Explanation:
Okay, so I need to find the smallest distance between the origin and a point on the graph of $ y = \frac{1}{\sqrt{2}}(x^2 - 3) $. The answer should be in the form $ \sqrt{a}/b $, where $ a $ and $ b $ are positive integers, and $ a $ is square-free. Then I have to find $ a + b $. Alright, let me think about how to approach this.

First, I remember that the distance between a point $ (x, y) $ and the origin $ (0, 0) $ is given by the distance formula: $ D = \sqrt{x^2 + y^2} $. Since we want the smallest distance, maybe I can minimize this distance function. However, dealing with square roots can be a bit messy, so maybe I can minimize the square of the distance instead, which should have its minimum at the same point. That might simplify calculations.

So, let me define $ D^2 = x^2 + y^2 $. Since $ y $ is given in terms of $ x $, I can substitute that into the equation. Let me do that:

$ D^2 = x^2 + \left( \frac{1}{\sqrt{2}}(x^2 - 3) \right)^2 $.

Let me simplify this expression. First, square the $ y $ term:

$ \left( \frac{1}{\sqrt{2}}(x^2 - 3) \right)^2 = \frac{1}{2}(x^2 - 3)^2 $.

So, substituting back into $ D^2 $:

$ D^2 = x^2 + \frac{1}{2}(x^2 - 3)^2 $.

Now, I need to find the minimum value of this function. To do that, I can take the derivative with respect to $ x $, set it equal to zero, and solve for $ x $. Then check if it's a minimum using the second derivative or some other method.

Let me compute the derivative. Let me denote $ f(x) = x^2 + \frac{1}{2}(x^2 - 3)^2 $.

First, expand $ (x^2 - 3)^2 $:

$ (x^2 - 3)^2 = x^4 - 6x^2 + 9 $.

Therefore, $ f(x) = x^2 + \frac{1}{2}(x^4 - 6x^2 + 9) $.

Let me simplify this:

$ f(x) = x^2 + \frac{1}{2}x^4 - 3x^2 + \frac{9}{2} $.

Combine like terms:

$ f(x) = \frac{1}{2}x^4 - 2x^2 + \frac{9}{2} $.

Now, take the derivative $ f'(x) $:

$ f'(x) = 2x^3 - 4x $.

Wait, let me check that again. The derivative of $ \frac{1}{2}x^4 $ is $ 2x^3 $, right? Because $ \frac{d}{dx} x^n = n x^{n-1} $, so $ \frac{1}{2} \times 4x^3 = 2x^3 $. Then the derivative of $ -2x^2 $ is $ -4x $. And the derivative of $ \frac{9}{2} $ is 0. So yes, $ f'(x) = 2x^3 - 4x $.

To find critical points, set $ f'(x) = 0 $:

$ 2x^3 - 4x = 0 $.

Factor out 2x:

$ 2x(x^2 - 2) = 0 $.

So, the critical points are when $ 2x = 0 $ or $ x^2 - 2 = 0 $.

Therefore, $ x = 0 $ or $ x = \sqrt{2} $ or $ x = -\sqrt{2} $.

Now, I need to check which of these critical points gives the minimum distance. Since the function $ f(x) $ is even? Wait, let me check. The original function $ y = \frac{1}{\sqrt{2}}(x^2 - 3) $ is even, because replacing $ x $ with $ -x $ doesn't change the value. Therefore, the graph is symmetric about the y-axis. Therefore, the distance from the origin should also be symmetric. Therefore, the minimum distance should occur at $ x = \sqrt{2} $ and $ x = -\sqrt{2} $, or maybe at $ x = 0 $. Let me check the value of $ f(x) $ at these points.

First, at $ x = 0 $:

$ f(0) = \frac{1}{2}(0)^4 - 2(0)^2 + \frac{9}{2} = 0 - 0 + \frac{9}{2} = \frac{9}{2} $.

So, the distance squared is $ 9/2 $, so the distance is $ \sqrt{9/2} = 3/\sqrt{2} $, which is approximately 2.121.

Now, let's check at $ x = \sqrt{2} $:

First compute $ x^2 = (\sqrt{2})^2 = 2 $.

So, $ f(\sqrt{2}) = \frac{1}{2}(2)^2 - 2(2) + \frac{9}{2} $.

Wait, hold on. Wait, earlier I had simplified $ f(x) $ as $ \frac{1}{2}x^4 - 2x^2 + \frac{9}{2} $. So, substituting $ x = \sqrt{2} $:

First, $ x^4 = (x^2)^2 = (2)^2 = 4 $.

Therefore, $ f(\sqrt{2}) = \frac{1}{2}(4) - 2(2) + \frac{9}{2} = 2 - 4 + 4.5 = 2.5 $.

Which is $ 5/2 $. Therefore, the distance squared is $ 5/2 $, so the distance is $ \sqrt{5/2} = \sqrt{10}/2 $. Wait, because $ \sqrt{5/2} = \sqrt{10}/2 $. Let me check that:

$ \sqrt{5/2} = \sqrt{5}/\sqrt{2} = (\sqrt{5} \times \sqrt{2}) / (\sqrt{2} \times \sqrt{2}) ) = \sqrt{10}/2 $. Yes, that's correct.

Similarly, at $ x = -\sqrt{2} $, the value will be the same because of the even function. So, the distance squared is 5/2, which is less than 9/2. Therefore, the minimum distance occurs at $ x = \pm \sqrt{2} $, and the minimal distance is $ \sqrt{10}/2 $.

Wait, but the problem says the answer can be expressed as $ \sqrt{a}/b $. So, $ \sqrt{10}/2 $ is already in that form, with $ a = 10 $ and $ b = 2 $. Then $ a + b = 12 $. But wait, let me check if 10 is square-free. 10 factors into 2 * 5, neither of which are squares, so yes, 10 is square-free. Therefore, the answer is 10 + 2 = 12. Is that correct?

Wait, but let me double-check my calculations because sometimes when you do calculus, you might make a mistake. Let me verify the steps again.

First, the distance squared is $ x^2 + y^2 $, with $ y = \frac{1}{\sqrt{2}}(x^2 - 3) $. So, substituting:

$ D^2 = x^2 + \frac{1}{2}(x^2 - 3)^2 $. Then expanding $ (x^2 - 3)^2 $ gives $ x^4 - 6x^2 + 9 $, so:

$ D^2 = x^2 + \frac{1}{2}x^4 - 3x^2 + \frac{9}{2} $, which simplifies to $ \frac{1}{2}x^4 - 2x^2 + \frac{9}{2} $. Then derivative is $ 2x^3 - 4x $, set to zero gives critical points at x=0, x=±√2. Then evaluating D² at x=√2:

x² = 2, x⁴ = 4. So D² = (1/2)(4) - 2*(2) + 9/2 = 2 - 4 + 4.5 = 2.5, which is 5/2. So D = sqrt(5/2) = sqrt(10)/2. So that seems correct.

But let me check if there's another way to approach this problem, maybe using geometry or something else, to confirm.

Alternatively, maybe using Lagrange multipliers? Let me try that.

We need to minimize the distance squared function $ f(x, y) = x^2 + y^2 $ subject to the constraint $ g(x, y) = y - \frac{1}{\sqrt{2}}(x^2 - 3) = 0 $.

The Lagrangian is $ \mathcal{L} = x^2 + y^2 - \lambda \left( y - \frac{1}{\sqrt{2}}(x^2 - 3) \right) $.

Taking partial derivatives:

∂L/∂x = 2x - λ*(-2x / sqrt(2)) = 0

Wait, let me compute that again. The constraint is $ g(x, y) = y - \frac{1}{\sqrt{2}}(x^2 - 3) = 0 $. Therefore, the partial derivatives of g:

∂g/∂x = - (2x)/sqrt(2) = - sqrt(2) x

∂g/∂y = 1

So, the Lagrangian equations are:

∂L/∂x = 2x - λ*(-sqrt(2) x) = 0

∂L/∂y = 2y - λ*(1) = 0

∂L/∂λ = - [ y - (1/sqrt(2))(x² - 3) ] = 0

So, equations:

1) 2x + λ*sqrt(2) x = 0

2) 2y - λ = 0

3) y = (1/sqrt(2))(x² - 3)

From equation 1: 2x + λ*sqrt(2) x = 0. Let's factor out x:

x(2 + λ*sqrt(2)) = 0.

Therefore, either x = 0 or 2 + λ*sqrt(2) = 0 => λ = -2 / sqrt(2) = -sqrt(2).

Case 1: x = 0.

