Explanation:
Okay, so I need to find the smallest distance between the origin and a point on the graph of y equals (1 over square root of 2) times (x squared minus 3). Then express that distance as sqrt(a)/b, where a and b are positive integers with a square-free, and then find a + b. Alright, let's start by recalling that the distance from a point (x, y) to the origin is given by the formula sqrt(x² + y²). Since the point (x, y) is on the graph of the given function, y is equal to (x² - 3)/sqrt(2). So maybe I can substitute that into the distance formula and then minimize the resulting function?

Let me write that down. The distance squared from the origin to a point (x, y) on the graph would be x² + y². If I substitute y, then the distance squared becomes x² + [(x² - 3)/sqrt(2)]². Since the square root is a monotonic function, minimizing the distance squared will also minimize the distance itself. That should simplify calculations, right? Let me check that.

Yes, because if I minimize distance squared, it's equivalent to minimizing distance. So let's compute that.

First, compute [(x² - 3)/sqrt(2)]². That's (x² - 3)² divided by 2. So the distance squared is x² + (x² - 3)² / 2. Let me write that out:

D(x) = x² + [(x² - 3)²]/2

Now, to find the minimum distance, I need to find the value of x that minimizes D(x). To do this, I can take the derivative of D(x) with respect to x, set it equal to zero, and solve for x. Then check if that critical point is indeed a minimum.

So, let's compute D'(x). Let's first expand the expression if needed, but maybe it's easier to differentiate term by term.

First term: d/dx [x²] = 2x.

Second term: d/dx [ (x² - 3)² / 2 ].

Let me compute that derivative. Let me denote u = x² - 3, so the second term is u² / 2. The derivative is (2u * u') / 2 = u * u'. So u' is 2x. Therefore, the derivative of the second term is (x² - 3)(2x). Therefore, putting it all together:

D'(x) = 2x + 2x(x² - 3)

Let me factor out 2x:

D'(x) = 2x [1 + (x² - 3)] = 2x [x² - 2]

Wait, let me check that again. Wait:

Wait, original derivative was 2x + 2x(x² - 3). Let me compute that again.

Wait, 2x + 2x(x² - 3). Let me factor out 2x:

Yes, 2x[1 + (x² - 3)] = 2x(x² - 2). Because 1 + x² - 3 is x² - 2. Correct.

So D'(x) = 2x(x² - 2). To find critical points, set D'(x) = 0.

So 2x(x² - 2) = 0. Therefore, the critical points are x = 0 or x² - 2 = 0 => x = sqrt(2) or x = -sqrt(2). So there are three critical points: 0, sqrt(2), -sqrt(2).

Now, I need to check which of these gives the minimum distance. Since the distance function is even? Wait, because if we replace x with -x, the distance squared remains the same. Because x² is same as (-x)^2, and y would be same as well because (x² - 3) is same for x and -x. So the graph is symmetric with respect to the y-axis. Therefore, the distance from the origin to points (x, y) and (-x, y) are same. Therefore, the minimum distance should occur at x = sqrt(2) or x = -sqrt(2), or maybe at x = 0. Let me check the value of D(x) at these critical points.

First, let's check x = 0:

At x = 0, y = (0 - 3)/sqrt(2) = -3/sqrt(2). So distance squared is 0² + (-3/sqrt(2))² = 9/2. So distance is sqrt(9/2) = 3/sqrt(2) = 3*sqrt(2)/2 ≈ 2.121.

Now, check at x = sqrt(2):

First, compute y: y = [(sqrt(2))² - 3]/sqrt(2) = [2 - 3]/sqrt(2) = (-1)/sqrt(2). So distance squared is (sqrt(2))² + [(-1)/sqrt(2)]² = 2 + 1/2 = 5/2. Therefore, distance is sqrt(5/2) = sqrt(10)/2 ≈ 1.581. That's smaller than previous value. Similarly for x = -sqrt(2), same result. So seems like the minimum distance occurs at x = sqrt(2) or -sqrt(2), giving distance sqrt(5/2). But let me make sure there's no other critical points or maybe check second derivative or something? Wait, since we have only three critical points, and we checked the ones at sqrt(2) and -sqrt(2) give smaller distance than x=0, so maybe that's the minimum. But just to be thorough, maybe check the second derivative at x = sqrt(2) to confirm it's a minimum.

Alternatively, since there are only three critical points, and distance tends to infinity as x approaches infinity, so the function D(x) has a minimum somewhere. Since D(x) is smooth, so the critical points we found must be the extrema. So since at x = sqrt(2) we get lower distance than at x=0, that is likely the minimum.

Therefore, the minimal distance is sqrt(5/2). But the problem says it can be expressed as sqrt(a)/b. Let me check.

sqrt(5/2) can be written as sqrt(5)/sqrt(2) which is rationalized as sqrt(10)/2. Because multiplying numerator and denominator by sqrt(2):

sqrt(5)/sqrt(2) = (sqrt(5)*sqrt(2))/ (sqrt(2)*sqrt(2)) ) = sqrt(10)/2. Therefore, the distance is sqrt(10)/2. So a is 10 and b is 2. Then a + b is 12. Wait, but let me check again. Because sqrt(10)/2 is indeed sqrt(10)/2. So a is 10, which is square-free (since 10 factors into 2*5, neither of which are squares), and b is 2. So a + b = 12. Therefore, the answer is 12? Wait, but let me check if I made a mistake here. Because maybe there is another point where the distance is smaller?

Wait, let me double-check my calculations. Let me compute D(x) at x = sqrt(2):

x = sqrt(2), so x² = 2. Then y = (2 - 3)/sqrt(2) = -1/sqrt(2). Then distance squared is x² + y² = 2 + (1/2) = 2.5 = 5/2. So distance is sqrt(5/2). Which is sqrt(10)/2. So sqrt(10)/2. So that is correct. Therefore, a is 10, b is 2, so a + b is 12. However, the problem says "the smallest distance... can be expressed as sqrt(a)/b", so sqrt(10)/2 is in that form, so 10 + 2 = 12. But I need to check if that is the correct answer. Wait, maybe I made a mistake when taking derivative?

Let me go back through the steps again.

Starting with D(x) = x² + [(x² - 3)^2]/2.

Taking derivative:

First term derivative is 2x. Second term derivative:

Let me compute derivative of (x² - 3)^2 / 2. Let me use chain rule. Let me think of it as 1/2*(x² -3)^2. The derivative is 1/2 * 2*(x² - 3)*(2x) = (x² - 3)*(2x). Which is same as 2x(x² - 3). Therefore, total derivative is 2x + 2x(x² - 3) = 2x[1 + x² - 3] = 2x(x² - 2). So that's correct. So critical points at x = 0, sqrt(2), -sqrt(2). So calculations are correct.

So then, if I compute D(x) at x = sqrt(2), which is 5/2. Therefore, sqrt(5/2) is the minimal distance. Which is sqrt(10)/2. So expressed as sqrt(a)/b, that would be sqrt(10)/2. Therefore, a = 10, b = 2, so a + b = 12. Wait, but let me check if maybe there is a mistake here. Because sometimes when you minimize distance squared, sometimes there are other points. Wait, but maybe I can also check with another method. For example, maybe using Lagrange multipliers? Or parametrize the curve and minimize.

