Explanation:
Okay, so I need to find the inverse function values for f inverse of 0 and f inverse of 6, and then add them together. Let me start by recalling what an inverse function is. The inverse function, f⁻¹(y), gives me the x-value such that f(x) = y. So, essentially, I need to find all x such that f(x) = 0 and f(x) = 6, then add those x-values together.

First, let me write down the function f(x) again to make sure I have it right:

f(x) is defined piecewise. For x ≤ 3, it's 3 - x, and for x > 3, it's -x³ + 2x² + 3x.

So, to find f⁻¹(0), I need to solve f(x) = 0. Similarly, for f⁻¹(6), solve f(x) = 6. Then add the solutions.

Let me start with f⁻¹(0). Let's solve f(x) = 0.

Case 1: x ≤ 3. Then f(x) = 3 - x. So set 3 - x = 0. Solving for x: 3 - x = 0 ⇒ x = 3.

But wait, x = 3 is in the domain of the first piece (since x ≤ 3). So that's valid.

Case 2: x > 3. Then f(x) = -x³ + 2x² + 3x. So set that equal to 0: -x³ + 2x² + 3x = 0. Let me factor this equation.

First, factor out an x: x(-x² + 2x + 3) = 0. So, either x = 0 or -x² + 2x + 3 = 0.

But wait, in this case, x > 3. So x = 0 is not in the domain here. So we need to solve -x² + 2x + 3 = 0. Let me multiply both sides by -1 to make it easier: x² - 2x - 3 = 0.

Factor this quadratic: looking for two numbers that multiply to -3 and add to -2. Let me think...  -3 and +1? Yes: (-3)(1) = -3, and -3 + 1 = -2. So, the factors are (x - 3)(x + 1) = 0. Therefore, solutions are x = 3 and x = -1.

But again, in this case, x > 3. So x = 3 is not in the domain (since it's x > 3), and x = -1 is also not in the domain. Therefore, there are no solutions in the second case.

Therefore, the only solution for f(x) = 0 is x = 3. Therefore, f⁻¹(0) = 3.

Okay, that seems straightforward. Now moving on to f⁻¹(6). So solve f(x) = 6. Again, check both cases.

Case 1: x ≤ 3. Then f(x) = 3 - x. So set 3 - x = 6. Solving for x: 3 - x = 6 ⇒ -x = 6 - 3 ⇒ -x = 3 ⇒ x = -3.

Check if x = -3 is in the domain x ≤ 3. Yes, it is. So that's a valid solution.

Case 2: x > 3. Then f(x) = -x³ + 2x² + 3x. Set that equal to 6: -x³ + 2x² + 3x = 6. Let me rearrange the equation: -x³ + 2x² + 3x - 6 = 0. Multiply both sides by -1 to make the leading coefficient positive: x³ - 2x² - 3x + 6 = 0.

So, need to solve x³ - 2x² - 3x + 6 = 0. Let me try to factor this cubic equation. Maybe there are rational roots. By Rational Root Theorem, possible roots are ±1, ±2, ±3, ±6. Let me test them.

First, test x = 1: 1 - 2 - 3 + 6 = 2 ≠ 0.

x = 2: 8 - 8 - 6 + 6 = 0. Oh, x = 2 is a root.

So, (x - 2) is a factor. Let me perform polynomial division or use synthetic division to factor this cubic.

Using synthetic division with root 2:

Coefficients: 1 (x³), -2 (x²), -3 (x), 6 (constant).

Bring down the 1. Multiply by 2: 1*2 = 2. Add to next coefficient: -2 + 2 = 0. Multiply by 2: 0*2 = 0. Add to next coefficient: -3 + 0 = -3. Multiply by 2: -3*2 = -6. Add to last coefficient: 6 + (-6) = 0. Perfect, so the cubic factors as (x - 2)(x² + 0x - 3) = (x - 2)(x² - 3).

Therefore, the equation factors to (x - 2)(x² - 3) = 0. Therefore, the roots are x = 2, x = √3, x = -√3.

But in this case, we are considering x > 3. So check which of these roots are greater than 3.

x = 2: Not greater than 3.

x = √3 ≈ 1.732: Not greater than 3.

x = -√3 ≈ -1.732: Also not.

Therefore, there are no solutions in the second case.

Wait, but hold on. The original equation was for x > 3. So even though we found roots, none of them are in the domain x > 3. Therefore, the only solution for f(x) = 6 is x = -3.

Wait, but that seems odd. Let me check my work again.