Then from equation 3: y = (1/sqrt(2))(0 - 3) = -3 / sqrt(2). Then from equation 2: 2y - λ = 0 => λ = 2y = 2*(-3)/sqrt(2) = -6 / sqrt(2) = -3*sqrt(2). But from equation 1, if x=0, then equation 1 is satisfied regardless of λ. So that's consistent. So this gives a critical point at (0, -3/sqrt(2)).

Case 2: λ = -sqrt(2). Then from equation 2: 2y - (-sqrt(2)) = 0 => 2y + sqrt(2) = 0 => y = -sqrt(2)/2.

Then from equation 3: y = (1/sqrt(2))(x² - 3). So:

-sqrt(2)/2 = (1/sqrt(2))(x² - 3)

Multiply both sides by sqrt(2):

- (sqrt(2) * sqrt(2))/2 = x² - 3 => -2/2 = x² - 3 => -1 = x² - 3 => x² = 2 => x = sqrt(2) or -sqrt(2). So that gives the points (sqrt(2), -sqrt(2)/2) and (-sqrt(2), -sqrt(2)/2).

So, we have three critical points: (0, -3/sqrt(2)), (sqrt(2), -sqrt(2)/2), (-sqrt(2), -sqrt(2)/2). Now, let's compute the distance squared for each.

First, for (0, -3/sqrt(2)):

Distance squared is 0² + (-3/sqrt(2))² = 9/2, same as before.

For (sqrt(2), -sqrt(2)/2):

Distance squared is (sqrt(2))² + (-sqrt(2)/2)^2 = 2 + (2/4) = 2 + 1/2 = 5/2, same as before. So that's consistent with the previous result. Therefore, the minimal distance squared is 5/2, so minimal distance is sqrt(5/2) = sqrt(10)/2. Therefore, the answer is sqrt(10)/2, so a = 10, b = 2, so a + b = 12.

But wait, let me check if there's a possibility that there's a smaller distance. For example, maybe the minimal distance occurs somewhere else? But according to both calculus methods, we found critical points and checked them. Also, since the function tends to infinity as x approaches infinity, the minimal must be at one of the critical points. Therefore, I think this is correct.

But let me check with another approach. Maybe parametric equations or something else.

Alternatively, think geometrically. The graph is a parabola. The origin is a point outside the parabola. The minimal distance from the origin to the parabola would be along the line connecting the origin to the closest point on the parabola. So, the line connecting the origin to the closest point on the parabola should be perpendicular to the tangent of the parabola at that point.

So, maybe we can use that condition. Let me try that.

Let me denote the point on the parabola as (x, y) = (x, (1/sqrt(2))(x² - 3)). The slope of the tangent line at this point can be found by taking the derivative of y with respect to x.

Given y = (1/sqrt(2))(x² - 3), so dy/dx = (2x)/sqrt(2) = sqrt(2) x.

Therefore, the slope of the tangent line at (x, y) is m_tangent = sqrt(2) x.

The slope of the line connecting the origin to the point (x, y) is m_line = (y - 0)/(x - 0) = y/x = [ (1/sqrt(2))(x² - 3) ] / x = (x² - 3)/(x sqrt(2)).

Since the line connecting the origin to the point is perpendicular to the tangent line, their slopes should multiply to -1. Therefore:

m_line * m_tangent = -1.

So:

[ (x² - 3)/(x sqrt(2)) ] * [ sqrt(2) x ] = -1.

Let me compute that:

The sqrt(2) in the denominator and the sqrt(2) in the numerator cancel out, and the x in the denominator and the x in the numerator cancel out. So we have:

(x² - 3) * 1 = -1 => x² - 3 = -1 => x² = 2 => x = sqrt(2) or x = -sqrt(2). Which is exactly the same result as before! Therefore, this confirms that the critical points are at x = ±sqrt(2). Therefore, the minimal distance is indeed sqrt(10)/2, so a + b = 10 + 2 = 12.

But just to be thorough, let me check if there's any possible mistake in assuming that the minimal distance occurs at those points. For example, maybe the parabola is opening upwards, and the origin is below it? Let me sketch mentally. The equation is y = (1/sqrt(2))(x² - 3). So when x = 0, y = -3/sqrt(2) ≈ -2.12. As |x| increases, y increases. So the vertex of the parabola is at (0, -3/sqrt(2)), and it opens upwards. The origin is at (0, 0), which is above the vertex. Therefore, the closest point on the parabola to the origin is likely somewhere on the right or left side of the parabola, not at the vertex. Which matches our previous result where the minimal distance is achieved at x = ±sqrt(2), which are points above the vertex. So that makes sense.

Alternatively, if I plug in x = sqrt(2), then y = (1/sqrt(2))((sqrt(2))² - 3) = (1/sqrt(2))(2 - 3) = -1/sqrt(2). Therefore, the point is (sqrt(2), -1/sqrt(2)). The distance from the origin is sqrt( (sqrt(2))² + (-1/sqrt(2))² ) = sqrt(2 + 1/2) = sqrt(5/2) = sqrt(10)/2. Which is the same as before.

Alternatively, if I think about the point (sqrt(2), -1/sqrt(2)), is that the closest point? Let me check another point. For example, take x = 1. Then y = (1/sqrt(2))(1 - 3) = (-2)/sqrt(2) = -sqrt(2). The distance squared is 1 + (sqrt(2))² = 1 + 2 = 3, which is 3, which is larger than 5/2 = 2.5. So yes, larger. If I take x = 2, then y = (1/sqrt(2))(4 - 3) = 1/sqrt(2). The distance squared is 4 + (1/2) = 4.5, which is larger. If I take x = sqrt(2), distance squared is 2.5. If I take x approaching infinity, distance squared approaches infinity. So yes, seems like 2.5 is the minimum.

Alternatively, maybe check x = something else. Let me try x = 1.5. Then x² = 2.25. Then y = (1/sqrt(2))(2.25 - 3) = (-0.75)/sqrt(2). Then distance squared is (1.5)^2 + [ (-0.75)/sqrt(2) ]^2 = 2.25 + (0.5625)/2 = 2.25 + 0.28125 = 2.53125, which is approximately 2.53, which is larger than 2.5. So yes, 2.5 is smaller. Therefore, seems like our answer is correct.

Therefore, I think the answer is 10 + 2 = 12. However, just to make sure, let me check if the problem says "the smallest distance between the origin and a point on the graph". So, the minimal distance is sqrt(10)/2, which is equal to sqrt(10)/2. The problem says "expressed as sqrt(a)/b", so sqrt(10)/2 is already in that form. Since 10 is square-free (its prime factors are 2 and 5, neither of which are squared), so a = 10, b = 2, so a + b = 12. Therefore, the answer is 12.

But just to be 100% sure, let me check if there's an alternative expression. For example, sometimes problems might have different forms. Let me see:

sqrt(10)/2 is the same as sqrt(10)/2. If I rationalize something, but no, that's already simplified. So yes, that's the form. Therefore, the answer is 12.

**Final Answer**
\boxed{12}

To find the smallest distance between the origin and a point on the graph of the function
$$
y = \frac{1}{\sqrt{2}}(x^2 - 3),
$$
we begin by using the **distance formula** between a point $(x, y)$ and the origin $(0, 0)$:

$$
D = \sqrt{x^2 + y^2}.
$$

To simplify the minimization process, we minimize the **square of the distance**, which is:

$$
D^2 = x^2 + y^2.
$$

Substituting the given expression for $ y $, we get:

$$
D^2 = x^2 + \left( \frac{1}{\sqrt{2}}(x^2 - 3) \right)^2 = x^2 + \frac{1}{2}(x^2 - 3)^2.
$$

Expanding the square:

$$
(x^2 - 3)^2 = x^4 - 6x^2 + 9,
$$
so
$$
D^2 = x^2 + \frac{1}{2}(x^4 - 6x^2 + 9) = \frac{1}{2}x^4 - 2x^2 + \frac{9}{2}.
$$

Now, we find the critical points by taking the derivative of $ D^2 $ with respect to $ x $:

$$
\frac{d}{dx} \left( \frac{1}{2}x^4 - 2x^2 + \frac{9}{2} \right) = 2x^3 - 4x.
$$

Setting the derivative equal to zero:

$$
2x^3 - 4x = 0 \Rightarrow 2x(x^2 - 2) = 0.
$$

This gives the critical points:

$$
x = 0, \quad x = \pm \sqrt{2}.
$$

We now evaluate $ D^2 $ at these points:

- At $ x = 0 $:
  $$
  D^2 = \frac{1}{2}(0)^4 - 2(0)^2 + \frac{9}{2} = \frac{9}{2}.
  $$

- At $ x = \pm \sqrt{2} $:
  $$
  x^2 = 2, \quad x^4 = 4, \quad \text{so} \quad D^2 = \frac{1}{2}(4) - 2(2) + \frac{9}{2} = 2 - 4 + \frac{9}{2} = \frac{5}{2}.
  $$

Thus, the **minimum distance squared** is $ \frac{5}{2} $, and the **minimum distance** is:

$$
D = \sqrt{\frac{5}{2}} = \frac{\sqrt{10}}{2}.
$$

This is in the form $ \frac{\sqrt{a}}{b} $, where $ a = 10 $ and $ b = 2 $. Since 10 is square-free (its prime factorization is $ 2 \times 5 $), the conditions of the problem are satisfied.