Alternatively, maybe parametrize the distance squared function. Let me think again. Let me write D(x) = x² + [(x² - 3)^2]/2. Let me expand that:

First, expand (x² - 3)^2: x^4 - 6x² + 9. Then divide by 2: (x^4)/2 - 3x² + 9/2. Then add x²: total D(x) is (x^4)/2 - 3x² + 9/2 + x² = (x^4)/2 - 2x² + 9/2. So D(x) = (1/2)x^4 - 2x² + 9/2. Then taking derivative is 2x³ - 4x. Which is same as 2x(x² - 2). Which is same as before. So that's same derivative. So critical points are same. So calculations are correct.

Alternatively, maybe I can set derivative to zero and solve 2x(x² - 2) = 0. So x = 0, or x = sqrt(2), -sqrt(2). So same result. Therefore, that is correct.

So then, the minimal distance is sqrt(5/2) which is sqrt(10)/2. Therefore, answer is 10 + 2 = 12. However, I feel that maybe the answer is different. Wait, maybe I made a mistake in the calculation of D(x) at x = sqrt(2). Let me check again:

At x = sqrt(2):

x² is 2. Then y is (x² - 3)/sqrt(2) = (2 - 3)/sqrt(2) = (-1)/sqrt(2). Therefore, y² is 1/2. Then distance squared is x² + y² = 2 + 1/2 = 5/2. Therefore, distance is sqrt(5/2) which is sqrt(10)/2. So that is correct. Therefore, the expression sqrt(a)/b is sqrt(10)/2, so a is 10, b is 2. Therefore, a + b is 12. So maybe 12 is the answer? Wait, but let me check if the problem says "the smallest distance between the origin and a point on the graph". Maybe there is a closer point?

Alternatively, maybe I can visualize the graph of y = (x² - 3)/sqrt(2). It's a parabola opening upwards, with vertex at (0, -3/sqrt(2)). The origin is at (0, 0). So the closest point would be somewhere on the parabola. Since the parabola is opening upwards, and the vertex is below the origin. So maybe the closest point is where the line connecting the origin to the parabola is perpendicular to the tangent of the parabola at that point. That's another way to think about it. Maybe using calculus was the right approach, but let me try this method to verify.

So, the line from the origin to the point (x, y) on the parabola has slope y/x. The slope of the tangent line to the parabola at that point can be found by taking the derivative of y with respect to x. Let me compute that. The derivative dy/dx is derivative of (x² - 3)/sqrt(2) which is (2x)/sqrt(2) = sqrt(2)*x. Therefore, the slope of tangent line is sqrt(2)*x. For the line connecting the origin to the point (x, y) to be perpendicular to the tangent line, their slopes should multiply to -1. Therefore:

(y/x) * (sqrt(2)x) = -1 => y * sqrt(2) = -1 => y = -1/sqrt(2)

But y is also equal to (x² - 3)/sqrt(2). Therefore, set them equal:

(x² - 3)/sqrt(2) = -1/sqrt(2) => x² - 3 = -1 => x² = 2 => x = sqrt(2) or -sqrt(2). Which is the same result as before! Therefore, confirming that the closest points are at x = sqrt(2) and -sqrt(2). Therefore, the minimal distance is indeed sqrt(5/2) = sqrt(10)/2. Therefore, answer is 10 + 2 = 12.

Wait, but why does the problem specify that a is not divisible by the square of any integer greater than one? Well, 10 factors into 2 * 5, both primes, so it's square-free. Therefore, that's okay. Therefore, the answer is 12. But just to be 100% sure, let me check with another approach. Maybe parametric equations or something else?

Alternatively, let me think of the distance squared as a function D(x) = x² + [(x² - 3)/sqrt(2)]². Let me compute D(x):

D(x) = x² + (x^4 - 6x² + 9)/2 = (2x² + x^4 - 6x² + 9)/2 = (x^4 - 4x² + 9)/2. Therefore, D(x) = (x^4 - 4x² + 9)/2. To find its minimum, take derivative:

dD/dx = (4x³ - 8x)/2 = 2x³ - 4x = 2x(x² - 2). Which again gives critical points at x = 0, sqrt(2), -sqrt(2). So same as before. Therefore, calculations are consistent. Therefore, minimal distance squared is D(sqrt(2)) = [( (sqrt(2))² )^2 - 4*(sqrt(2))² + 9]/2 = [ (4 - 8 + 9)] / 2 = (5)/2. Therefore, same result. So distance is sqrt(5/2) = sqrt(10)/2. So answer is 12.

Alternatively, maybe I can check with specific numbers. Let me compute the distance at x=1. Let me see:

At x=1, y = (1 - 3)/sqrt(2) = (-2)/sqrt(2) = -sqrt(2). Then distance squared is 1 + (sqrt(2))² = 1 + 2 = 3. So distance is sqrt(3) ≈ 1.732, which is larger than sqrt(10)/2 ≈ 1.581. So yes, sqrt(10)/2 is smaller. What about at x = sqrt(2) ≈ 1.414? Then distance squared is 5/2 = 2.5, which is less than 3. So that is correct.

Alternatively, check x = 2. Then y = (4 - 3)/sqrt(2) = 1/sqrt(2). Then distance squared is 4 + (1/2) = 4.5, which is bigger. So yes, seems like the minimal distance is indeed at x = sqrt(2). So I think my answer is correct. Therefore, the answer is 12.

Wait, but let me check again the problem statement to make sure I didn't misread anything. The function is y = (1/sqrt(2))*(x² - 3). Yes. The distance is from origin. Yes. The answer is expressed as sqrt(a)/b, with a square-free. So sqrt(10)/2, which is already simplified, so a is 10 and b is 2, so 10 + 2 is 12. Therefore, I think 12 is correct.

But just to be thorough, let me check if sqrt(10)/2 is indeed the minimal distance. Suppose I think of the graph. The parabola y = (x² - 3)/sqrt(2). So when x is zero, y is -3/sqrt(2) ≈ -2.121. So the vertex is at (0, -2.121). The origin is at (0, 0). So the closest distance would be from the origin to this parabola. The closest point is where the line from the origin is perpendicular to the tangent. Which we already found. So sqrt(10)/2 is approximately 1.581, which is less than the distance from origin to vertex, which is sqrt(0² + (-3/sqrt(2))²) = 3/sqrt(2) ≈ 2.121. So yes, that makes sense. Therefore, 1.581 is smaller. Therefore, seems correct.

Alternatively, maybe I can use calculus to check the second derivative at x = sqrt(2) to confirm it's a minimum. Let's compute D''(x). From earlier, D'(x) = 2x(x² - 2). Therefore, D''(x) is derivative of 2x³ - 4x, which is 6x² - 4. At x = sqrt(2), D''(sqrt(2)) = 6*(sqrt(2))² - 4 = 6*2 - 4 = 12 - 4 = 8, which is positive. Therefore, the function is concave upward at x = sqrt(2), so that critical point is a local minimum. Similarly, at x = 0, D''(0) = 0 - 4 = -4, which is concave downward, so a local maximum. Therefore, that confirms that the minimum occurs at x = sqrt(2) and -sqrt(2). Therefore, the minimal distance is sqrt(10)/2. Therefore, answer is 12.