Wait, when I solved f(x) = 6 for x > 3, I had the equation -x³ + 2x² + 3x = 6. Then rearranged to -x³ + 2x² + 3x - 6 = 0. Then multiplied by -1 to get x³ - 2x² - 3x + 6 = 0. Then found that x = 2 is a root, factored into (x - 2)(x² - 3). So the roots are 2, sqrt(3), -sqrt(3).

But none of these are greater than 3. Therefore, there are no solutions in the second case. Therefore, the only solution is x = -3.

But wait, let me check if maybe I made a mistake in the algebra. Let me verify with x = -3.

Original function for x ≤ 3: f(-3) = 3 - (-3) = 3 + 3 = 6. Yes, that works.

But maybe there's another solution in the second case? Let me check with x = 4, for example. Let me compute f(4): -4³ + 2*(4)^2 + 3*4 = -64 + 32 + 12 = -64 + 44 = -20. Which is not 6. How about x = 3? Wait, x = 3 is in the first case. f(3) = 3 - 3 = 0.

Wait, maybe there's a mistake in the factoring? Let me check again.

Original equation for x > 3: -x³ + 2x² + 3x = 6.

So, moving 6 to the left: -x³ + 2x² + 3x - 6 = 0.

Multiply by -1: x³ - 2x² - 3x + 6 = 0.

Testing x = 2: 8 - 8 - 6 + 6 = 0. Correct.

Then factoring (x - 2) out.

Using polynomial division: divide x³ - 2x² - 3x + 6 by (x - 2).

Alternatively, write it as x³ - 2x² - 3x + 6.

Divide by (x - 2).

Using long division:

Divide x³ by x: x². Multiply (x - 2) by x²: x³ - 2x². Subtract from the original polynomial:

(x³ - 2x² - 3x + 6) - (x³ - 2x²) = 0x³ + 0x² - 3x + 6.

Now divide -3x by x: -3. Multiply (x - 2) by -3: -3x + 6. Subtract:

(-3x + 6) - (-3x + 6) = 0.

So the division gives x² - 3. Therefore, the cubic factors as (x - 2)(x² - 3). So that's correct.

Therefore, the roots are indeed 2, sqrt(3), -sqrt(3). None of which are greater than 3. Therefore, no solution in the second case.

Therefore, the only solution for f(x) = 6 is x = -3.

Wait, but that seems strange. Let me check if maybe there's another solution. Maybe I made a mistake in assuming that there are no solutions in the second case. Let me try plugging in some numbers greater than 3.

For example, let me try x = 4: f(4) = -64 + 32 + 12 = -20.

x = 3.5: f(3.5) = -(3.5)^3 + 2*(3.5)^2 + 3*(3.5). Let me compute that:

First, 3.5^3 = 42.875, so -42.875.

2*(3.5)^2 = 2*(12.25) = 24.5.

3*(3.5) = 10.5.

Adding them up: -42.875 + 24.5 + 10.5 = (-42.875 + 35) = -7.875.

So f(3.5) = -7.875, which is less than 6.

How about x approaching infinity? The leading term is -x³, so as x increases, f(x) tends to negative infinity. Therefore, the function for x > 3 is decreasing? Let me check the derivative.

Wait, maybe I can check if the function is increasing or decreasing.

For x > 3, f(x) = -x³ + 2x² + 3x. The derivative is f’(x) = -3x² + 4x + 3.

Let me check the sign of the derivative. For x > 3, let's plug in x = 4: f’(4) = -3*(16) + 4*(4) + 3 = -48 + 16 + 3 = -29 < 0. So the function is decreasing for x > 3.

Therefore, since it's decreasing, and at x = 3, the function is f(3) = 0 (from the first case). Wait, but for x approaching 3 from the right, what is the value? Let me compute the limit as x approaches 3 from the right.

f(3+) = - (3)^3 + 2*(3)^2 + 3*(3) = -27 + 18 + 9 = 0. So the function is continuous at x = 3? Wait, because from the left, f(3) = 3 - 3 = 0, and from the right, it's also 0. So the function is continuous at x = 3.

But for x > 3, the function is decreasing from 0 to negative infinity. Therefore, the maximum value of f(x) for x > 3 is 0, achieved at x = 3, and then it decreases. Therefore, there are no solutions to f(x) = 6 in the x > 3 case.

Therefore, the only solution is x = -3.

Therefore, f⁻¹(6) = -3.

Wait, but then f⁻¹(0) is 3, and f⁻¹(6) is -3. Therefore, adding them together: 3 + (-3) = 0.

But that seems too easy. Let me check again.

