * (1 - 3x + 3x²) = 9x(1 - x)(1 - 3x + 3x²).

Set equal to x:

9x(1 - x)(1 - 3x + 3x²) = x.

Assuming x ≠ 0, divide both sides by x:

9(1 - x)(1 - 3x + 3x²) = 1.

Expand the left-hand side:

First, expand (1 - x)(1 - 3x + 3x²):

Multiply term by term:

1*(1 - 3x + 3x²) = 1 - 3x + 3x².

- x*(1 - 3x + 3x²) = -x + 3x² - 3x³.

Adding together:

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

Multiply by 9:

9 - 36x + 54x² - 27x³ = 1.

Subtract 1:

8 - 36x + 54x² - 27x³ = 0 ⇒ -27x³ + 54x² - 36x + 8 = 0 ⇒ 27x³ - 54x² + 36x - 8 = 0.

Earlier, I factored this as (x - 2/3)(27x² - 36x + 12) = 0.

But let me check if this is correct.

Let me perform polynomial division of 27x³ - 54x² + 36x - 8 by (x - 2/3).

Using synthetic division with root 2/3:

Coefficients: 27 | -54 | 36 | -8

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.

So, the quotient is 27x² - 36x + 12.

Then, 27x² - 36x + 12 = 0 ⇒ divide by 3: 9x² - 12x + 4 = 0 ⇒ discriminant D = 144 - 144 = 0 ⇒ repeated root x = 12/(2*9) = 12/18 = 2/3.

Therefore, the equation factors as (x - 2/3)^3 * 27 = 0? Wait, no.

Wait, the original cubic was 27x³ - 54x² + 36x - 8 = 0, which factors as (x - 2/3)(27x² - 36x + 12) = 0. And the quadratic factors as 3(9x² - 12x + 4) = 3(3x - 2)^2.

Therefore, the equation is (x - 2/3) * 3(3x - 2)^2 = 0 ⇒ 3(x - 2/3)(3x - 2)^2 = 0.

Therefore, the solutions are x = 2/3 (with multiplicity 1) and x = 2/3 (with multiplicity 2 from the quadratic). Wait, no.

Wait, 3x - 2 = 0 ⇒ x = 2/3. So, the quadratic is (3x - 2)^2, so the equation is (x - 2/3) * 3*(3x - 2)^2 = 0 ⇒ (x - 2/3) * 3*(x - 2/3)^2 = 0 ⇒ 3(x - 2/3)^3 = 0.

Therefore, the only real solution is x = 2/3 with multiplicity 3.

Therefore, for λ = 3, the equation f(f(x)) = x has only x = 0 and x = 2/3 as solutions.

But according to bifurcation theory, at λ = 3, the period-2 cycle is born. So, why isn't there a period-2 cycle for λ = 3?

Wait, maybe the period-2 cycle is born at λ = 3, but it's a double root? Or maybe the period-2 points coincide with the fixed points?

Wait, for λ = 3, the fixed point at x = 2/3 is unstable, and the period-2 points might be at the same location?

Wait, no. If the period-2 cycle is born at λ = 3, then for λ > 3, there are period-2 points.

But for λ = 3, the period-2 points might be at the same location as the fixed point?

Wait, let me check.

If we have a period-2 cycle, then the points x and y satisfy f(x) = y and f(y) = x.

If x = y, then it's a fixed point. So, for λ = 3, the equation f(f(x)) = x has only the fixed points, but the period-2 points are tangent to the fixed points?

This suggests that at λ = 3, the period-2 cycle is born with a tangency, i.e., the period-2 points coincide with the fixed points.

Therefore, for λ > 3, the period-2 points exist.

But earlier, for λ = 4, we saw that there are period-2 points.

Therefore, the answer might be λ > 3.

But let me check for λ = 3.5.

Let me try λ = 3.5.

For λ = 3.5, fixed points are x = 0 and x = 1 - 1/3.5 = 1 - 2/7 = 5/7 ≈ 0.714.

Let me try solving f(f(x)) = x.

But this might be tedious. Alternatively, let me consider the equation we had earlier:

For λ ≥ 1, the equation is:

λ³x(x - 1)^2 + λ²(x - 1) + 1 = 0.

Let me denote this as:

λ³x(x - 1)^2 + λ²(x - 1) + 1 = 0.

Let me make a substitution. Let me set t = x - 1, so x = t + 1, where t ∈ [-1, 0].

Substituting into the equation:

λ³(t + 1)(t)^2 + λ²(t) + 1 = 0.

Expanding:

λ³(t + 1)t² + λ²t + 1 = 0.

Which is:

λ³t³ + λ³t² + λ²t + 1 = 0.

So:

λ³t³ + λ³t² + λ²t + 1 = 0.

This is a cubic equation in t.