Therefore, I think the answer is \boxed{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 analyze this geometrically using the **distance formula** from the origin $(0, 0)$ to a point $(x, y)$ on the graph. This distance is given by
$$
d = \sqrt{x^2 + y^2}.
$$
Substituting the expression for $y$, we have:
$$
d = \sqrt{x^2 + \left(\frac{x^2 - 3}{\sqrt{2}}\right)^2}.
$$
Simplifying the squared terms, we get:
$$
d^2 = x^2 + \frac{(x^2 - 3)^2}{2}.
$$
To minimize the distance, it's sufficient to minimize $d^2$, which avoids unnecessary square roots during optimization. Let's define
$$
D(x) = x^2 + \frac{(x^2 - 3)^2}{2}.
$$

---

### Step 1: Find Critical Points

We differentiate $D(x)$:
$$
D(x) = x^2 + \frac{x^4 - 6x^2 + 9}{2}.
$$
$$
D(x) = \frac{x^4 - 4x^2 + 9}{2}.
$$
$$
D'(x) = \frac{4x^3 - 8x}{2} = 2x^3 - 4x = 2x(x^2 - 2).
$$
Setting $D'(x) = 0$, we get:
$$
x = 0, \quad x = \sqrt{2}, \quad x = -\sqrt{2}.
$$

---

### Step 2: Evaluate $d^2$ at Critical Points

Evaluate $D(x)$ at these critical points:

- At $x = 0$:
  $$
  D(0) = \frac{0^4 - 4(0)^2 + 9}{2} = \frac{9}{2} \Rightarrow d = \sqrt{\frac{9}{2}} = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}.
  $$

- At $x = \sqrt{2}$:
  $$
  D(\sqrt{2}) = \frac{(\sqrt{2})^4 - 4(\sqrt{2})^2 + 9}{2} = \frac{4 - 8 + 9}{2} = \frac{5}{2} \Rightarrow d = \sqrt{\frac{5}{2}}.
  $$

Now simplify the irrational expression:

$$
\sqrt{\frac{5}{2}} = \frac{\sqrt{5}}{\sqrt{2}} = \frac{\sqrt{10}}{2},
$$
by rationalizing the denominator. So the minimal distance is $\frac{\sqrt{10}}{2}$.

---

### Step 3: Express in the Form $\frac{\sqrt{a}}{b}$

We write $d = \frac{\sqrt{10}}{2}$, so:
- $a = 10$ (square-free since 10 = 2 × 5, no square factor > 1),
- $b = 2$.

Therefore,
$$
a + b = 10 + 2 = \boxed{12}.
$$

---

### 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 equals (1 over square root of 2) times (x squared minus 3). Then express that distance as sqrt(a)/b, where a and b are positive integers with a square-free, and then find a + b. Alright, let's start by recalling that the distance from a point (x, y) to the origin is given by the formula sqrt(x² + y²). Since the point (x, y) is on the graph of the given function, y is equal to (x² - 3)/sqrt(2). So maybe I can substitute that into the distance formula and then minimize the resulting function?

Let me write that down. The distance squared from the origin to a point (x, y) on the graph would be x² + y². If I substitute y, then the distance squared becomes x² + [(x² - 3)/sqrt(2)]². Since the square root is a monotonic function, minimizing the distance squared will also minimize the distance itself. That should simplify calculations, right? Let me check that.

Yes, because if I minimize distance squared, it's equivalent to minimizing distance. So let's compute that.

First, compute [(x² - 3)/sqrt(2)]². That's (x² - 3)² divided by 2. So the distance squared is x² + (x² - 3)² / 2. Let me write that out:

D(x) = x² + [(x² - 3)²]/2

Now, to find the minimum distance, I need to find the value of x that minimizes D(x). To do this, I can take the derivative of D(x) with respect to x, set it equal to zero, and solve for x. Then check if that critical point is indeed a minimum.

So, let's compute D'(x). Let's first expand the expression if needed, but maybe it's easier to differentiate term by term.

First term: d/dx [x²] = 2x.

Second term: d/dx [ (x² - 3)² / 2 ].

Let me compute that derivative. Let me denote u = x² - 3, so the second term is u² / 2. The derivative is (2u * u') / 2 = u * u'. So u' is 2x. Therefore, the derivative of the second term is (x² - 3)(2x). Therefore, putting it all together:

D'(x) = 2x + 2x(x² - 3)

Let me factor out 2x:

D'(x) = 2x [1 + (x² - 3)] = 2x [x² - 2]

Wait, let me check that again. Wait:

Wait, original derivative was 2x + 2x(x² - 3). Let me compute that again.

Wait, 2x + 2x(x² - 3). Let me factor out 2x:

Yes, 2x[1 + (x² - 3)] = 2x(x² - 2). Because 1 + x² - 3 is x² - 2. Correct.

So D'(x) = 2x(x² - 2). To find critical points, set D'(x) = 0.

So 2x(x² - 2) = 0. Therefore, the critical points are x = 0 or x² - 2 = 0 => x = sqrt(2) or x = -sqrt(2). So there are three critical points: 0, sqrt(2), -sqrt(2).

Now, I need to check which of these gives the minimum distance. Since the distance function is even? Wait, because if we replace x with -x, the distance squared remains the same. Because x² is same as (-x)^2, and y would be same as well because (x² - 3) is same for x and -x. So the graph is symmetric with respect to the y-axis. Therefore, the distance from the origin to points (x, y) and (-x, y) are same. Therefore, the minimum distance should occur at x = sqrt(2) or x = -sqrt(2), or maybe at x = 0. Let me check the value of D(x) at these critical points.

First, let's check x = 0:

At x = 0, y = (0 - 3)/sqrt(2) = -3/sqrt(2). So distance squared is 0² + (-3/sqrt(2))² = 9/2. So distance is sqrt(9/2) = 3/sqrt(2) = 3*sqrt(2)/2 ≈ 2.121.

Now, check at x = sqrt(2):

First, compute y: y = [(sqrt(2))² - 3]/sqrt(2) = [2 - 3]/sqrt(2) = (-1)/sqrt(2). So distance squared is (sqrt(2))² + [(-1)/sqrt(2)]² = 2 + 1/2 = 5/2. Therefore, distance is sqrt(5/2) = sqrt(10)/2 ≈ 1.581. That's smaller than previous value. Similarly for x = -sqrt(2), same result. So seems like the minimum distance occurs at x = sqrt(2) or -sqrt(2), giving distance sqrt(5/2). But let me make sure there's no other critical points or maybe check second derivative or something? Wait, since we have only three critical points, and we checked the ones at sqrt(2) and -sqrt(2) give smaller distance than x=0, so maybe that's the minimum. But just to be thorough, maybe check the second derivative at x = sqrt(2) to confirm it's a minimum.