Wait, maybe I made a mistake in solving f(x) = 6? Let me check again.

For x ≤ 3, solving 3 - x = 6 gives x = -3. Correct.

For x > 3, solving -x³ + 2x² + 3x = 6. Which led to x³ - 2x² - 3x + 6 = 0, which factors as (x - 2)(x² - 3) = 0. So roots at 2, sqrt(3), -sqrt(3). None of which are greater than 3. Therefore, no solution.

Therefore, f⁻¹(6) is indeed -3.

Therefore, adding f⁻¹(0) + f⁻¹(6) = 3 + (-3) = 0.

But wait, the problem says "Calculate f⁻¹(0) + f⁻¹(6)". Is there a possibility that there are multiple inverses? Wait, but inverse functions are only defined if the original function is bijective. However, since the function f(x) is piecewise defined, we need to check if it's one-to-one.

Wait, let me check if f(x) is one-to-one. For the first piece, x ≤ 3, f(x) = 3 - x, which is a linear function with slope -1, so it's strictly decreasing. For the second piece, x > 3, f(x) = -x³ + 2x² + 3x. Let me check if it's strictly decreasing. Earlier, we saw that the derivative is f’(x) = -3x² + 4x + 3. Let me check if this derivative is always negative for x > 3.

At x = 3, f’(3) = -3*(9) + 4*(3) + 3 = -27 + 12 + 3 = -12 < 0.

As x increases beyond 3, the derivative becomes more negative because the -3x² term dominates. Therefore, the function is strictly decreasing on x > 3.

Therefore, the entire function f(x) is strictly decreasing on both intervals. However, we need to check if there's any overlap in the function's values between the two pieces.

At x = 3, f(x) = 0. For x < 3, f(x) = 3 - x, which is greater than 0 (since x ≤ 3, so 3 - x ≥ 0). For x > 3, f(x) is decreasing from 0 to negative infinity. Therefore, the function is strictly decreasing over the entire domain, with the left piece (x ≤ 3) going from +infty (as x approaches -infty) down to 0 at x = 3, and the right piece going from 0 down to -infty as x approaches +infty. Therefore, the function is strictly decreasing overall, hence injective (one-to-one), so the inverse function exists.

Therefore, each y-value corresponds to exactly one x-value. Therefore, f⁻¹(0) is 3, and f⁻¹(6) is -3, so their sum is 0.

But let me check once more for f⁻¹(6). Maybe I missed something?

Suppose someone might think that maybe there's another solution in the second case. Let me try solving -x³ + 2x² + 3x = 6 again.

Let me rearrange: -x³ + 2x² + 3x - 6 = 0. Let me try plugging in x = 4: -64 + 32 + 12 - 6 = -64 + 38 = -26 ≠ 0.

x = 3: -27 + 18 + 9 - 6 = -27 + 21 = -6 ≠ 0.

x = 2: -8 + 8 + 6 - 6 = 0. Which is the root we found.

x = 1: -1 + 2 + 3 - 6 = -2 ≠ 0.

x = 0: 0 + 0 + 0 - 6 = -6 ≠ 0.

x = -1: -(-1)^3 + 2(-1)^2 + 3(-1) = 1 + 2 - 3 = 0. Wait, but that's for the equation -x³ + 2x² + 3x = 0.

But in our case, we have -x³ + 2x² + 3x = 6. So, for x = -1, left side is 0, which is not 6.

So, no, there are no solutions in the second case.

Therefore, I think my conclusion is correct.

Therefore, f⁻¹(0) + f⁻¹(6) = 3 + (-3) = 0.

But wait, the problem says "Calculate f⁻¹(0) + f⁻¹(6)". Maybe I need to check if there are multiple pre-images? But since the function is strictly decreasing, each y has exactly one pre-image.

But let me check for f⁻¹(0). The function is 0 at x = 3. Is there another x where f(x) = 0? For x > 3, we saw that the equation had solutions x = 3 and x = -1, but x = 3 is not in the domain of the second case. So only x = 3.

Therefore, yes, only one pre-image.

Therefore, the answer is 0.

But let me check once more. Maybe I made a mistake in the inverse function?

Wait, if f(3) = 0, then f⁻¹(0) = 3.

If f(-3) = 3 - (-3) = 6, so f⁻¹(6) = -3.

Therefore, 3 + (-3) = 0.

Yes, that seems correct.

Alternatively, maybe the problem expects different answers? Let me check once again.

Wait, maybe I made a mistake in solving f(x) = 6 for x ≤ 3?