---

### Final Answer:

$$
\boxed{12}
$$

[Assistant]
Explanation:
Okay, so I need to find the smallest distance between the origin and a point on the graph of $ y = \frac{1}{\sqrt{2}}(x^2 - 3) $. The answer should be in the form $ \sqrt{a}/b $, where $ a $ and $ b $ are positive integers, and $ a $ is square-free. Then I have to find $ a + b $. Alright, let me think about how to approach this.

First, I remember that the distance between a point $ (x, y) $ and the origin $ (0, 0) $ is given by the distance formula: $ D = \sqrt{x^2 + y^2} $. Since we want the smallest distance, maybe I can minimize this distance function. However, dealing with square roots can be a bit messy, so maybe I can minimize the square of the distance instead, which should have its minimum at the same point. That might simplify calculations.

So, let me define $ D^2 = x^2 + y^2 $. Since $ y $ is given in terms of $ x $, I can substitute that into the equation. Let me do that:

$ D^2 = x^2 + \left( \frac{1}{\sqrt{2}}(x^2 - 3) \right)^2 $.

Let me simplify this expression. First, square the $ y $ term:

$ \left( \frac{1}{\sqrt{2}}(x^2 - 3) \right)^2 = \frac{1}{2}(x^2 - 3)^2 $.

So, substituting back into $ D^2 $:

$ D^2 = x^2 + \frac{1}{2}(x^2 - 3)^2 $.

Now, I need to find the minimum value of this function. To do that, I can take the derivative with respect to $ x $, set it equal to zero, and solve for $ x $. Then check if it's a minimum using the second derivative or some other method.

Let me compute the derivative. Let me denote $ f(x) = x^2 + \frac{1}{2}(x^2 - 3)^2 $.

First, expand $ (x^2 - 3)^2 $:

$ (x^2 - 3)^2 = x^4 - 6x^2 + 9 $.

Therefore, $ f(x) = x^2 + \frac{1}{2}(x^4 - 6x^2 + 9) $.

Let me simplify this:

$ f(x) = x^2 + \frac{1}{2}x^4 - 3x^2 + \frac{9}{2} $.

Combine like terms:

$ f(x) = \frac{1}{2}x^4 - 2x^2 + \frac{9}{2} $.

Now, take the derivative $ f'(x) $:

$ f'(x) = 2x^3 - 4x $.

Wait, let me check that again. The derivative of $ \frac{1}{2}x^4 $ is $ 2x^3 $, right? Because $ \frac{d}{dx} x^n = n x^{n-1} $, so $ \frac{1}{2} \times 4x^3 = 2x^3 $. Then the derivative of $ -2x^2 $ is $ -4x $. And the derivative of $ \frac{9}{2} $ is 0. So yes, $ f'(x) = 2x^3 - 4x $.

To find critical points, set $ f'(x) = 0 $:

$ 2x^3 - 4x = 0 $.

Factor out 2x:

$ 2x(x^2 - 2) = 0 $.

So, the critical points are when $ 2x = 0 $ or $ x^2 - 2 = 0 $.

Therefore, $ x = 0 $ or $ x = \sqrt{2} $ or $ x = -\sqrt{2} $.

Now, I need to check which of these critical points gives the minimum distance. Since the function $ f(x) $ is even? Wait, let me check. The original function $ y = \frac{1}{\sqrt{2}}(x^2 - 3) $ is even, because replacing $ x $ with $ -x $ doesn't change the value. Therefore, the graph is symmetric about the y-axis. Therefore, the distance from the origin should also be symmetric. Therefore, the minimum distance should occur at $ x = \sqrt{2} $ and $ x = -\sqrt{2} $, or maybe at $ x = 0 $. Let me check the value of $ f(x) $ at these points.

First, at $ x = 0 $:

$ f(0) = \frac{1}{2}(0)^4 - 2(0)^2 + \frac{9}{2} = 0 - 0 + \frac{9}{2} = \frac{9}{2} $.

So, the distance squared is $ 9/2 $, so the distance is $ \sqrt{9/2} = 3/\sqrt{2} $, which is approximately 2.121.

Now, let's check at $ x = \sqrt{2} $:

First compute $ x^2 = (\sqrt{2})^2 = 2 $.

So, $ f(\sqrt{2}) = \frac{1}{2}(2)^2 - 2(2) + \frac{9}{2} $.

Wait, hold on. Wait, earlier I had simplified $ f(x) $ as $ \frac{1}{2}x^4 - 2x^2 + \frac{9}{2} $. So, substituting $ x = \sqrt{2} $:

First, $ x^4 = (x^2)^2 = (2)^2 = 4 $.

Therefore, $ f(\sqrt{2}) = \frac{1}{2}(4) - 2(2) + \frac{9}{2} = 2 - 4 + 4.5 = 2.5 $.

Which is $ 5/2 $. Therefore, the distance squared is $ 5/2 $, so the distance is $ \sqrt{5/2} = \sqrt{10}/2 $. Wait, because $ \sqrt{5/2} = \sqrt{10}/2 $. Let me check that:

$ \sqrt{5/2} = \sqrt{5}/\sqrt{2} = (\sqrt{5} \times \sqrt{2}) / (\sqrt{2} \times \sqrt{2}) ) = \sqrt{10}/2 $. Yes, that's correct.

Similarly, at $ x = -\sqrt{2} $, the value will be the same because of the even function. So, the distance squared is 5/2, which is less than 9/2. Therefore, the minimum distance occurs at $ x = \pm \sqrt{2} $, and the minimal distance is $ \sqrt{10}/2 $.

Wait, but the problem says the answer can be expressed as $ \sqrt{a}/b $. So, $ \sqrt{10}/2 $ is already in that form, with $ a = 10 $ and $ b = 2 $. Then $ a + b = 12 $. But wait, let me check if 10 is square-free. 10 factors into 2 * 5, neither of which are squares, so yes, 10 is square-free. Therefore, the answer is 10 + 2 = 12. Is that correct?

Wait, but let me double-check my calculations because sometimes when you do calculus, you might make a mistake. Let me verify the steps again.

First, the distance squared is $ x^2 + y^2 $, with $ y = \frac{1}{\sqrt{2}}(x^2 - 3) $. So, substituting:

$ D^2 = x^2 + \frac{1}{2}(x^2 - 3)^2 $. Then expanding $ (x^2 - 3)^2 $ gives $ x^4 - 6x^2 + 9 $, so:

$ D^2 = x^2 + \frac{1}{2}x^4 - 3x^2 + \frac{9}{2} $, which simplifies to $ \frac{1}{2}x^4 - 2x^2 + \frac{9}{2} $. Then derivative is $ 2x^3 - 4x $, set to zero gives critical points at x=0, x=±√2. Then evaluating D² at x=√2:

x² = 2, x⁴ = 4. So D² = (1/2)(4) - 2*(2) + 9/2 = 2 - 4 + 4.5 = 2.5, which is 5/2. So D = sqrt(5/2) = sqrt(10)/2. So that seems correct.

But let me check if there's another way to approach this problem, maybe using geometry or something else, to confirm.

Alternatively, maybe using Lagrange multipliers? Let me try that.

We need to minimize the distance squared function $ f(x, y) = x^2 + y^2 $ subject to the constraint $ g(x, y) = y - \frac{1}{\sqrt{2}}(x^2 - 3) = 0 $.

The Lagrangian is $ \mathcal{L} = x^2 + y^2 - \lambda \left( y - \frac{1}{\sqrt{2}}(x^2 - 3) \right) $.