Alternatively, since there are only three critical points, and distance tends to infinity as x approaches infinity, so the function D(x) has a minimum somewhere. Since D(x) is smooth, so the critical points we found must be the extrema. So since at x = sqrt(2) we get lower distance than at x=0, that is likely the minimum.

Therefore, the minimal distance is sqrt(5/2). But the problem says it can be expressed as sqrt(a)/b. Let me check.

sqrt(5/2) can be written as sqrt(5)/sqrt(2) which is rationalized as sqrt(10)/2. Because multiplying numerator and denominator by sqrt(2):

sqrt(5)/sqrt(2) = (sqrt(5)*sqrt(2))/ (sqrt(2)*sqrt(2)) ) = sqrt(10)/2. Therefore, the distance is sqrt(10)/2. So a is 10 and b is 2. Then a + b is 12. Wait, but let me check again. Because sqrt(10)/2 is indeed sqrt(10)/2. So a is 10, which is square-free (since 10 factors into 2*5, neither of which are squares), and b is 2. So a + b = 12. Therefore, the answer is 12? Wait, but let me check if I made a mistake here. Because maybe there is another point where the distance is smaller?

Wait, let me double-check my calculations. Let me compute D(x) at x = sqrt(2):

x = sqrt(2), so x² = 2. Then y = (2 - 3)/sqrt(2) = -1/sqrt(2). Then distance squared is x² + y² = 2 + (1/2) = 2.5 = 5/2. So distance is sqrt(5/2). Which is sqrt(10)/2. So sqrt(10)/2. So that is correct. Therefore, a is 10, b is 2, so a + b is 12. However, the problem says "the smallest distance... can be expressed as sqrt(a)/b", so sqrt(10)/2 is in that form, so 10 + 2 = 12. But I need to check if that is the correct answer. Wait, maybe I made a mistake when taking derivative?

Let me go back through the steps again.

Starting with D(x) = x² + [(x² - 3)^2]/2.

Taking derivative:

First term derivative is 2x. Second term derivative:

Let me compute derivative of (x² - 3)^2 / 2. Let me use chain rule. Let me think of it as 1/2*(x² -3)^2. The derivative is 1/2 * 2*(x² - 3)*(2x) = (x² - 3)*(2x). Which is same as 2x(x² - 3). Therefore, total derivative is 2x + 2x(x² - 3) = 2x[1 + x² - 3] = 2x(x² - 2). So that's correct. So critical points at x = 0, sqrt(2), -sqrt(2). So calculations are correct.

So then, if I compute D(x) at x = sqrt(2), which is 5/2. Therefore, sqrt(5/2) is the minimal distance. Which is sqrt(10)/2. So expressed as sqrt(a)/b, that would be sqrt(10)/2. Therefore, a = 10, b = 2, so a + b = 12. Wait, but let me check if maybe there is a mistake here. Because sometimes when you minimize distance squared, sometimes there are other points. Wait, but maybe I can also check with another method. For example, maybe using Lagrange multipliers? Or parametrize the curve and minimize.

Alternatively, maybe parametrize the distance squared function. Let me think again. Let me write D(x) = x² + [(x² - 3)^2]/2. Let me expand that:

First, expand (x² - 3)^2: x^4 - 6x² + 9. Then divide by 2: (x^4)/2 - 3x² + 9/2. Then add x²: total D(x) is (x^4)/2 - 3x² + 9/2 + x² = (x^4)/2 - 2x² + 9/2. So D(x) = (1/2)x^4 - 2x² + 9/2. Then taking derivative is 2x³ - 4x. Which is same as 2x(x² - 2). Which is same as before. So that's same derivative. So critical points are same. So calculations are correct.

Alternatively, maybe I can set derivative to zero and solve 2x(x² - 2) = 0. So x = 0, or x = sqrt(2), -sqrt(2). So same result. Therefore, that is correct.

So then, the minimal distance is sqrt(5/2) which is sqrt(10)/2. Therefore, answer is 10 + 2 = 12. However, I feel that maybe the answer is different. Wait, maybe I made a mistake in the calculation of D(x) at x = sqrt(2). Let me check again:

At x = sqrt(2):

x² is 2. Then y is (x² - 3)/sqrt(2) = (2 - 3)/sqrt(2) = (-1)/sqrt(2). Therefore, y² is 1/2. Then distance squared is x² + y² = 2 + 1/2 = 5/2. Therefore, distance is sqrt(5/2) which is sqrt(10)/2. So that is correct. Therefore, the expression sqrt(a)/b is sqrt(10)/2, so a is 10, b is 2. Therefore, a + b is 12. So maybe 12 is the answer? Wait, but let me check if the problem says "the smallest distance between the origin and a point on the graph". Maybe there is a closer point?

Alternatively, maybe I can visualize the graph of y = (x² - 3)/sqrt(2). It's a parabola opening upwards, with vertex at (0, -3/sqrt(2)). The origin is at (0, 0). So the closest point would be somewhere on the parabola. Since the parabola is opening upwards, and the vertex is below the origin. So maybe the closest point is where the line connecting the origin to the parabola is perpendicular to the tangent of the parabola at that point. That's another way to think about it. Maybe using calculus was the right approach, but let me try this method to verify.

So, the line from the origin to the point (x, y) on the parabola has slope y/x. The slope of the tangent line to the parabola at that point can be found by taking the derivative of y with respect to x. Let me compute that. The derivative dy/dx is derivative of (x² - 3)/sqrt(2) which is (2x)/sqrt(2) = sqrt(2)*x. Therefore, the slope of tangent line is sqrt(2)*x. For the line connecting the origin to the point (x, y) to be perpendicular to the tangent line, their slopes should multiply to -1. Therefore:

(y/x) * (sqrt(2)x) = -1 => y * sqrt(2) = -1 => y = -1/sqrt(2)

But y is also equal to (x² - 3)/sqrt(2). Therefore, set them equal:

(x² - 3)/sqrt(2) = -1/sqrt(2) => x² - 3 = -1 => x² = 2 => x = sqrt(2) or -sqrt(2). Which is the same result as before! Therefore, confirming that the closest points are at x = sqrt(2) and -sqrt(2). Therefore, the minimal distance is indeed sqrt(5/2) = sqrt(10)/2. Therefore, answer is 10 + 2 = 12.

Wait, but why does the problem specify that a is not divisible by the square of any integer greater than one? Well, 10 factors into 2 * 5, both primes, so it's square-free. Therefore, that's okay. Therefore, the answer is 12. But just to be 100% sure, let me check with another approach. Maybe parametric equations or something else?