For x ≤ 3, f(x) = 3 - x. So 3 - x = 6 ⇒ x = 3 - 6 = -3. Correct.

Yes, that's right.

Alternatively, maybe there's another solution in the second case?

But we saw that the equation for x > 3 leads to no solutions.

Alternatively, maybe I need to consider that the inverse function could have multiple values? But since the function is strictly decreasing, it's one-to-one, so inverse is well-defined.

Therefore, I think my answer is correct.

**Final Answer**
\boxed{0}

To solve the problem, we are given a piecewise function:

$$
f(x) = \begin{cases}
3 - x & \text{if } x \leq 3 \\
-x^3 + 2x^2 + 3x & \text{if } x > 3
\end{cases}
$$

We are asked to compute:

$$
f^{-1}(0) + f^{-1}(6)
$$

---

### Step 1: Find $ f^{-1}(0) $

We need to find all $ x $ such that $ f(x) = 0 $.

#### Case 1: $ x \leq 3 $

$$
f(x) = 3 - x = 0 \Rightarrow x = 3
$$

This is valid since $ x = 3 \leq 3 $.

#### Case 2: $ x > 3 $

$$
f(x) = -x^3 + 2x^2 + 3x = 0
$$

Factor:

$$
x(-x^2 + 2x + 3) = 0
\Rightarrow x = 0 \quad \text{or} \quad -x^2 + 2x + 3 = 0
$$

Solving the quadratic:

$$
-x^2 + 2x + 3 = 0 \Rightarrow x^2 - 2x - 3 = 0 \Rightarrow (x - 3)(x + 1) = 0
\Rightarrow x = 3 \text{ or } x = -1
$$

But in this case, $ x > 3 $, so only $ x = 3 $ is a solution, but it is **not** in the domain $ x > 3 $. Therefore, **no solution** in this case.

**Conclusion:** The only solution is $ x = 3 $, so:

$$
f^{-1}(0) = 3
$$

---

### Step 2: Find $ f^{-1}(6) $

We need to find all $ x $ such that $ f(x) = 6 $.

#### Case 1: $ x \leq 3 $

$$
f(x) = 3 - x = 6 \Rightarrow x = -3
$$

This is valid since $ x = -3 \leq 3 $.

#### Case 2: $ x > 3 $

$$
f(x) = -x^3 + 2x^2 + 3x = 6
\Rightarrow -x^3 + 2x^2 + 3x - 6 = 0
$$

Multiply by $ -1 $:

$$
x^3 - 2x^2 - 3x + 6 = 0
$$

Try rational root $ x = 2 $:

$$
2^3 - 2(2^2) - 3(2) + 6 = 8 - 8 - 6 + 6 = 0
$$

So $ x = 2 $ is a root. Factor:

$$
x^3 - 2x^2 - 3x + 6 = (x - 2)(x^2 - 3)
$$

Solve $ x^2 - 3 = 0 \Rightarrow x = \pm \sqrt{3} $

So the roots are $ x = 2, \sqrt{3}, -\sqrt{3} $

But in this case, $ x > 3 $, so none of these roots are valid.

**Conclusion:** The only solution is $ x = -3 $, so:

$$
f^{-1}(6) = -3
$$

---

### Step 3: Add the Results

$$
f^{-1}(0) + f^{-1}(6) = 3 + (-3) = \boxed{0}
$$

[Assistant]
Explanation:
Okay, so I need to find the inverse function values for f inverse of 0 and f inverse of 6, and then add them together. Let me start by recalling what an inverse function is. The inverse function, f⁻¹(y), gives me the x-value such that f(x) = y. So, essentially, I need to find all x such that f(x) = 0 and f(x) = 6, then add those x-values together.

First, let me write down the function f(x) again to make sure I have it right:

f(x) is defined piecewise. For x ≤ 3, it's 3 - x, and for x > 3, it's -x³ + 2x² + 3x.

So, to find f⁻¹(0), I need to solve f(x) = 0. Similarly, for f⁻¹(6), solve f(x) = 6. Then add the solutions.

Let me start with f⁻¹(0). Let's solve f(x) = 0.

Case 1: x ≤ 3. Then f(x) = 3 - x. So set 3 - x = 0. Solving for x: 3 - x = 0 ⇒ x = 3.

But wait, x = 3 is in the domain of the first piece (since x ≤ 3). So that's valid.

Case 2: x > 3. Then f(x) = -x³ + 2x² + 3x. So set that equal to 0: -x³ + 2x² + 3x = 0. Let me factor this equation.