Taking partial derivatives:

∂L/∂x = 2x - λ*(-2x / sqrt(2)) = 0

Wait, let me compute that again. The constraint is $ g(x, y) = y - \frac{1}{\sqrt{2}}(x^2 - 3) = 0 $. Therefore, the partial derivatives of g:

∂g/∂x = - (2x)/sqrt(2) = - sqrt(2) x

∂g/∂y = 1

So, the Lagrangian equations are:

∂L/∂x = 2x - λ*(-sqrt(2) x) = 0

∂L/∂y = 2y - λ*(1) = 0

∂L/∂λ = - [ y - (1/sqrt(2))(x² - 3) ] = 0

So, equations:

1) 2x + λ*sqrt(2) x = 0

2) 2y - λ = 0

3) y = (1/sqrt(2))(x² - 3)

From equation 1: 2x + λ*sqrt(2) x = 0. Let's factor out x:

x(2 + λ*sqrt(2)) = 0.

Therefore, either x = 0 or 2 + λ*sqrt(2) = 0 => λ = -2 / sqrt(2) = -sqrt(2).

Case 1: x = 0.

Then from equation 3: y = (1/sqrt(2))(0 - 3) = -3 / sqrt(2). Then from equation 2: 2y - λ = 0 => λ = 2y = 2*(-3)/sqrt(2) = -6 / sqrt(2) = -3*sqrt(2). But from equation 1, if x=0, then equation 1 is satisfied regardless of λ. So that's consistent. So this gives a critical point at (0, -3/sqrt(2)).

Case 2: λ = -sqrt(2). Then from equation 2: 2y - (-sqrt(2)) = 0 => 2y + sqrt(2) = 0 => y = -sqrt(2)/2.

Then from equation 3: y = (1/sqrt(2))(x² - 3). So:

-sqrt(2)/2 = (1/sqrt(2))(x² - 3)

Multiply both sides by sqrt(2):

- (sqrt(2) * sqrt(2))/2 = x² - 3 => -2/2 = x² - 3 => -1 = x² - 3 => x² = 2 => x = sqrt(2) or -sqrt(2). So that gives the points (sqrt(2), -sqrt(2)/2) and (-sqrt(2), -sqrt(2)/2).

So, we have three critical points: (0, -3/sqrt(2)), (sqrt(2), -sqrt(2)/2), (-sqrt(2), -sqrt(2)/2). Now, let's compute the distance squared for each.

First, for (0, -3/sqrt(2)):

Distance squared is 0² + (-3/sqrt(2))² = 9/2, same as before.

For (sqrt(2), -sqrt(2)/2):

Distance squared is (sqrt(2))² + (-sqrt(2)/2)^2 = 2 + (2/4) = 2 + 1/2 = 5/2, same as before. So that's consistent with the previous result. Therefore, the minimal distance squared is 5/2, so minimal distance is sqrt(5/2) = sqrt(10)/2. Therefore, the answer is sqrt(10)/2, so a = 10, b = 2, so a + b = 12.

But wait, let me check if there's a possibility that there's a smaller distance. For example, maybe the minimal distance occurs somewhere else? But according to both calculus methods, we found critical points and checked them. Also, since the function tends to infinity as x approaches infinity, the minimal must be at one of the critical points. Therefore, I think this is correct.

But let me check with another approach. Maybe parametric equations or something else.

Alternatively, think geometrically. The graph is a parabola. The origin is a point outside the parabola. The minimal distance from the origin to the parabola would be along the line connecting the origin to the closest point on the parabola. So, the line connecting the origin to the closest point on the parabola should be perpendicular to the tangent of the parabola at that point.

So, maybe we can use that condition. Let me try that.

Let me denote the point on the parabola as (x, y) = (x, (1/sqrt(2))(x² - 3)). The slope of the tangent line at this point can be found by taking the derivative of y with respect to x.

Given y = (1/sqrt(2))(x² - 3), so dy/dx = (2x)/sqrt(2) = sqrt(2) x.

Therefore, the slope of the tangent line at (x, y) is m_tangent = sqrt(2) x.

The slope of the line connecting the origin to the point (x, y) is m_line = (y - 0)/(x - 0) = y/x = [ (1/sqrt(2))(x² - 3) ] / x = (x² - 3)/(x sqrt(2)).

Since the line connecting the origin to the point is perpendicular to the tangent line, their slopes should multiply to -1. Therefore:

m_line * m_tangent = -1.

So:

[ (x² - 3)/(x sqrt(2)) ] * [ sqrt(2) x ] = -1.

Let me compute that:

The sqrt(2) in the denominator and the sqrt(2) in the numerator cancel out, and the x in the denominator and the x in the numerator cancel out. So we have:

(x² - 3) * 1 = -1 => x² - 3 = -1 => x² = 2 => x = sqrt(2) or x = -sqrt(2). Which is exactly the same result as before! Therefore, this confirms that the critical points are at x = ±sqrt(2). Therefore, the minimal distance is indeed sqrt(10)/2, so a + b = 10 + 2 = 12.

But just to be thorough, let me check if there's any possible mistake in assuming that the minimal distance occurs at those points. For example, maybe the parabola is opening upwards, and the origin is below it? Let me sketch mentally. The equation is y = (1/sqrt(2))(x² - 3). So when x = 0, y = -3/sqrt(2) ≈ -2.12. As |x| increases, y increases. So the vertex of the parabola is at (0, -3/sqrt(2)), and it opens upwards. The origin is at (0, 0), which is above the vertex. Therefore, the closest point on the parabola to the origin is likely somewhere on the right or left side of the parabola, not at the vertex. Which matches our previous result where the minimal distance is achieved at x = ±sqrt(2), which are points above the vertex. So that makes sense.

Alternatively, if I plug in x = sqrt(2), then y = (1/sqrt(2))((sqrt(2))² - 3) = (1/sqrt(2))(2 - 3) = -1/sqrt(2). Therefore, the point is (sqrt(2), -1/sqrt(2)). The distance from the origin is sqrt( (sqrt(2))² + (-1/sqrt(2))² ) = sqrt(2 + 1/2) = sqrt(5/2) = sqrt(10)/2. Which is the same as before.

Alternatively, if I think about the point (sqrt(2), -1/sqrt(2)), is that the closest point? Let me check another point. For example, take x = 1. Then y = (1/sqrt(2))(1 - 3) = (-2)/sqrt(2) = -sqrt(2). The distance squared is 1 + (sqrt(2))² = 1 + 2 = 3, which is 3, which is larger than 5/2 = 2.5. So yes, larger. If I take x = 2, then y = (1/sqrt(2))(4 - 3) = 1/sqrt(2). The distance squared is 4 + (1/2) = 4.5, which is larger. If I take x = sqrt(2), distance squared is 2.5. If I take x approaching infinity, distance squared approaches infinity. So yes, seems like 2.5 is the minimum.

Alternatively, maybe check x = something else. Let me try x = 1.5. Then x² = 2.25. Then y = (1/sqrt(2))(2.25 - 3) = (-0.75)/sqrt(2). Then distance squared is (1.5)^2 + [ (-0.75)/sqrt(2) ]^2 = 2.25 + (0.5625)/2 = 2.25 + 0.28125 = 2.53125, which is approximately 2.53, which is larger than 2.5. So yes, 2.5 is smaller. Therefore, seems like our answer is correct.

Therefore, I think the answer is 10 + 2 = 12. However, just to make sure, let me check if the problem says "the smallest distance between the origin and a point on the graph". So, the minimal distance is sqrt(10)/2, which is equal to sqrt(10)/2. The problem says "expressed as sqrt(a)/b", so sqrt(10)/2 is already in that form. Since 10 is square-free (its prime factors are 2 and 5, neither of which are squared), so a = 10, b = 2, so a + b = 12. Therefore, the answer is 12.

But just to be 100% sure, let me check if there's an alternative expression. For example, sometimes problems might have different forms. Let me see:

sqrt(10)/2 is the same as sqrt(10)/2. If I rationalize something, but no, that's already simplified. So yes, that's the form. Therefore, the answer is 12.