Alternatively, let me think of the distance squared as a function D(x) = x² + [(x² - 3)/sqrt(2)]². Let me compute D(x):

D(x) = x² + (x^4 - 6x² + 9)/2 = (2x² + x^4 - 6x² + 9)/2 = (x^4 - 4x² + 9)/2. Therefore, D(x) = (x^4 - 4x² + 9)/2. To find its minimum, take derivative:

dD/dx = (4x³ - 8x)/2 = 2x³ - 4x = 2x(x² - 2). Which again gives critical points at x = 0, sqrt(2), -sqrt(2). So same as before. Therefore, calculations are consistent. Therefore, minimal distance squared is D(sqrt(2)) = [( (sqrt(2))² )^2 - 4*(sqrt(2))² + 9]/2 = [ (4 - 8 + 9)] / 2 = (5)/2. Therefore, same result. So distance is sqrt(5/2) = sqrt(10)/2. So answer is 12.

Alternatively, maybe I can check with specific numbers. Let me compute the distance at x=1. Let me see:

At x=1, y = (1 - 3)/sqrt(2) = (-2)/sqrt(2) = -sqrt(2). Then distance squared is 1 + (sqrt(2))² = 1 + 2 = 3. So distance is sqrt(3) ≈ 1.732, which is larger than sqrt(10)/2 ≈ 1.581. So yes, sqrt(10)/2 is smaller. What about at x = sqrt(2) ≈ 1.414? Then distance squared is 5/2 = 2.5, which is less than 3. So that is correct.

Alternatively, check x = 2. Then y = (4 - 3)/sqrt(2) = 1/sqrt(2). Then distance squared is 4 + (1/2) = 4.5, which is bigger. So yes, seems like the minimal distance is indeed at x = sqrt(2). So I think my answer is correct. Therefore, the answer is 12.

Wait, but let me check again the problem statement to make sure I didn't misread anything. The function is y = (1/sqrt(2))*(x² - 3). Yes. The distance is from origin. Yes. The answer is expressed as sqrt(a)/b, with a square-free. So sqrt(10)/2, which is already simplified, so a is 10 and b is 2, so 10 + 2 is 12. Therefore, I think 12 is correct.

But just to be thorough, let me check if sqrt(10)/2 is indeed the minimal distance. Suppose I think of the graph. The parabola y = (x² - 3)/sqrt(2). So when x is zero, y is -3/sqrt(2) ≈ -2.121. So the vertex is at (0, -2.121). The origin is at (0, 0). So the closest distance would be from the origin to this parabola. The closest point is where the line from the origin is perpendicular to the tangent. Which we already found. So sqrt(10)/2 is approximately 1.581, which is less than the distance from origin to vertex, which is sqrt(0² + (-3/sqrt(2))²) = 3/sqrt(2) ≈ 2.121. So yes, that makes sense. Therefore, 1.581 is smaller. Therefore, seems correct.

Alternatively, maybe I can use calculus to check the second derivative at x = sqrt(2) to confirm it's a minimum. Let's compute D''(x). From earlier, D'(x) = 2x(x² - 2). Therefore, D''(x) is derivative of 2x³ - 4x, which is 6x² - 4. At x = sqrt(2), D''(sqrt(2)) = 6*(sqrt(2))² - 4 = 6*2 - 4 = 12 - 4 = 8, which is positive. Therefore, the function is concave upward at x = sqrt(2), so that critical point is a local minimum. Similarly, at x = 0, D''(0) = 0 - 4 = -4, which is concave downward, so a local maximum. Therefore, that confirms that the minimum occurs at x = sqrt(2) and -sqrt(2). Therefore, the minimal distance is sqrt(10)/2. Therefore, answer is 12.

Therefore, I think the answer is \boxed{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 analyze this geometrically using the **distance formula** from the origin $(0, 0)$ to a point $(x, y)$ on the graph. This distance is given by
$$
d = \sqrt{x^2 + y^2}.
$$
Substituting the expression for $y$, we have:
$$
d = \sqrt{x^2 + \left(\frac{x^2 - 3}{\sqrt{2}}\right)^2}.
$$
Simplifying the squared terms, we get:
$$
d^2 = x^2 + \frac{(x^2 - 3)^2}{2}.
$$
To minimize the distance, it's sufficient to minimize $d^2$, which avoids unnecessary square roots during optimization. Let's define
$$
D(x) = x^2 + \frac{(x^2 - 3)^2}{2}.
$$

---

### Step 1: Find Critical Points

We differentiate $D(x)$:
$$
D(x) = x^2 + \frac{x^4 - 6x^2 + 9}{2}.
$$
$$
D(x) = \frac{x^4 - 4x^2 + 9}{2}.
$$
$$
D'(x) = \frac{4x^3 - 8x}{2} = 2x^3 - 4x = 2x(x^2 - 2).
$$
Setting $D'(x) = 0$, we get:
$$
x = 0, \quad x = \sqrt{2}, \quad x = -\sqrt{2}.
$$

---

### Step 2: Evaluate $d^2$ at Critical Points

Evaluate $D(x)$ at these critical points:

- At $x = 0$:
  $$
  D(0) = \frac{0^4 - 4(0)^2 + 9}{2} = \frac{9}{2} \Rightarrow d = \sqrt{\frac{9}{2}} = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}.
  $$

- At $x = \sqrt{2}$:
  $$
  D(\sqrt{2}) = \frac{(\sqrt{2})^4 - 4(\sqrt{2})^2 + 9}{2} = \frac{4 - 8 + 9}{2} = \frac{5}{2} \Rightarrow d = \sqrt{\frac{5}{2}}.
  $$

Now simplify the irrational expression:

$$
\sqrt{\frac{5}{2}} = \frac{\sqrt{5}}{\sqrt{2}} = \frac{\sqrt{10}}{2},
$$
by rationalizing the denominator. So the minimal distance is $\frac{\sqrt{10}}{2}$.

---

### Step 3: Express in the Form $\frac{\sqrt{a}}{b}$

We write $d = \frac{\sqrt{10}}{2}$, so:
- $a = 10$ (square-free since 10 = 2 × 5, no square factor > 1),
- $b = 2$.

Therefore,
$$
a + b = 10 + 2 = \boxed{12}.
$$

---

### 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 find all x-values where f(x) equals 0 and where f(x) equals 6, and then add those x-values together.

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

- For x ≤ 3, f(x) = 3 - x
- For x > 3, f(x) = -x³ + 2x² + 3x

So, to find f⁻¹(0), I need to solve the equation f(x) = 0. Similarly, for f⁻¹(6), I need to solve f(x) = 6. Let me tackle each one separately.

Starting with f⁻¹(0). Let's set f(x) = 0 and solve for x. Since the function is piecewise, I need to check both cases.

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

3 - x = 0
=> x = 3

So, x = 3 is a solution here. Now, check if this x is in the domain of this piece, which is x ≤ 3. Since 3 is equal to 3, it's included. So that's valid.

Case 2: x > 3. Then f(x) = -x³ + 2x² + 3x. Set this 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. Wait, but in this case, we are considering x > 3, so x = 0 is not in this domain. Let me solve -x² + 2x + 3 = 0. Let me multiply both sides by -1 to make it easier:

x² - 2x - 3 = 0

Now, factor or use quadratic formula. Let me try factoring. Looking for two numbers that multiply to -3 and add to -2. Hmm. Let me think: 1 and -3? 1 * (-3) = -3, and 1 + (-3) = -2. Yes! So, the equation factors as:

(x - 3)(x + 1) = 0

Therefore, the solutions are x = 3 and x = -1. But again, in this case, we are looking for x > 3. So x = 3 is not greater than 3, and x = -1 is not greater than 3. So, there are no solutions in this case.