First, factor out an x: x(-x² + 2x + 3) = 0. So, either x = 0 or -x² + 2x + 3 = 0.

But wait, in this case, x > 3. So x = 0 is not in the domain here. So we need to solve -x² + 2x + 3 = 0. Let me multiply both sides by -1 to make it easier: x² - 2x - 3 = 0.

Factor this quadratic: looking for two numbers that multiply to -3 and add to -2. Let me think...  -3 and +1? Yes: (-3)(1) = -3, and -3 + 1 = -2. So, the factors are (x - 3)(x + 1) = 0. Therefore, solutions are x = 3 and x = -1.

But again, in this case, x > 3. So x = 3 is not in the domain (since it's x > 3), and x = -1 is also not in the domain. Therefore, there are no solutions in the second case.

Therefore, the only solution for f(x) = 0 is x = 3. Therefore, f⁻¹(0) = 3.

Okay, that seems straightforward. Now moving on to f⁻¹(6). So solve f(x) = 6. Again, check both cases.

Case 1: x ≤ 3. Then f(x) = 3 - x. So set 3 - x = 6. Solving for x: 3 - x = 6 ⇒ -x = 6 - 3 ⇒ -x = 3 ⇒ x = -3.

Check if x = -3 is in the domain x ≤ 3. Yes, it is. So that's a valid solution.

Case 2: x > 3. Then f(x) = -x³ + 2x² + 3x. Set that equal to 6: -x³ + 2x² + 3x = 6. Let me rearrange the equation: -x³ + 2x² + 3x - 6 = 0. Multiply both sides by -1 to make the leading coefficient positive: x³ - 2x² - 3x + 6 = 0.

So, need to solve x³ - 2x² - 3x + 6 = 0. Let me try to factor this cubic equation. Maybe there are rational roots. By Rational Root Theorem, possible roots are ±1, ±2, ±3, ±6. Let me test them.

First, test x = 1: 1 - 2 - 3 + 6 = 2 ≠ 0.

x = 2: 8 - 8 - 6 + 6 = 0. Oh, x = 2 is a root.

So, (x - 2) is a factor. Let me perform polynomial division or use synthetic division to factor this cubic.

Using synthetic division with root 2:

Coefficients: 1 (x³), -2 (x²), -3 (x), 6 (constant).

Bring down the 1. Multiply by 2: 1*2 = 2. Add to next coefficient: -2 + 2 = 0. Multiply by 2: 0*2 = 0. Add to next coefficient: -3 + 0 = -3. Multiply by 2: -3*2 = -6. Add to last coefficient: 6 + (-6) = 0. Perfect, so the cubic factors as (x - 2)(x² + 0x - 3) = (x - 2)(x² - 3).

Therefore, the equation factors to (x - 2)(x² - 3) = 0. Therefore, the roots are x = 2, x = √3, x = -√3.

But in this case, we are considering x > 3. So check which of these roots are greater than 3.

x = 2: Not greater than 3.

x = √3 ≈ 1.732: Not greater than 3.

x = -√3 ≈ -1.732: Also not.

Therefore, there are no solutions in the second case.

Wait, but hold on. The original equation was for x > 3. So even though we found roots, none of them are in the domain x > 3. Therefore, the only solution for f(x) = 6 is x = -3.

Wait, but that seems odd. Let me check my work again.

Wait, when I solved f(x) = 6 for x > 3, I had the equation -x³ + 2x² + 3x = 6. Then rearranged to -x³ + 2x² + 3x - 6 = 0. Then multiplied by -1 to get x³ - 2x² - 3x + 6 = 0. Then found that x = 2 is a root, factored into (x - 2)(x² - 3). So the roots are 2, sqrt(3), -sqrt(3).

But none of these are greater than 3. Therefore, there are no solutions in the second case. Therefore, the only solution is x = -3.

But wait, let me check if maybe I made a mistake in the algebra. Let me verify with x = -3.

Original function for x ≤ 3: f(-3) = 3 - (-3) = 3 + 3 = 6. Yes, that works.

But maybe there's another solution in the second case? Let me check with x = 4, for example. Let me compute f(4): -4³ + 2*(4)^2 + 3*4 = -64 + 32 + 12 = -64 + 44 = -20. Which is not 6. How about x = 3? Wait, x = 3 is in the first case. f(3) = 3 - 3 = 0.

Wait, maybe there's a mistake in the factoring? Let me check again.

Original equation for x > 3: -x³ + 2x² + 3x = 6.

So, moving 6 to the left: -x³ + 2x² + 3x - 6 = 0.