Let me factor this equation.

Let me try to factor it.

Let me write it as:

λ³t³ + λ³t² + λ²t + 1 = 0.

Group terms:

(λ³t³ + λ³t²) + (λ²t + 1) = 0 ⇒ λ³t²(t + 1) + (λ²t + 1) = 0.

Not helpful.

Alternatively, try to factor as (at + b)(ct² + dt + e) = 0.

Assume it factors as (at + b)(ct² + dt + e) = 0.

Expanding: act³ + (ad + bc)t² + (ae + bd)t + be = 0.

Comparing coefficients:

act³ + (ad + bc)t² + (ae + bd)t + be = λ³t³ + λ³t² + λ²t + 1.

Therefore:

ac = λ³,

ad + bc = λ³,

ae + bd = λ²,

be = 1.

Assume that b and e are constants such that be = 1. Let me assume b = 1, e = 1.

Then:

ac = λ³,

ad + c = λ³,

a*1 + d*1 = λ²,

be = 1.

From the third equation: a + d = λ².

From the second equation: ad + c = λ³.

From the first equation: ac = λ³.

Let me solve for c from the first equation: c = λ³ / a.

Substitute into the second equation: ad + λ³ / a = λ³ ⇒ ad = λ³ - λ³ / a = λ³(1 - 1/a).

From the third equation: d = λ² - a.

Substitute into the above: a(λ² - a) = λ³(1 - 1/a).

Multiply both sides by a:

a²(λ² - a) = λ³(a - 1).

Expand left side: a²λ² - a³.

Right side: λ³a - λ³.

Bring all terms to left:

a²λ² - a³ - λ³a + λ³ = 0.

Factor:

- a³ + a²λ² - λ³a + λ³ = 0.

Factor by grouping:

(-a³ + a²λ²) + (-λ³a + λ³) = 0 ⇒ -a²(a - λ²) - λ³(a - 1) = 0.

Not sure.

Alternatively, let me try specific values.

Assume that a = λ. Then:

From ac = λ³ ⇒ c = λ³ / a = λ³ / λ = λ².

From third equation: d = λ² - a = λ² - λ.

From second equation: ad + c = λ³ ⇒ λ*(λ² - λ) + λ² = λ³ - λ² + λ² = λ³, which matches.

Therefore, this works!

Therefore, the cubic factors as (at + b)(ct² + dt + e) = (λt + 1)(λ²t² + (λ² - λ)t + 1).

Let me check:

(λt + 1)(λ²t² + (λ² - λ)t + 1) = λt * λ²t² + λt * (λ² - λ)t + λt * 1 + 1 * λ²t² + 1 * (λ² - λ)t + 1 * 1.

Compute term by term:

λ³t³ + λ(λ² - λ)t² + λt + λ²t² + (λ² - λ)t + 1.

Simplify:

λ³t³ + [λ³ - λ² + λ²]t² + [λ + λ² - λ]t + 1 = λ³t³ + λ³t² + λ²t + 1.

Which matches the original equation.

Therefore, the equation factors as:

(λt + 1)(λ²t² + (λ² - λ)t + 1) = 0.

Therefore, the solutions are:

1. λt + 1 = 0 ⇒ t = -1/λ.

2. λ²t² + (λ² - λ)t + 1 = 0.

So, for the first factor, t = -1/λ. Recall that t = x - 1, so x = t + 1 = 1 - 1/λ. Which is the fixed point we already knew.

For the second factor, solve λ²t² + (λ² - λ)t + 1 = 0.

Let me compute the discriminant D:

D = (λ² - λ)^2 - 4 * λ² * 1 = λ^4 - 2λ³ + λ² - 4λ² = λ^4 - 2λ³ - 3λ².

Factor:

D = λ²(λ² - 2λ - 3) = λ²(λ - 3)(λ + 1).

Since λ ≥ 1, and λ + 1 > 0, the discriminant is non-negative when λ - 3 ≥ 0 ⇒ λ ≥ 3.

Therefore, for λ ≥ 3, the quadratic equation has real roots.

Therefore, for λ ≥ 3, the equation f(f(x)) = x has solutions from the second factor.

Therefore, the solutions for t are:

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

Simplify:

√(λ²(λ - 3)(λ + 1)) = λ√[(λ - 3)(λ + 1)].

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

Therefore, the solutions for t are:

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

Recall that t = x - 1, so x = t + 1 = 1 + t.

Therefore, the solutions for x are:

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

Simplify:

x = [λ + 1 ± √((λ - 3)(λ + 1))]/(2λ).

Let me simplify the expression under the square root:

(λ - 3)(λ + 1) = λ² - 2λ - 3.

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

Now, since x must be in [0,1], we need to check if these solutions are in [0,1].

Let me analyze this expression.

First, for λ ≥ 3, the discriminant is non-negative.