**Final Answer**
\boxed{12}

To find the smallest distance between the origin and a point on the graph of the function
$$
y = \frac{1}{\sqrt{2}}(x^2 - 3),
$$
we begin by using the **distance formula** between a point $(x, y)$ and the origin $(0, 0)$:

$$
D = \sqrt{x^2 + y^2}.
$$

To simplify the minimization process, we minimize the **square of the distance**, which is:

$$
D^2 = x^2 + y^2.
$$

Substituting the given expression for $ y $, we get:

$$
D^2 = x^2 + \left( \frac{1}{\sqrt{2}}(x^2 - 3) \right)^2 = x^2 + \frac{1}{2}(x^2 - 3)^2.
$$

Expanding the square:

$$
(x^2 - 3)^2 = x^4 - 6x^2 + 9,
$$
so
$$
D^2 = x^2 + \frac{1}{2}(x^4 - 6x^2 + 9) = \frac{1}{2}x^4 - 2x^2 + \frac{9}{2}.
$$

Now, we find the critical points by taking the derivative of $ D^2 $ with respect to $ x $:

$$
\frac{d}{dx} \left( \frac{1}{2}x^4 - 2x^2 + \frac{9}{2} \right) = 2x^3 - 4x.
$$

Setting the derivative equal to zero:

$$
2x^3 - 4x = 0 \Rightarrow 2x(x^2 - 2) = 0.
$$

This gives the critical points:

$$
x = 0, \quad x = \pm \sqrt{2}.
$$

We now evaluate $ D^2 $ at these points:

- At $ x = 0 $:
  $$
  D^2 = \frac{1}{2}(0)^4 - 2(0)^2 + \frac{9}{2} = \frac{9}{2}.
  $$

- At $ x = \pm \sqrt{2} $:
  $$
  x^2 = 2, \quad x^4 = 4, \quad \text{so} \quad D^2 = \frac{1}{2}(4) - 2(2) + \frac{9}{2} = 2 - 4 + \frac{9}{2} = \frac{5}{2}.
  $$

Thus, the **minimum distance squared** is $ \frac{5}{2} $, and the **minimum distance** is:

$$
D = \sqrt{\frac{5}{2}} = \frac{\sqrt{10}}{2}.
$$

This is in the form $ \frac{\sqrt{a}}{b} $, where $ a = 10 $ and $ b = 2 $. Since 10 is square-free (its prime factorization is $ 2 \times 5 $), the conditions of the problem are satisfied.

---

### Final Answer:

$$
\boxed{12}
$$

Solution:
By the distance formula, we are trying to minimize $\sqrt{x^2+y^2}=\sqrt{x^2+(1/2)(x^4-6x^2+9)}$. In general, minimization problems like this require calculus, but one optimization method that sometimes works is to try to complete the square.  Pulling out a factor of $1/2$ from under the radical, we have \begin{align*}
\frac{1}{\sqrt{2}}\sqrt{2x^2+x^4-6x^2+9}&=\frac{1}{\sqrt{2}}\sqrt{(x^4-4x^2+4)+5} \\
&= \frac{1}{\sqrt{2}}\sqrt{(x^2-2)^2+5}.
\end{align*}This last expression is minimized when the square equals $0$, i.e. when $x=\sqrt{2}$. Then the distance is $\sqrt{5}/\sqrt{2}=\sqrt{10}/2$.  Hence the desired answer is $\boxed{12}$.


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 solve the equation f(x) = 0 and f(x) = 6, then add the solutions together.

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

f(x) = 3 - x if x ≤ 3,

and

f(x) = -x³ + 2x² + 3x if x > 3.

So, depending on whether x is less than or equal to 3 or greater than 3, I use different expressions. Therefore, when solving for f(x) = 0 and f(x) = 6, I need to check both cases for each equation.

Let me start with finding f⁻¹(0). That means I need to find all x such that f(x) = 0. Since the function is piecewise, I should check both pieces.

First, check the case where x ≤ 3. So, set 3 - x = 0. Solving for x:

3 - x = 0 ⇒ x = 3.

So, x = 3 is a solution here. Now, check the other case where x > 3. So, set -x³ + 2x² + 3x = 0. Let me write that equation:

-x³ + 2x² + 3x = 0.

Hmm, maybe I can factor this equation. Let me factor out an x first:

x(-x² + 2x + 3) = 0.

So, that gives me x = 0 or -x² + 2x + 3 = 0. But wait, since we are in the case where x > 3, x = 0 is not in that interval. So, let me solve -x² + 2x + 3 = 0. Multiply both sides by -1 to make it easier:

x² - 2x - 3 = 0.

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

But again, we are in the case where x > 3. So, x = 3 is not in this interval (since x > 3), and x = -1 is also not in this interval. Therefore, there are no solutions in the x > 3 case.

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

Wait, but let me check if x = 3 is included in the first case. Yes, because the first case is x ≤ 3, so x = 3 is included. So, that's correct.

Now, moving on to f⁻¹(6). So, need to find x such that f(x) = 6. Again, check both cases.

First, check x ≤ 3. So, set 3 - x = 6. Solving for x:

3 - x = 6 ⇒ -x = 6 - 3 ⇒ -x = 3 ⇒ x = -3.

So, x = -3 is a solution here. Now, check the other case where x > 3. So, set -x³ + 2x² + 3x = 6. Let me write that equation:

-x³ + 2x² + 3x = 6.

Let me rearrange it to bring all terms to one side:

-x³ + 2x² + 3x - 6 = 0.

Alternatively, 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 for x > 3.

Hmm, solving a cubic equation. Maybe I can factor this. Let me try rational root theorem. The possible rational roots are factors of 6 over factors of 1, so ±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.

Divide x³ - 2x² - 3x + 6 by (x - 2).

Using synthetic division:

Coefficients: 1 | -2 | -3 | 6

Write down 2 (the root):

Bring down the 1.

Multiply 1 by 2: 2. Add to next coefficient: -2 + 2 = 0.

Multiply 0 by 2: 0. Add to next coefficient: -3 + 0 = -3.

Multiply -3 by 2: -6. Add to last coefficient: 6 + (-6) = 0. Perfect, no remainder.

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, and x = -√3.

But remember, we are in the case where x > 3. So, check which of these roots are greater than 3.

x = 2 is not, x = √3 ≈ 1.732, also not, and x = -√3 ≈ -1.732, also not. Therefore, none of the roots of the cubic equation are in the x > 3 interval.

Wait, so does that mean there is no solution in the x > 3 case? But then, the only solution for f(x) = 6 is x = -3?

But let me check if that's correct. Let me verify f(-3). Since -3 ≤ 3, use the first case: f(-3) = 3 - (-3) = 3 + 3 = 6. Yes, that works.

But wait, is there another solution in the x > 3 case? Because maybe I made a mistake in solving the equation. Let me check again.

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

So, rearranged: -x³ + 2x² + 3x - 6 = 0.

I multiplied by -1 to get x³ - 2x² - 3x + 6 = 0. Then factored as (x - 2)(x² - 3) = 0.

So, the roots are x = 2, x = sqrt(3), x = -sqrt(3). All of which are less than 3. Therefore, there are no solutions in x > 3.

Therefore, the only solution for f(x) = 6 is x = -3. Therefore, f⁻¹(6) = -3.

Wait, but that would mean f⁻¹(0) + f⁻¹(6) = 3 + (-3) = 0? But that seems too straightforward. Let me check again.

Wait, maybe I made a mistake in solving for f(x) = 6. Let me check if there's another solution in the x > 3 case. Maybe I need to check if the cubic equation has another root?

Wait, the cubic equation was x³ - 2x² - 3x + 6 = 0. We found x = 2 is a root, and then factored it as (x - 2)(x² - 3). So, the quadratic factor is x² - 3, which gives roots sqrt(3) and -sqrt(3). So, no other roots. Therefore, indeed, there are no solutions for x > 3.

But let me check with another approach. Let me consider the function for x > 3: f(x) = -x³ + 2x² + 3x. Let me analyze its behavior.

As x approaches infinity, the -x³ term dominates, so f(x) tends to negative infinity. At x = 3, the function is defined by the second piece, but since x > 3, let me check the value at x = 3. Wait, actually, the function is defined as -x³ + 2x² + 3x for x > 3. So, at x approaching 3 from the right, what is the value? Let me compute f(3) using the first piece: 3 - 3 = 0. But for x approaching 3 from the right, f(x) approaches - (3)^3 + 2*(3)^2 + 3*(3) = -27 + 18 + 9 = 0. So, the function is continuous at x = 3.

So, the function for x > 3 starts at 0 when x approaches 3 from the right, and then decreases to negative infinity as x increases. Therefore, the function f(x) for x > 3 is decreasing from 0 to negative infinity. Therefore, it can only take values from negative infinity up to 0. Therefore, the equation f(x) = 6 would have no solution in x > 3, since the maximum value there is 0.

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

Therefore, f⁻¹(0) is 3, and f⁻¹(6) is -3, so their sum is 0.

Wait, but the problem says "Calculate f⁻¹(0) + f⁻¹(6)". So, 3 + (-3) = 0? That seems surprisingly simple. Let me check once more.

First, for f⁻¹(0): solving f(x) = 0.

Case 1: x ≤ 3: 3 - x = 0 ⇒ x = 3.

Case 2: x > 3: -x³ + 2x² + 3x = 0. As before, solutions x = 0, 3, -1. None of which are in x > 3. So, only x = 3.

For f⁻¹(6):

Case 1: x ≤ 3: 3 - x = 6 ⇒ x = -3.