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

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

Case 1: x ≤ 3. Then f(x) = 3 - x. Set this equal to 6:

3 - x = 6
=> -x = 6 - 3
=> -x = 3
=> x = -3

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

Case 2: x > 3. Then f(x) = -x³ + 2x² + 3x. Set this 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

Now, I need to solve this cubic equation. Let me try to factor it. Maybe there's a rational root. By Rational Root Theorem, possible rational roots are factors of 6 over factors of 1, so ±1, ±2, ±3, ±6.

Let me test x = 1: 1 - 2 - 3 + 6 = 2 ≠ 0

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

Therefore, (x - 2) is a factor. Let me perform polynomial division or use synthetic division to factor it out.

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

So the roots are x = 2, x = √3, and x = -√3.

But remember, in Case 2, we are looking for x > 3. Let's check these roots:

x = 2: Not greater than 3. Disregard.

x = √3 ≈ 1.732: Not greater than 3. Disregard.

x = -√3 ≈ -1.732: Not greater than 3. Disregard.

So, none of these roots satisfy x > 3. Wait, so that means there are no solutions in this case?

But wait, that seems odd. If so, then the only solution for f(x) = 6 is x = -3. But let me check if I did everything correctly.

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

I rearranged to -x³ + 2x² + 3x - 6 = 0, then multiplied by -1 to get x³ - 2x² - 3x + 6 = 0. Then tested x=2, which worked. Factored into (x - 2)(x² - 3). So, the roots are 2, sqrt(3), -sqrt(3). All less than or equal to 3. Therefore, indeed, there are no solutions in x > 3.

Wait, so does that mean that f(x) = 6 only has solution x = -3? But let me verify by plugging in some x > 3 into the second piece to see if maybe there's a solution there. Let's check for example x = 4:

f(4) = -64 + 32 + 12 = (-64 + 32) = -32 + 12 = -20. So f(4) = -20, which is less than 6. What about x approaching infinity? The leading term is -x³, so as x increases, f(x) tends to negative infinity. So maybe there's no solution for x > 3?

But then, if x approaches 3 from the right, let's compute f(3+) which would be approaching -27 + 18 + 9 = 0. So f(x) approaches 0 as x approaches 3 from the right. Wait, but f(3) from the left is 3 - 3 = 0. So the function is continuous at x = 3? Let me check:

Left limit as x approaches 3: 3 - 3 = 0.

Right limit as x approaches 3: -27 + 18 + 9 = 0. So yes, continuous at x = 3. Therefore, the function is continuous at x = 3, and for x > 3, the function is decreasing from 0 to negative infinity? Wait, but when x approaches 3 from the right, it's approaching 0, and then decreasing? So maybe the function is decreasing for x > 3?

Wait, let me check the derivative for the second piece to see its behavior. The derivative of f(x) for x > 3 is f'(x) = -3x² + 4x + 3. Let me check its sign. For example, at x = 3, f'(3) = -27 + 12 + 3 = -12. So negative. Let me check at x = 4, derivative is -48 + 16 + 3 = -29. Still negative. So derivative is negative for x > 3, meaning the function is decreasing on x > 3. Therefore, since at x = 3, the function value is 0, and it's decreasing for x > 3, then for x > 3, f(x) is less than 0. Hence, there are no solutions for f(x) = 6 in x > 3. Therefore, only solution is x = -3 for f(x) = 6.

Therefore, f⁻¹(6) = -3.

Wait, but hold on. Let me check another value for x > 3. For example, x = 0. Wait, but x must be greater than 3. How about x = 3. Let me check again. Wait, f(3) is 0, and as we go to higher x, the function decreases. So the maximum value of f(x) for x > 3 is 0, achieved at x = 3, and then it goes to negative infinity. Therefore, there's no x > 3 where f(x) = 6. Hence, only solution is x = -3.

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

Wait, but before I conclude, let me verify that there are no other solutions for f(x) = 0. For the first case, x ≤ 3, we found x = 3. But maybe there's another solution in that case? Let me check.

For x ≤ 3, f(x) = 3 - x. Setting equal to 0 gives x = 3. Is there another x? Let me solve 3 - x = 0, which is only x=3. So no other solutions there. So f⁻¹(0) = 3 is correct.

For f(x)=6, we found x=-3. Is there another solution?

In Case 1, x ≤ 3, solving 3 - x = 6 gives x = -3. Is that the only solution? Yes. Then in Case 2, there are no solutions, so only x = -3. So f⁻¹(6) = -3. Therefore, adding them gives 3 + (-3) = 0. So the answer is 0? Hmm.

But let me check again for f(x) = 6. Maybe I made a mistake in solving the cubic equation? Let me go back.

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

- x³ + 2x² + 3x - 6 = 0. Multiply by -1: x³ - 2x² - 3x + 6 = 0. Then factoring, we found that (x - 2)(x² - 3) = 0, so roots at 2, sqrt(3), -sqrt(3). But since x > 3, none of these roots work. So no solution. So that seems correct.

But let me try plugging x = -3 into the second piece. Wait, x = -3 is not in x > 3. But when solving for f(x) = 6 in Case 1, x ≤ 3, we found x = -3. So that's correct. So f(-3) = 3 - (-3) = 6. Yes, that works. So f(-3) = 6. So that's correct.

Therefore, f⁻¹(6) = -3. Then f⁻¹(0) = 3. Then 3 + (-3) = 0. Therefore, the answer is 0. But before I finish, let me check if there's another solution for f(x) = 0. For example, maybe in the second piece, even though we solved and found no solutions?

Wait, for f(x) = 0, in the second piece, we had the equation -x³ + 2x² + 3x = 0. Which we factored as x(-x² + 2x + 3) = 0, leading to x = 0, x = 3, x = -1. But x > 3, so none of these. So only x = 3 in the first piece. So that's correct.

Therefore, the answer is 0. However, just to be thorough, let me check another value for f(x) in case I made a mistake. Let me check f(3) = 3 - 3 = 0. Correct. f(-3) = 3 - (-3) = 6. Correct. And for f(x)=6, only x=-3. For f(x)=0, only x=3. Therefore, their sum is 0. Therefore, the answer is \boxed{0}.

Wait, but let me check if there's a possible mistake in assuming that inverse functions have only one value? Wait, inverse functions are only defined if the original function is bijective. However, the problem says "Calculate f⁻¹(0) + f⁻¹(6)". So, perhaps in some cases, there might be multiple inverses, but since the question says f inverse, not f inverse of something, maybe they are referring to the principal value? But in this case, both 0 and 6 have only one preimage. For 0, only x=3. For 6, only x=-3. So the sum is 0.

Alternatively, maybe there's an error in my thinking about the inverse function. Let me check.