Multiply by -1: x³ - 2x² - 3x + 6 = 0.

Testing x = 2: 8 - 8 - 6 + 6 = 0. Correct.

Then factoring (x - 2) out.

Using polynomial division: divide x³ - 2x² - 3x + 6 by (x - 2).

Alternatively, write it as x³ - 2x² - 3x + 6.

Divide by (x - 2).

Using long division:

Divide x³ by x: x². Multiply (x - 2) by x²: x³ - 2x². Subtract from the original polynomial:

(x³ - 2x² - 3x + 6) - (x³ - 2x²) = 0x³ + 0x² - 3x + 6.

Now divide -3x by x: -3. Multiply (x - 2) by -3: -3x + 6. Subtract:

(-3x + 6) - (-3x + 6) = 0.

So the division gives x² - 3. Therefore, the cubic factors as (x - 2)(x² - 3). So that's correct.

Therefore, the roots are indeed 2, sqrt(3), -sqrt(3). None of which are greater than 3. Therefore, no solution in the second case.

Therefore, the only solution for f(x) = 6 is x = -3.

Wait, but that seems strange. Let me check if maybe there's another solution. Maybe I made a mistake in assuming that there are no solutions in the second case. Let me try plugging in some numbers greater than 3.

For example, let me try x = 4: f(4) = -64 + 32 + 12 = -20.

x = 3.5: f(3.5) = -(3.5)^3 + 2*(3.5)^2 + 3*(3.5). Let me compute that:

First, 3.5^3 = 42.875, so -42.875.

2*(3.5)^2 = 2*(12.25) = 24.5.

3*(3.5) = 10.5.

Adding them up: -42.875 + 24.5 + 10.5 = (-42.875 + 35) = -7.875.

So f(3.5) = -7.875, which is less than 6.

How about x approaching infinity? The leading term is -x³, so as x increases, f(x) tends to negative infinity. Therefore, the function for x > 3 is decreasing? Let me check the derivative.

Wait, maybe I can check if the function is increasing or decreasing.

For x > 3, f(x) = -x³ + 2x² + 3x. The derivative is f’(x) = -3x² + 4x + 3.

Let me check the sign of the derivative. For x > 3, let's plug in x = 4: f’(4) = -3*(16) + 4*(4) + 3 = -48 + 16 + 3 = -29 < 0. So the function is decreasing for x > 3.

Therefore, since it's decreasing, and at x = 3, the function is f(3) = 0 (from the first case). Wait, but for x approaching 3 from the right, what is the value? Let me compute the limit as x approaches 3 from the right.

f(3+) = - (3)^3 + 2*(3)^2 + 3*(3) = -27 + 18 + 9 = 0. So the function is continuous at x = 3? Wait, because from the left, f(3) = 3 - 3 = 0, and from the right, it's also 0. So the function is continuous at x = 3.

But for x > 3, the function is decreasing from 0 to negative infinity. Therefore, the maximum value of f(x) for x > 3 is 0, achieved at x = 3, and then it decreases. Therefore, there are no solutions to f(x) = 6 in the x > 3 case.

Therefore, the only solution is x = -3.

Therefore, f⁻¹(6) = -3.

Wait, but then f⁻¹(0) is 3, and f⁻¹(6) is -3. Therefore, adding them together: 3 + (-3) = 0.

But that seems too easy. Let me check again.

Wait, maybe I made a mistake in solving f(x) = 6? Let me check again.

For x ≤ 3, solving 3 - x = 6 gives x = -3. Correct.

For x > 3, solving -x³ + 2x² + 3x = 6. Which led to x³ - 2x² - 3x + 6 = 0, which factors as (x - 2)(x² - 3) = 0. So roots at 2, sqrt(3), -sqrt(3). None of which are greater than 3. Therefore, no solution.

Therefore, f⁻¹(6) is indeed -3.

Therefore, adding f⁻¹(0) + f⁻¹(6) = 3 + (-3) = 0.

But wait, the problem says "Calculate f⁻¹(0) + f⁻¹(6)". Is there a possibility that there are multiple inverses? Wait, but inverse functions are only defined if the original function is bijective. However, since the function f(x) is piecewise defined, we need to check if it's one-to-one.

Wait, let me check if f(x) is one-to-one. For the first piece, x ≤ 3, f(x) = 3 - x, which is a linear function with slope -1, so it's strictly decreasing. For the second piece, x > 3, f(x) = -x³ + 2x² + 3x. Let me check if it's strictly decreasing. Earlier, we saw that the derivative is f’(x) = -3x² + 4x + 3. Let me check if this derivative is always negative for x > 3.