Let me consider the two roots:

x₁ = [λ + 1 + √(λ² - 2λ - 3)]/(2λ)

x₂ = [λ + 1 - √(λ² - 2λ - 3)]/(2λ)

Let me check if these are in [0,1].

First, note that for λ ≥ 3, the term under the square root is λ² - 2λ - 3. Let me compute it for λ = 3: 9 - 6 - 3 = 0. For λ > 3, it's positive.

Let me check x₁ and x₂ for λ = 3:

At λ = 3, √(9 - 6 - 3) = √0 = 0.

So, x₁ = [3 + 1 + 0]/(6) = 4/6 = 2/3, which is the fixed point.

x₂ = [3 + 1 - 0]/6 = 4/6 = 2/3. So, both roots coincide with the fixed point.

For λ > 3, let's take λ = 4:

x₁ = [4 + 1 + √(16 - 8 - 3)]/(8) = [5 + √5]/8 ≈ [5 + 2.236]/8 ≈ 7.236/8 ≈ 0.9045, which is in [0,1].

x₂ = [5 - √5]/8 ≈ [5 - 2.236]/8 ≈ 2.764/8 ≈ 0.3455, which is also in [0,1].

Therefore, for λ > 3, we have two distinct solutions x₁ and x₂ in [0,1], which are period-2 points.

For λ = 3, they coincide with the fixed point.

Therefore, the period-2 points exist for λ > 3.

Therefore, the values of λ for which there exists an x ∈ [0,1] such that f(x) ≠ x but f(f(x)) = x are λ ∈ (3, 4].

But wait, the problem says 0 ≤ λ ≤ 4.

But we need to check if for λ = 3, there are such points. At λ = 3, the solutions x₁ and x₂ coincide with the fixed point x = 2/3. Therefore, for λ = 3, the only solutions are the fixed points, so there are no period-2 points.

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

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

For λ = 3.1, the discriminant is (3.1)^2 - 2*3.1 - 3 = 9.61 - 6.2 - 3 = 0.41 > 0.

Therefore, there are two distinct solutions x₁ and x₂, which are different from the fixed points.

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

Therefore, the answer is λ ∈ (3, 4].

But the problem says "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."

Therefore, the answer is λ ∈ (3, 4].

But let me check for λ = 3.

At λ = 3, the period-2 points coincide with the fixed point, so there are no such x.

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

But the problem says "values of λ, 0 ≤ λ ≤ 4", so including 4.

But let me check if for λ = 4, there are such points. As we saw earlier, yes, there are two period-2 points.

Therefore, the answer is λ ∈ (3, 4].

But the problem might expect a closed interval or something else.

Wait, but let me check for λ = 3.

At λ = 3, the equation f(f(x)) = x has only fixed points. Therefore, no period-2 points.

Therefore, the values of λ are those strictly greater than 3 and up to 4.

But the problem says "values of λ, 0 ≤ λ ≤ 4", so the answer is λ ∈ (3, 4].

But the question says "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."

Therefore, the answer is λ ∈ (3, 4].

But let me check if there are any other λ values where period-2 points exist.

For example, for λ between 1 and 3, we saw that for λ = 2 and λ = 3, there are no period-2 points.

But what about for some λ between 1 and 3?

Suppose λ = 2.5.

Let me try solving the equation for λ = 2.5.

But this might be time-consuming.

Alternatively, recall that the discriminant of the quadratic factor is D = λ²(λ - 3)(λ + 1). For λ < 3, D is negative, so no real solutions. Therefore, for λ < 3, the only real solutions to f(f(x)) = x are the fixed points.

Therefore, for λ < 3, there are no period-2 points.

Therefore, the answer is λ ∈ (3, 4].

But the problem says "values of λ, 0 ≤ λ ≤ 4", so including 4.

But the original question says "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."

Therefore, the answer is λ ∈ (3, 4].

But the problem might expect the answer in terms of intervals.

But let me check if for λ = 3, there are any such points.

At λ = 3, the equation f(f(x)) = x has only fixed points. Therefore, no such x.

Therefore, the answer is λ ∈ (3, 4].

But the problem might want it in boxed notation.

But let me check if there's a mistake.

Wait, the original equation for the period-2 points is derived from the quadratic factor, which has real solutions only when λ ≥ 3. For λ = 3, the solutions coincide with the fixed point. For λ > 3, they are distinct.

Therefore, the answer is λ ∈ (3, 4].

But the problem says "values of λ, 0 ≤ λ ≤ 4", so including 4.

But the question is from a math competition or similar, and often such problems expect answers in specific forms.

But let me check if there's a possibility that for λ = 3, there are period-2 points.

At λ = 3, the fixed point x = 2/3 is unstable. The period-2 points might be at the same location?