Case 2: x > 3: f(x) = -x³ + 2x² + 3x. As discussed, this function for x > 3 is decreasing from 0 to -infty, so never reaches 6.

Therefore, only x = -3.

Therefore, adding them: 3 + (-3) = 0.

But maybe I need to check if there are multiple inverses? Wait, inverse functions are only defined if the original function is bijective. However, since the problem says "Calculate f⁻¹(0) + f⁻¹(6)", it's possible that there are multiple pre-images, but maybe the question assumes that there is only one?

Wait, for f⁻¹(0), we have only one pre-image, x = 3. For f⁻¹(6), only x = -3. So, adding them gives 0.

But maybe I made a mistake in assuming that f is invertible? Wait, the problem didn't specify that f is invertible, but it's asking for f⁻¹(0) and f⁻¹(6), so presumably, there is a unique inverse value for each.

Alternatively, maybe there are multiple solutions, but the question is expecting the sum of all solutions? But in that case, for f⁻¹(0), there is only one solution, and for f⁻¹(6), only one solution.

Alternatively, maybe I made a mistake in solving for f(x) = 6? Let me check with x = -3.

f(-3) = 3 - (-3) = 6. Correct.

Is there another x where f(x) = 6? Let me check for x > 3. Suppose I pick x = 4. Then f(4) = -64 + 32 + 12 = -64 + 44 = -20. Which is less than 6. How about x = 0? Wait, x = 0 is in the first case. f(0) = 3 - 0 = 3. Not 6.

How about x = 1? f(1) = 3 - 1 = 2. Not 6.

x = -1? f(-1) = 3 - (-1) = 4. Still not 6.

x = -2? f(-2) = 3 - (-2) = 5.

x = -3? 3 - (-3) = 6.

So, only x = -3.

Therefore, seems correct.

But maybe I need to check if there's another solution in the second case? Let me think. Suppose I graph the function. For x ≤ 3, it's a line decreasing from infinity to 0. For x > 3, it's a cubic function that starts at 0 when x approaches 3 from the right, then decreases. So, the function is decreasing on both intervals? Wait, for x ≤ 3, f(x) = 3 - x is decreasing. For x > 3, f(x) = -x³ + 2x² + 3x. Let me check its derivative.

f'(x) for x > 3 is derivative of -x³ + 2x² + 3x, which is -3x² + 4x + 3. Let me check if this derivative is always negative for x > 3.

Let me compute f'(x) = -3x² + 4x + 3. Let me check at x = 3: f'(3) = -27 + 12 + 3 = -12. Negative. For x > 3, since the derivative is a quadratic opening downward (coefficient -3), and since at x = 3 it's -12, and as x increases, the derivative becomes more negative. Therefore, the function is decreasing for all x > 3. Therefore, the function is decreasing on both intervals.

Therefore, the function is decreasing on (-infty, 3] and decreasing on (3, ∞). However, since at x = 3, the function is continuous (as we saw earlier, both sides approach 0). Therefore, the entire function is decreasing on (-infty, ∞). Therefore, it is one-to-one, hence invertible.

Therefore, there is exactly one pre-image for each y in the range of f.

Therefore, f⁻¹(0) is 3, and f⁻¹(6) is -3, so their sum is 0.

But let me check once more. If the function is strictly decreasing, then for each y, there is exactly one x. Therefore, f⁻¹(0) is 3, and f⁻¹(6) is -3.

But let me check if f(-3) = 6 and f(3) = 0. Since the function is decreasing, then as x increases from -infty to ∞, f(x) decreases from ∞ to -infty. Therefore, 6 is achieved at x = -3, and 0 is achieved at x = 3. Therefore, their sum is indeed 0.

Therefore, the answer is 0.

But wait, the problem says "Calculate f⁻¹(0) + f⁻¹(6)". So, 3 + (-3) = 0.

But maybe I need to check if there's another solution for f(x) = 6? For example, maybe there's another x > 3 where f(x) = 6? But according to the analysis, the function for x > 3 is decreasing from 0 to -infty, so it can't reach 6.

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

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

Let me rearrange it: -x³ + 2x² + 3x - 6 = 0.

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

Factored as (x - 2)(x² - 3) = 0.

So, roots at x = 2, sqrt(3), -sqrt(3).

But x > 3, so none of these roots are in that interval. Therefore, no solution.

Therefore, yes, only x = -3.

Therefore, the answer is 0.

But maybe the problem expects something else? Let me check once again.

Alternatively, maybe I made a mistake in the inverse function? For example, maybe f⁻¹(0) is not 3? Let me check f(3).

For x = 3, since x ≤ 3, f(3) = 3 - 3 = 0. Correct.

For f(-3), as before, 3 - (-3) = 6. Correct.

Therefore, the inverse function values are correct.

Therefore, the answer is 0.

But just to be thorough, let me check if there's another x in the second case where f(x) = 6. Suppose I try to solve -x³ + 2x² + 3x = 6. Let me plug in x = 4: -64 + 32 + 12 = -20 ≠ 6. x = 2: -8 + 8 + 6 = 6. Wait, x = 2. But x = 2 is not in the x > 3 case. So, x = 2 is in the first case.

Wait, but if x = 2, which is ≤ 3, then f(2) = 3 - 2 = 1 ≠ 6. Wait, but earlier when I solved the equation for x > 3, I found x = 2 as a root? Wait, no. Wait, when I solved the equation for x > 3, I had the equation -x³ + 2x² + 3x = 6, which led to x³ - 2x² - 3x + 6 = 0, which had roots x = 2, sqrt(3), -sqrt(3). But x = 2 is not in x > 3. However, if I plug x = 2 into the original equation for x > 3, which is -x³ + 2x² + 3x, I get -8 + 8 + 6 = 6. But x = 2 is not in the domain of the second case. Therefore, even though algebraically x = 2 is a solution to the equation, it's not in the domain of the second case. Therefore, it's not a valid solution.

Therefore, there are no solutions in the second case.

Therefore, the conclusion is correct.

Therefore, the answer is 0.

But just to make sure, let me check if there's a possibility that the function is not strictly decreasing? For example, maybe there's a local maximum or minimum in the second case?

Earlier, I calculated the derivative for x > 3: f'(x) = -3x² + 4x + 3. Let me check if this derivative ever becomes positive for x > 3.

Set -3x² + 4x + 3 > 0 ⇒ 3x² - 4x - 3 < 0.

Solving 3x² - 4x - 3 = 0.

Discriminant D = 16 + 36 = 52.

Solutions: [4 ± sqrt(52)] / 6 = [4 ± 2*sqrt(13)] / 6 = [2 ± sqrt(13)] / 3.

sqrt(13) ≈ 3.606, so the positive root is [2 + 3.606]/3 ≈ 5.606 / 3 ≈ 1.868.

The other root is negative.

Therefore, the inequality 3x² - 4x - 3 < 0 holds between the two roots. Since one root is ≈1.868 and the other is negative. Therefore, for x between ≈-0.6 and ≈1.868, the derivative is positive. But for x > 3, which is outside of that interval, the derivative is negative. Therefore, the function is decreasing for x > 3.

Therefore, the function is indeed decreasing on x > 3. Therefore, no local maxima or minima there.

Therefore, the function is strictly decreasing everywhere, so invertible.

Therefore, the answer is 0.

But just to make sure, let me check if there's a possible mistake in the problem statement? The function is defined as 3 - x for x ≤ 3, and -x³ + 2x² + 3x for x > 3.

Yes, that's what was given.

Alternatively, maybe I misread the function? Let me check again.

Original problem:

f(x) is defined as:

3 - x if x ≤ 3,

and

-x³ + 2x² + 3x if x > 3.

Yes.

Therefore, calculations are correct.

Therefore, the answer is 0.

But just to be 100% sure, let me check with another approach.

Suppose I consider the inverse function. Since the function is strictly decreasing, then for any y in the range of f, there is exactly one x.

The range of f:

For x ≤ 3, f(x) = 3 - x. As x approaches -infty, f(x) approaches infinity. At x = 3, f(x) = 0.

For x > 3, f(x) = -x³ + 2x² + 3x. As x approaches infinity, f(x) approaches -infty. At x approaching 3 from the right, f(x) approaches 0.

Therefore, the range of f is (-infty, ∞). Therefore, the function is surjective onto ℝ. Therefore, for any real number y, there is exactly one x such that f(x) = y.

Therefore, f⁻¹(0) is 3, and f⁻¹(6) is -3.

Therefore, their sum is 0.

Therefore, the answer is 0.

I think that's solid.