But f(x) is a piecewise function. Let me check if the function is one-to-one. Because if it's not one-to-one, then the inverse isn't a function. However, since the problem is asking for f⁻¹(0) and f⁻¹(6), it's implied that those values exist and are unique. So, maybe the function is one-to-one?

Wait, let me check if f(x) is one-to-one. For the first piece, f(x) = 3 - x, which is a decreasing linear function. For x ≤ 3, it's decreasing. For the second piece, f(x) = -x³ + 2x² + 3x, which we saw earlier has derivative f’(x) = -3x² + 4x + 3. Let me check if this is always decreasing for x > 3. At x = 3, derivative is -27 + 12 + 3 = -12. As x increases beyond 3, since the derivative is a quadratic function opening downward (coefficient -3), so it's decreasing for x > something. Wait, actually, the derivative f’(x) = -3x² + 4x + 3. The derivative of this derivative is -6x + 4, which is negative for x > 4/6 = 2/3. Therefore, for x > 3, the derivative f’(x) is decreasing. Since at x = 3, it's -12, and it keeps decreasing as x increases. Therefore, for x > 3, the derivative is always negative, meaning the function is strictly decreasing on x > 3. So, on the entire domain, the function is decreasing on (-infty, 3], decreasing on (3, ∞), and continuous at x = 3. Therefore, the function is strictly decreasing overall. Therefore, it's one-to-one, so the inverse function exists and is well-defined. Therefore, each y-value has exactly one preimage. Therefore, f⁻¹(0) = 3 and f⁻¹(6) = -3. Hence their sum is 0. Therefore, the answer is 0.

But wait, let me verify with another example. Suppose someone thinks that maybe for f(x) = 6, there's another solution in the second piece. Let me check x = something else. For example, maybe there's a mistake in my factoring step?

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

Which is equivalent to -x³ + 2x² + 3x - 6 = 0. Let me try plugging x = 1: -1 + 2 + 3 -6 = -2 ≠ 0. x = 2: -8 + 8 + 6 -6 = 0. So yes, x=2 is a root. Then factoring as (x - 2)(quadratic). Then quadratic is x² - 3. So that's correct. Therefore, roots at 2, sqrt(3), -sqrt(3). All less than 3. Thus, no solution for x > 3.

Alternatively, maybe I can graph the function mentally. For x ≤ 3, it's a line decreasing from infinity to 0 at x = 3. For x > 3, it's a cubic decreasing from 0 to negative infinity. Therefore, the entire function is strictly decreasing. Therefore, 6 is achieved only once at x = -3, and 0 is achieved only once at x = 3. Therefore, their sum is 0.

Therefore, I think my answer is correct.

**Final Answer**
\boxed{0}

To solve the problem, we are given the 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) $, which means we must find all $ x $ such that $ f(x) = 0 $ and $ f(x) = 6 $, and then sum the corresponding values.

---

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

We search for values of $ x $ such that $ f(x) = 0 $.

**Case 1: $ x \leq 3 $**
$$
f(x) = 3 - x = 0 \Rightarrow x = 3
$$
This value lies in the domain $ x \leq 3 $, so it's valid.

**Case 2: $ x > 3 $**
$$
f(x) = -x^3 + 2x^2 + 3x = 0
$$
Factor the expression:
$$
x(-x^2 + 2x + 3) = 0 \Rightarrow x = 0 \text{ or } -x^2 + 2x + 3 = 0
$$
Solving the quadratic:
$$
-x^2 + 2x + 3 = 0 \Rightarrow x^2 - 2x - 3 = 0 \Rightarrow (x - 3)(x + 1) = 0
\Rightarrow x = 3, -1
$$
All these roots (0, 3, -1) are **not greater than 3**, so no solutions in this case.

**Conclusion:** The only solution to $ f(x) = 0 $ is $ x = 3 $, so:
$$
f^{-1}(0) = 3
$$

---

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

Now we find $ x $ such that $ f(x) = 6 $.

**Case 1: $ x \leq 3 $**
$$
f(x) = 3 - x = 6 \Rightarrow x = -3
$$
This value lies in the domain $ x \leq 3 $, so it's valid.

**Case 2: $ x > 3 $**
$$
f(x) = -x^3 + 2x^2 + 3x = 6 \Rightarrow -x^3 + 2x^2 + 3x - 6 = 0
$$
Multiply by $ -1 $:
$$
x^3 - 2x^2 - 3x + 6 = 0
$$
Use Rational Root Theorem: Try $ x = 2 $, plug in:
$$
2^3 - 2(2)^2 - 3(2) + 6 = 8 - 8 - 6 + 6 = 0 \Rightarrow \text{Valid root}
$$
So, $ x = 2 $ is a root. Factor:
$$
x^3 - 2x^2 - 3x + 6 = (x - 2)(x^2 - 3) = 0
\Rightarrow x = 2, \sqrt{3}, -\sqrt{3}
$$
None of these values are greater than 3, so no solutions in this case.

**Conclusion:** The only solution to $ f(x) = 6 $ is $ x = -3 $, so:
$$
f^{-1}(6) = -3
$$

---

### Final Step: Compute the Sum

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

[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-values where f(x) equals 0 and where f(x) equals 6, and then add those x-values together.

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

- For x ≤ 3, f(x) = 3 - x
- For x > 3, f(x) = -x³ + 2x² + 3x

So, to find f⁻¹(0), I need to solve the equation f(x) = 0. Similarly, for f⁻¹(6), I need to solve f(x) = 6. Let me tackle each one separately.

Starting with f⁻¹(0). Let's set f(x) = 0 and solve for x. Since the function is piecewise, I need to check both cases.

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

3 - x = 0
=> x = 3

So, x = 3 is a solution here. Now, check if this x is in the domain of this piece, which is x ≤ 3. Since 3 is equal to 3, it's included. So that's valid.

Case 2: x > 3. Then f(x) = -x³ + 2x² + 3x. Set this 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. Wait, but in this case, we are considering x > 3, so x = 0 is not in this domain. Let me solve -x² + 2x + 3 = 0. Let me multiply both sides by -1 to make it easier:

x² - 2x - 3 = 0

Now, factor or use quadratic formula. Let me try factoring. Looking for two numbers that multiply to -3 and add to -2. Hmm. Let me think: 1 and -3? 1 * (-3) = -3, and 1 + (-3) = -2. Yes! So, the equation factors as:

(x - 3)(x + 1) = 0

Therefore, the solutions are x = 3 and x = -1. But again, in this case, we are looking for x > 3. So x = 3 is not greater than 3, and x = -1 is not greater than 3. So, there are no solutions in this case.

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

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

Case 1: x ≤ 3. Then f(x) = 3 - x. Set this equal to 6:

3 - x = 6
=> -x = 6 - 3
=> -x = 3
=> x = -3

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

Case 2: x > 3. Then f(x) = -x³ + 2x² + 3x. Set this 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

Now, I need to solve this cubic equation. Let me try to factor it. Maybe there's a rational root. By Rational Root Theorem, possible rational roots are factors of 6 over factors of 1, so ±1, ±2, ±3, ±6.