But according to our calculation, the period-2 points coincide with the fixed point.

Alternatively, maybe there are other solutions?

Wait, for λ = 3, the equation f(f(x)) = x has solutions x = 0, x = 2/3 (with multiplicity 3).

But x = 0 is a fixed point.

Therefore, there are no other solutions.

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

Therefore, the answer is λ ∈ (3, 4].

But the problem says "values of λ, 0 ≤ λ ≤ 4", so the answer is all λ in (3, 4].

But the problem might expect the answer as λ > 3, but since λ is in [0,4], it's λ ∈ (3, 4].

But let me check if for λ = 3, there are any points where f(f(x)) = x but f(x) ≠ x.

At λ = 3, suppose there is such a point. Then, it would have to satisfy f(f(x)) = x and f(x) ≠ x.

But according to our previous calculation, the only solutions are x = 0 and x = 2/3.

At x = 0, f(x) = 0, so it's a fixed point.

At x = 2/3, f(x) = 3*(2/3)*(1 - 2/3) = 3*(2/3)*(1/3) = 2/3. So, also a fixed point.

Therefore, no such points.

Therefore, the answer is λ ∈ (3, 4].

But the problem says "values of λ, 0 ≤ λ ≤ 4", so the answer is λ in (3, 4].

But in some contexts, the interval might be closed, but since at λ = 3 there are no such points, it's open at 3.

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

But the problem might expect it written as 3 < λ ≤ 4.

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

So, boxed answer would be \boxed{(3, 4]}.

But let me check if the problem expects a specific format.

The original problem says "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."

Since the answer is an interval, and in many math competitions, they might expect it written as 3 < λ ≤ 4, so boxed as \boxed{(3, 4]}.

But sometimes, depending on the convention, they might write it as 3 < λ ≤ 4, but since the user asked for the answer in boxed notation, likely \boxed{(3, 4]}.

But let me check if there's a possibility that λ = 3 is included.

But as discussed, at λ = 3, there are no such points.

Therefore, the answer is λ ∈ (3, 4].

**Final Answer**
The values of $\lambda$ are $\boxed{(3, 4]}$.

To solve the problem, we are given a function:

$$
f(x) = \lambda x(1 - x), \quad \text{with } 0 \leq \lambda \leq 4
$$

and we are asked to find the values of $\lambda$ for which there exists an $x \in [0,1]$ such that:

$$
f(x) \neq x \quad \text{but} \quad f(f(x)) = x
$$

---

### Step 1: Understanding the Condition

We are looking for **period-2 points** of the function $f$, i.e., points $x$ such that:

- $f(x) \neq x$ (not fixed points),
- $f(f(x)) = x$ (period-2 behavior).

This means that $x$ is not a fixed point, but it is part of a 2-cycle: $x \to f(x) \to x$.

---

### Step 2: Fixed Points of $f(x)$

To find fixed points, solve:

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

This simplifies to:

$$
x(\lambda(1 - x) - 1) = 0
$$

So the fixed points are:

- $x = 0$
- $x = 1 - \frac{1}{\lambda}$, provided $\lambda \geq 1$

---

### Step 3: Period-2 Points

To find period-2 points, we solve:

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

This is a quartic equation in $x$, and it includes all fixed points as solutions. To find **non-fixed** solutions, we subtract the fixed points and look for other real roots.

After algebraic manipulation and factoring, we find that the equation $f(f(x)) = x$ can be factored as:

$$
f(f(x)) - x = (x)(x - (1 - \frac{1}{\lambda})) \cdot \text{(quadratic)}
$$

The quadratic factor has real roots **only when** the discriminant is non-negative. This leads to the condition:

$$
\lambda^2(\lambda - 3)(\lambda + 1) \geq 0
$$

Since $\lambda \in [0,4]$, and $\lambda + 1 > 0$, the inequality reduces to:

$$
\lambda - 3 \geq 0 \Rightarrow \lambda \geq 3
$$

---

### Step 4: Behavior at $\lambda = 3$

At $\lambda = 3$, the discriminant is zero, and the quadratic factor has a **double root**. This means the period-2 points **coincide** with the fixed point $x = 1 - \frac{1}{3} = \frac{2}{3}$. Thus, **no new period-2 points** are introduced at $\lambda = 3$.

---

### Step 5: Behavior for $\lambda > 3$

For $\lambda > 3$, the discriminant is positive, and the quadratic factor has **two distinct real roots**. These roots correspond to **period-2 points** that are **not fixed points**, satisfying the condition $f(x) \neq x$ but $f(f(x)) = x$.

---

### Final Conclusion

The values of $\lambda$ for which there exists an $x \in [0,1]$ such that $f(x) \neq x$ but $f(f(x)) = x$ are:

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

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 $\lambda \in \boxed{(3,4]}.$