**Final Answer**
\boxed{0}

To solve the problem, we are given a piecewise-defined 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: Understanding the Inverse Function

The inverse function $ f^{-1}(y) $ gives the value of $ x $ such that $ f(x) = y $. Since the function is piecewise, we must consider both cases when solving for $ f(x) = y $.

---

### Step 2: Find $ f^{-1}(0) $

We solve $ f(x) = 0 $.

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

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

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

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

Factor:

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

Solve the quadratic:

$$
-x^2 + 2x + 3 = 0 \Rightarrow x^2 - 2x - 3 = 0 \Rightarrow (x - 3)(x + 1) = 0
$$

Solutions: $ x = 3 $, $ x = -1 $

But since we are in the case $ x > 3 $, none of these are valid.

✅ So, the only solution is $ x = 3 $

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

---

### Step 3: Find $ f^{-1}(6) $

We solve $ f(x) = 6 $.

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

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

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

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

Rearranging:

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

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:

$$
(x - 2)(x^2 - 3) = 0 \Rightarrow x = 2, \sqrt{3}, -\sqrt{3}
$$

But $ x > 3 $, so none of these are valid.

✅ So, the only solution is $ x = -3 $

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

---

### Step 4: Add the Results

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

---

### Final Answer

$$
\boxed{0}
$$

[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 solve the equation f(x) = 0 and f(x) = 6, then add the solutions together.

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

f(x) = 3 - x if x ≤ 3,

and

f(x) = -x³ + 2x² + 3x if x > 3.

So, depending on whether x is less than or equal to 3 or greater than 3, I use different expressions. Therefore, when solving for f(x) = 0 and f(x) = 6, I need to check both cases for each equation.

Let me start with finding f⁻¹(0). That means I need to find all x such that f(x) = 0. Since the function is piecewise, I should check both pieces.

First, check the case where x ≤ 3. So, set 3 - x = 0. Solving for x:

3 - x = 0 ⇒ x = 3.

So, x = 3 is a solution here. Now, check the other case where x > 3. So, set -x³ + 2x² + 3x = 0. Let me write that equation:

-x³ + 2x² + 3x = 0.

Hmm, maybe I can factor this equation. Let me factor out an x first:

x(-x² + 2x + 3) = 0.

So, that gives me x = 0 or -x² + 2x + 3 = 0. But wait, since we are in the case where x > 3, x = 0 is not in that interval. So, let me solve -x² + 2x + 3 = 0. Multiply both sides by -1 to make it easier:

x² - 2x - 3 = 0.

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

But again, we are in the case where x > 3. So, x = 3 is not in this interval (since x > 3), and x = -1 is also not in this interval. Therefore, there are no solutions in the x > 3 case.

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

Wait, but let me check if x = 3 is included in the first case. Yes, because the first case is x ≤ 3, so x = 3 is included. So, that's correct.

Now, moving on to f⁻¹(6). So, need to find x such that f(x) = 6. Again, check both cases.

First, check x ≤ 3. So, set 3 - x = 6. Solving for x:

3 - x = 6 ⇒ -x = 6 - 3 ⇒ -x = 3 ⇒ x = -3.

So, x = -3 is a solution here. Now, check the other case where x > 3. So, set -x³ + 2x² + 3x = 6. Let me write that equation:

-x³ + 2x² + 3x = 6.

Let me rearrange it to bring all terms to one side:

-x³ + 2x² + 3x - 6 = 0.

Alternatively, 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 for x > 3.

Hmm, solving a cubic equation. Maybe I can factor this. Let me try rational root theorem. The possible rational roots are factors of 6 over factors of 1, so ±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.

Divide x³ - 2x² - 3x + 6 by (x - 2).

Using synthetic division:

Coefficients: 1 | -2 | -3 | 6

Write down 2 (the root):

Bring down the 1.

Multiply 1 by 2: 2. Add to next coefficient: -2 + 2 = 0.

Multiply 0 by 2: 0. Add to next coefficient: -3 + 0 = -3.

Multiply -3 by 2: -6. Add to last coefficient: 6 + (-6) = 0. Perfect, no remainder.

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, and x = -√3.

But remember, we are in the case where x > 3. So, check which of these roots are greater than 3.

x = 2 is not, x = √3 ≈ 1.732, also not, and x = -√3 ≈ -1.732, also not. Therefore, none of the roots of the cubic equation are in the x > 3 interval.

Wait, so does that mean there is no solution in the x > 3 case? But then, the only solution for f(x) = 6 is x = -3?

But let me check if that's correct. Let me verify f(-3). Since -3 ≤ 3, use the first case: f(-3) = 3 - (-3) = 3 + 3 = 6. Yes, that works.

But wait, is there another solution in the x > 3 case? Because maybe I made a mistake in solving the equation. Let me check again.

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

So, rearranged: -x³ + 2x² + 3x - 6 = 0.

I multiplied by -1 to get x³ - 2x² - 3x + 6 = 0. Then factored as (x - 2)(x² - 3) = 0.

So, the roots are x = 2, x = sqrt(3), x = -sqrt(3). All of which are less than 3. Therefore, there are no solutions in x > 3.

Therefore, the only solution for f(x) = 6 is x = -3. Therefore, f⁻¹(6) = -3.

Wait, but that would mean f⁻¹(0) + f⁻¹(6) = 3 + (-3) = 0? But that seems too straightforward. Let me check again.

Wait, maybe I made a mistake in solving for f(x) = 6. Let me check if there's another solution in the x > 3 case. Maybe I need to check if the cubic equation has another root?

Wait, the cubic equation was x³ - 2x² - 3x + 6 = 0. We found x = 2 is a root, and then factored it as (x - 2)(x² - 3). So, the quadratic factor is x² - 3, which gives roots sqrt(3) and -sqrt(3). So, no other roots. Therefore, indeed, there are no solutions for x > 3.

But let me check with another approach. Let me consider the function for x > 3: f(x) = -x³ + 2x² + 3x. Let me analyze its behavior.

As x approaches infinity, the -x³ term dominates, so f(x) tends to negative infinity. At x = 3, the function is defined by the second piece, but since x > 3, let me check the value at x = 3. Wait, actually, the function is defined as -x³ + 2x² + 3x for x > 3. So, at x approaching 3 from the right, what is the value? Let me compute f(3) using the first piece: 3 - 3 = 0. But for x approaching 3 from the right, f(x) approaches - (3)^3 + 2*(3)^2 + 3*(3) = -27 + 18 + 9 = 0. So, the function is continuous at x = 3.

So, the function for x > 3 starts at 0 when x approaches 3 from the right, and then decreases to negative infinity as x increases. Therefore, the function f(x) for x > 3 is decreasing from 0 to negative infinity. Therefore, it can only take values from negative infinity up to 0. Therefore, the equation f(x) = 6 would have no solution in x > 3, since the maximum value there is 0.

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

Therefore, f⁻¹(0) is 3, and f⁻¹(6) is -3, so their sum is 0.

Wait, but the problem says "Calculate f⁻¹(0) + f⁻¹(6)". So, 3 + (-3) = 0? That seems surprisingly simple. Let me check once more.

First, for f⁻¹(0): solving f(x) = 0.

Case 1: x ≤ 3: 3 - x = 0 ⇒ x = 3.

Case 2: x > 3: -x³ + 2x² + 3x = 0. As before, solutions x = 0, 3, -1. None of which are in x > 3. So, only x = 3.

For f⁻¹(6):

Case 1: x ≤ 3: 3 - x = 6 ⇒ x = -3.

Case 2: x > 3: f(x) = -x³ + 2x² + 3x. As discussed, this function for x > 3 is decreasing from 0 to -infty, so never reaches 6.

Therefore, only x = -3.

Therefore, adding them: 3 + (-3) = 0.

But maybe I need to check if there are multiple inverses? Wait, inverse functions are only defined if the original function is bijective. However, since the problem says "Calculate f⁻¹(0) + f⁻¹(6)", it's possible that there are multiple pre-images, but maybe the question assumes that there is only one?

Wait, for f⁻¹(0), we have only one pre-image, x = 3. For f⁻¹(6), only x = -3. So, adding them gives 0.

But maybe I made a mistake in assuming that f is invertible? Wait, the problem didn't specify that f is invertible, but it's asking for f⁻¹(0) and f⁻¹(6), so presumably, there is a unique inverse value for each.

Alternatively, maybe there are multiple solutions, but the question is expecting the sum of all solutions? But in that case, for f⁻¹(0), there is only one solution, and for f⁻¹(6), only one solution.

Alternatively, maybe I made a mistake in solving for f(x) = 6? Let me check with x = -3.

f(-3) = 3 - (-3) = 6. Correct.