At x = 3, f’(3) = -3*(9) + 4*(3) + 3 = -27 + 12 + 3 = -12 < 0.

As x increases beyond 3, the derivative becomes more negative because the -3x² term dominates. Therefore, the function is strictly decreasing on x > 3.

Therefore, the entire function f(x) is strictly decreasing on both intervals. However, we need to check if there's any overlap in the function's values between the two pieces.

At x = 3, f(x) = 0. For x < 3, f(x) = 3 - x, which is greater than 0 (since x ≤ 3, so 3 - x ≥ 0). For x > 3, f(x) is decreasing from 0 to negative infinity. Therefore, the function is strictly decreasing over the entire domain, with the left piece (x ≤ 3) going from +infty (as x approaches -infty) down to 0 at x = 3, and the right piece going from 0 down to -infty as x approaches +infty. Therefore, the function is strictly decreasing overall, hence injective (one-to-one), so the inverse function exists.

Therefore, each y-value corresponds to exactly one x-value. Therefore, f⁻¹(0) is 3, and f⁻¹(6) is -3, so their sum is 0.

But let me check once more for f⁻¹(6). Maybe I missed something?

Suppose someone might think that maybe there's another solution in the second case. Let me try solving -x³ + 2x² + 3x = 6 again.

Let me rearrange: -x³ + 2x² + 3x - 6 = 0. Let me try plugging in x = 4: -64 + 32 + 12 - 6 = -64 + 38 = -26 ≠ 0.

x = 3: -27 + 18 + 9 - 6 = -27 + 21 = -6 ≠ 0.

x = 2: -8 + 8 + 6 - 6 = 0. Which is the root we found.

x = 1: -1 + 2 + 3 - 6 = -2 ≠ 0.

x = 0: 0 + 0 + 0 - 6 = -6 ≠ 0.

x = -1: -(-1)^3 + 2(-1)^2 + 3(-1) = 1 + 2 - 3 = 0. Wait, but that's for the equation -x³ + 2x² + 3x = 0.

But in our case, we have -x³ + 2x² + 3x = 6. So, for x = -1, left side is 0, which is not 6.

So, no, there are no solutions in the second case.

Therefore, I think my conclusion is correct.

Therefore, f⁻¹(0) + f⁻¹(6) = 3 + (-3) = 0.

But wait, the problem says "Calculate f⁻¹(0) + f⁻¹(6)". Maybe I need to check if there are multiple pre-images? But since the function is strictly decreasing, each y has exactly one pre-image.

But let me check for f⁻¹(0). The function is 0 at x = 3. Is there another x where f(x) = 0? For x > 3, we saw that the equation had solutions x = 3 and x = -1, but x = 3 is not in the domain of the second case. So only x = 3.

Therefore, yes, only one pre-image.

Therefore, the answer is 0.

But let me check once more. Maybe I made a mistake in the inverse function?

Wait, if f(3) = 0, then f⁻¹(0) = 3.

If f(-3) = 3 - (-3) = 6, so f⁻¹(6) = -3.

Therefore, 3 + (-3) = 0.

Yes, that seems correct.

Alternatively, maybe the problem expects different answers? Let me check once again.

Wait, maybe I made a mistake in solving f(x) = 6 for x ≤ 3?

For x ≤ 3, f(x) = 3 - x. So 3 - x = 6 ⇒ x = 3 - 6 = -3. Correct.

Yes, that's right.

Alternatively, maybe there's another solution in the second case?

But we saw that the equation for x > 3 leads to no solutions.

Alternatively, maybe I need to consider that the inverse function could have multiple values? But since the function is strictly decreasing, it's one-to-one, so inverse is well-defined.

Therefore, I think my answer is correct.

**Final Answer**
\boxed{0}

To solve the problem, we are given a piecewise function:

$$
f(x) = \begin{cases}
3 - x & \text{if } x \leq 3 \\
-x^3 + 2x^2 + 3x & \text{if } x > 3
\end{cases}
$$

We are asked to compute:

$$
f^{-1}(0) + f^{-1}(6)
$$

---

### Step 1: Find $ f^{-1}(0) $

We need to find all $ x $ such that $ f(x) = 0 $.

#### Case 1: $ x \leq 3 $

$$
f(x) = 3 - x = 0 \Rightarrow x = 3
$$

This is valid since $ x = 3 \leq 3 $.

#### Case 2: $ x > 3 $

$$
f(x) = -x^3 + 2x^2 + 3x = 0
$$

Factor:

$$
x(-x^2 + 2x + 3) = 0
\Rightarrow x = 0 \quad \text{or} \quad -x^2 + 2x + 3 = 0
$$

Solving the quadratic:

$$
-x^2 + 2x + 3 = 0 \Rightarrow x^2 - 2x - 3 = 0 \Rightarrow (x - 3)(x + 1) = 0
\Rightarrow x = 3 \text{ or } x = -1
$$

But in this case, $ x > 3 $, so only $ x = 3 $ is a solution, but it is **not** in the domain $ x > 3 $. Therefore, **no solution** in this case.

**Conclusion:** The only solution is $ x = 3 $, so:

$$
f^{-1}(0) = 3
$$

---

### Step 2: Find $ f^{-1}(6) $

We need to find all $ x $ such that $ f(x) = 6 $.

#### Case 1: $ x \leq 3 $

$$
f(x) = 3 - x = 6 \Rightarrow x = -3
$$

This is valid since $ x = -3 \leq 3 $.

#### Case 2: $ x > 3 $

$$
f(x) = -x^3 + 2x^2 + 3x = 6
\Rightarrow -x^3 + 2x^2 + 3x - 6 = 0
$$

Multiply by $ -1 $:

$$
x^3 - 2x^2 - 3x + 6 = 0
$$

Try rational root $ x = 2 $:

$$
2^3 - 2(2^2) - 3(2) + 6 = 8 - 8 - 6 + 6 = 0
$$

So $ x = 2 $ is a root. Factor:

$$
x^3 - 2x^2 - 3x + 6 = (x - 2)(x^2 - 3)
$$

Solve $ x^2 - 3 = 0 \Rightarrow x = \pm \sqrt{3} $

So the roots are $ x = 2, \sqrt{3}, -\sqrt{3} $

But in this case, $ x > 3 $, so none of these roots are valid.

**Conclusion:** The only solution is $ x = -3 $, so:

$$
f^{-1}(6) = -3
$$

---

### Step 3: Add the Results

$$
f^{-1}(0) + f^{-1}(6) = 3 + (-3) = \boxed{0}
$$

Solution:
The number $f^{-1}(0)$ is the value of $x$ such that $f(x) = 0$.  Since the function $f$ is defined piecewise, to find this value, we must consider both cases $x \le 3$ and $x > 3$.

If $x \le 3$ and $f(x) = 0$, then $3 - x = 0$, which leads to $x = 3$.  Note that this value satisfies the condition $x \le 3$.  If $x > 3$ and $f(x) = 0$, then $-x^3 + 2x^2 + 3x = 0$.  This equation factors as $-x(x - 3)(x + 1) = 0$, so $x = 0$, $x = 3$, or $x = -1$.  But none of these values satisfies $x > 3$, so the solution is $x = 3$, which means $f^{-1}(0) = 3$.

Now we compute $f^{-1}(6)$, which is the value of $x$ such that $f(x) = 6$.

If $x \le 3$ and $f(x) = 6$, then $3 - x = 6$, which leads to $x = -3$.  Note that this value satisfies the condition $x \le 3$.  If $x > 3$ and $f(x) = 6$, then $-x^3 + 2x^2 + 3x = 6$, or $x^3 - 2x^2 - 3x + 6 = 0$.  This equation factors as $(x - 2)(x^2 - 3) = 0$, so $x = 2$, $x = \sqrt{3}$, or $x = -\sqrt{3}$.  But none of these values satisfies $x > 3$, so the solution is $x = -3$, which means $f^{-1}(6) = -3$.

Therefore, $f^{-1}(0)+f^{-1}(6) = 3 + (-3) = \boxed{0}$.

[asy]
unitsize(3mm);
defaultpen(linewidth(.7pt)+fontsize(8pt));
import graph;

draw((-20,0)--(20,0),Arrows(4));
draw((0,-20)--(0,20),Arrows(4));

real f(real x) {return 3-x;}
real g(real x) {return -x^3+2x^2+3x;}

real x;

draw(graph(f,-15,3),BeginArrow(4));
draw(graph(g,3,4),EndArrow(4));

real eps = 0.2;

draw((-eps,3)--(eps,3));
draw((-eps,0)--(eps,0));
draw((-eps,-3)--(eps,-3));

dot("$(-3,6)$",(-3,6),SW);
dot("$(3,0)$",(3,0),NE);

label("$f(x)$",(3,20.5));
label("$x$",(20.5,-1));
[/asy]