Is there another x where f(x) = 6? Let me check for x > 3. Suppose I pick x = 4. Then f(4) = -64 + 32 + 12 = -64 + 44 = -20. Which is less than 6. How about x = 0? Wait, x = 0 is in the first case. f(0) = 3 - 0 = 3. Not 6.

How about x = 1? f(1) = 3 - 1 = 2. Not 6.

x = -1? f(-1) = 3 - (-1) = 4. Still not 6.

x = -2? f(-2) = 3 - (-2) = 5.

x = -3? 3 - (-3) = 6.

So, only x = -3.

Therefore, seems correct.

But maybe I need to check if there's another solution in the second case? Let me think. Suppose I graph the function. For x ≤ 3, it's a line decreasing from infinity to 0. For x > 3, it's a cubic function that starts at 0 when x approaches 3 from the right, then decreases. So, the function is decreasing on both intervals? Wait, for x ≤ 3, f(x) = 3 - x is decreasing. For x > 3, f(x) = -x³ + 2x² + 3x. Let me check its derivative.

f'(x) for x > 3 is derivative of -x³ + 2x² + 3x, which is -3x² + 4x + 3. Let me check if this derivative is always negative for x > 3.

Let me compute f'(x) = -3x² + 4x + 3. Let me check at x = 3: f'(3) = -27 + 12 + 3 = -12. Negative. For x > 3, since the derivative is a quadratic opening downward (coefficient -3), and since at x = 3 it's -12, and as x increases, the derivative becomes more negative. Therefore, the function is decreasing for all x > 3. Therefore, the function is decreasing on both intervals.

Therefore, the function is decreasing on (-infty, 3] and decreasing on (3, ∞). However, since at x = 3, the function is continuous (as we saw earlier, both sides approach 0). Therefore, the entire function is decreasing on (-infty, ∞). Therefore, it is one-to-one, hence invertible.

Therefore, there is exactly one pre-image for each y in the range of f.

Therefore, f⁻¹(0) is 3, and f⁻¹(6) is -3, so their sum is 0.

But let me check once more. If the function is strictly decreasing, then for each y, there is exactly one x. Therefore, f⁻¹(0) is 3, and f⁻¹(6) is -3.

But let me check if f(-3) = 6 and f(3) = 0. Since the function is decreasing, then as x increases from -infty to ∞, f(x) decreases from ∞ to -infty. Therefore, 6 is achieved at x = -3, and 0 is achieved at x = 3. Therefore, their sum is indeed 0.

Therefore, the answer is 0.

But wait, the problem says "Calculate f⁻¹(0) + f⁻¹(6)". So, 3 + (-3) = 0.

But maybe I need to check if there's another solution for f(x) = 6? For example, maybe there's another x > 3 where f(x) = 6? But according to the analysis, the function for x > 3 is decreasing from 0 to -infty, so it can't reach 6.

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

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

Let me rearrange it: -x³ + 2x² + 3x - 6 = 0.

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

Factored as (x - 2)(x² - 3) = 0.

So, roots at x = 2, sqrt(3), -sqrt(3).

But x > 3, so none of these roots are in that interval. Therefore, no solution.

Therefore, yes, only x = -3.

Therefore, the answer is 0.

But maybe the problem expects something else? Let me check once again.

Alternatively, maybe I made a mistake in the inverse function? For example, maybe f⁻¹(0) is not 3? Let me check f(3).

For x = 3, since x ≤ 3, f(3) = 3 - 3 = 0. Correct.

For f(-3), as before, 3 - (-3) = 6. Correct.

Therefore, the inverse function values are correct.

Therefore, the answer is 0.

But just to be thorough, let me check if there's another x in the second case where f(x) = 6. Suppose I try to solve -x³ + 2x² + 3x = 6. Let me plug in x = 4: -64 + 32 + 12 = -20 ≠ 6. x = 2: -8 + 8 + 6 = 6. Wait, x = 2. But x = 2 is not in the x > 3 case. So, x = 2 is in the first case.

Wait, but if x = 2, which is ≤ 3, then f(2) = 3 - 2 = 1 ≠ 6. Wait, but earlier when I solved the equation for x > 3, I found x = 2 as a root? Wait, no. Wait, when I solved the equation for x > 3, I had the equation -x³ + 2x² + 3x = 6, which led to x³ - 2x² - 3x + 6 = 0, which had roots x = 2, sqrt(3), -sqrt(3). But x = 2 is not in x > 3. However, if I plug x = 2 into the original equation for x > 3, which is -x³ + 2x² + 3x, I get -8 + 8 + 6 = 6. But x = 2 is not in the domain of the second case. Therefore, even though algebraically x = 2 is a solution to the equation, it's not in the domain of the second case. Therefore, it's not a valid solution.

Therefore, there are no solutions in the second case.

Therefore, the conclusion is correct.

Therefore, the answer is 0.

But just to make sure, let me check if there's a possibility that the function is not strictly decreasing? For example, maybe there's a local maximum or minimum in the second case?

Earlier, I calculated the derivative for x > 3: f'(x) = -3x² + 4x + 3. Let me check if this derivative ever becomes positive for x > 3.

Set -3x² + 4x + 3 > 0 ⇒ 3x² - 4x - 3 < 0.

Solving 3x² - 4x - 3 = 0.

Discriminant D = 16 + 36 = 52.

Solutions: [4 ± sqrt(52)] / 6 = [4 ± 2*sqrt(13)] / 6 = [2 ± sqrt(13)] / 3.

sqrt(13) ≈ 3.606, so the positive root is [2 + 3.606]/3 ≈ 5.606 / 3 ≈ 1.868.

The other root is negative.

Therefore, the inequality 3x² - 4x - 3 < 0 holds between the two roots. Since one root is ≈1.868 and the other is negative. Therefore, for x between ≈-0.6 and ≈1.868, the derivative is positive. But for x > 3, which is outside of that interval, the derivative is negative. Therefore, the function is decreasing for x > 3.

Therefore, the function is indeed decreasing on x > 3. Therefore, no local maxima or minima there.

Therefore, the function is strictly decreasing everywhere, so invertible.

Therefore, the answer is 0.

But just to make sure, let me check if there's a possible mistake in the problem statement? The function is defined as 3 - x for x ≤ 3, and -x³ + 2x² + 3x for x > 3.

Yes, that's what was given.

Alternatively, maybe I misread the function? Let me check again.

Original problem:

f(x) is defined as:

3 - x if x ≤ 3,

and

-x³ + 2x² + 3x if x > 3.

Yes.

Therefore, calculations are correct.

Therefore, the answer is 0.

But just to be 100% sure, let me check with another approach.

Suppose I consider the inverse function. Since the function is strictly decreasing, then for any y in the range of f, there is exactly one x.

The range of f:

For x ≤ 3, f(x) = 3 - x. As x approaches -infty, f(x) approaches infinity. At x = 3, f(x) = 0.

For x > 3, f(x) = -x³ + 2x² + 3x. As x approaches infinity, f(x) approaches -infty. At x approaching 3 from the right, f(x) approaches 0.

Therefore, the range of f is (-infty, ∞). Therefore, the function is surjective onto ℝ. Therefore, for any real number y, there is exactly one x such that f(x) = y.

Therefore, f⁻¹(0) is 3, and f⁻¹(6) is -3.

Therefore, their sum is 0.

Therefore, the answer is 0.

I think that's solid.

**Final Answer**
\boxed{0}

To solve the problem, we are given a piecewise-defined 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: Understanding the Inverse Function

The inverse function $ f^{-1}(y) $ gives the value of $ x $ such that $ f(x) = y $. Since the function is piecewise, we must consider both cases when solving for $ f(x) = y $.

---

### Step 2: Find $ f^{-1}(0) $

We solve $ f(x) = 0 $.

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

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

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

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

Factor:

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

Solve the quadratic:

$$
-x^2 + 2x + 3 = 0 \Rightarrow x^2 - 2x - 3 = 0 \Rightarrow (x - 3)(x + 1) = 0
$$

Solutions: $ x = 3 $, $ x = -1 $

But since we are in the case $ x > 3 $, none of these are valid.

✅ So, the only solution is $ x = 3 $

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

---

### Step 3: Find $ f^{-1}(6) $

We solve $ f(x) = 6 $.

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

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

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

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

Rearranging:

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

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:

$$
(x - 2)(x^2 - 3) = 0 \Rightarrow x = 2, \sqrt{3}, -\sqrt{3}
$$

But $ x > 3 $, so none of these are valid.

✅ So, the only solution is $ x = -3 $

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

---

### Step 4: Add the Results

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

---

### Final Answer

$$
\boxed{0}
$$

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]