Let me test x = 1: 1 - 2 - 3 + 6 = 2 ≠ 0

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

Therefore, (x - 2) is a factor. Let me perform polynomial division or use synthetic division to factor it out.

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

So the roots are x = 2, x = √3, and x = -√3.

But remember, in Case 2, we are looking for x > 3. Let's check these roots:

x = 2: Not greater than 3. Disregard.

x = √3 ≈ 1.732: Not greater than 3. Disregard.

x = -√3 ≈ -1.732: Not greater than 3. Disregard.

So, none of these roots satisfy x > 3. Wait, so that means there are no solutions in this case?

But wait, that seems odd. If so, then the only solution for f(x) = 6 is x = -3. But let me check if I did everything correctly.

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

I rearranged to -x³ + 2x² + 3x - 6 = 0, then multiplied by -1 to get x³ - 2x² - 3x + 6 = 0. Then tested x=2, which worked. Factored into (x - 2)(x² - 3). So, the roots are 2, sqrt(3), -sqrt(3). All less than or equal to 3. Therefore, indeed, there are no solutions in x > 3.

Wait, so does that mean that f(x) = 6 only has solution x = -3? But let me verify by plugging in some x > 3 into the second piece to see if maybe there's a solution there. Let's check for example x = 4:

f(4) = -64 + 32 + 12 = (-64 + 32) = -32 + 12 = -20. So f(4) = -20, which is less than 6. What about x approaching infinity? The leading term is -x³, so as x increases, f(x) tends to negative infinity. So maybe there's no solution for x > 3?

But then, if x approaches 3 from the right, let's compute f(3+) which would be approaching -27 + 18 + 9 = 0. So f(x) approaches 0 as x approaches 3 from the right. Wait, but f(3) from the left is 3 - 3 = 0. So the function is continuous at x = 3? Let me check:

Left limit as x approaches 3: 3 - 3 = 0.

Right limit as x approaches 3: -27 + 18 + 9 = 0. So yes, continuous at x = 3. Therefore, the function is continuous at x = 3, and for x > 3, the function is decreasing from 0 to negative infinity? Wait, but when x approaches 3 from the right, it's approaching 0, and then decreasing? So maybe the function is decreasing for x > 3?

Wait, let me check the derivative for the second piece to see its behavior. The derivative of f(x) for x > 3 is f'(x) = -3x² + 4x + 3. Let me check its sign. For example, at x = 3, f'(3) = -27 + 12 + 3 = -12. So negative. Let me check at x = 4, derivative is -48 + 16 + 3 = -29. Still negative. So derivative is negative for x > 3, meaning the function is decreasing on x > 3. Therefore, since at x = 3, the function value is 0, and it's decreasing for x > 3, then for x > 3, f(x) is less than 0. Hence, there are no solutions for f(x) = 6 in x > 3. Therefore, only solution is x = -3 for f(x) = 6.

Therefore, f⁻¹(6) = -3.

Wait, but hold on. Let me check another value for x > 3. For example, x = 0. Wait, but x must be greater than 3. How about x = 3. Let me check again. Wait, f(3) is 0, and as we go to higher x, the function decreases. So the maximum value of f(x) for x > 3 is 0, achieved at x = 3, and then it goes to negative infinity. Therefore, there's no x > 3 where f(x) = 6. Hence, only solution is x = -3.

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

Wait, but before I conclude, let me verify that there are no other solutions for f(x) = 0. For the first case, x ≤ 3, we found x = 3. But maybe there's another solution in that case? Let me check.

For x ≤ 3, f(x) = 3 - x. Setting equal to 0 gives x = 3. Is there another x? Let me solve 3 - x = 0, which is only x=3. So no other solutions there. So f⁻¹(0) = 3 is correct.

For f(x)=6, we found x=-3. Is there another solution?

In Case 1, x ≤ 3, solving 3 - x = 6 gives x = -3. Is that the only solution? Yes. Then in Case 2, there are no solutions, so only x = -3. So f⁻¹(6) = -3. Therefore, adding them gives 3 + (-3) = 0. So the answer is 0? Hmm.

But let me check again for f(x) = 6. Maybe I made a mistake in solving the cubic equation? Let me go back.

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

- x³ + 2x² + 3x - 6 = 0. Multiply by -1: x³ - 2x² - 3x + 6 = 0. Then factoring, we found that (x - 2)(x² - 3) = 0, so roots at 2, sqrt(3), -sqrt(3). But since x > 3, none of these roots work. So no solution. So that seems correct.

But let me try plugging x = -3 into the second piece. Wait, x = -3 is not in x > 3. But when solving for f(x) = 6 in Case 1, x ≤ 3, we found x = -3. So that's correct. So f(-3) = 3 - (-3) = 6. Yes, that works. So f(-3) = 6. So that's correct.

Therefore, f⁻¹(6) = -3. Then f⁻¹(0) = 3. Then 3 + (-3) = 0. Therefore, the answer is 0. But before I finish, let me check if there's another solution for f(x) = 0. For example, maybe in the second piece, even though we solved and found no solutions?

Wait, for f(x) = 0, in the second piece, we had the equation -x³ + 2x² + 3x = 0. Which we factored as x(-x² + 2x + 3) = 0, leading to x = 0, x = 3, x = -1. But x > 3, so none of these. So only x = 3 in the first piece. So that's correct.

Therefore, the answer is 0. However, just to be thorough, let me check another value for f(x) in case I made a mistake. Let me check f(3) = 3 - 3 = 0. Correct. f(-3) = 3 - (-3) = 6. Correct. And for f(x)=6, only x=-3. For f(x)=0, only x=3. Therefore, their sum is 0. Therefore, the answer is \boxed{0}.

Wait, but let me check if there's a possible mistake in assuming that inverse functions have only one value? Wait, inverse functions are only defined if the original function is bijective. However, the problem says "Calculate f⁻¹(0) + f⁻¹(6)". So, perhaps in some cases, there might be multiple inverses, but since the question says f inverse, not f inverse of something, maybe they are referring to the principal value? But in this case, both 0 and 6 have only one preimage. For 0, only x=3. For 6, only x=-3. So the sum is 0.

Alternatively, maybe there's an error in my thinking about the inverse function. Let me check.

But f(x) is a piecewise function. Let me check if the function is one-to-one. Because if it's not one-to-one, then the inverse isn't a function. However, since the problem is asking for f⁻¹(0) and f⁻¹(6), it's implied that those values exist and are unique. So, maybe the function is one-to-one?

Wait, let me check if f(x) is one-to-one. For the first piece, f(x) = 3 - x, which is a decreasing linear function. For x ≤ 3, it's decreasing. For the second piece, f(x) = -x³ + 2x² + 3x, which we saw earlier has derivative f’(x) = -3x² + 4x + 3. Let me check if this is always decreasing for x > 3. At x = 3, derivative is -27 + 12 + 3 = -12. As x increases beyond 3, since the derivative is a quadratic function opening downward (coefficient -3), so it's decreasing for x > something. Wait, actually, the derivative f’(x) = -3x² + 4x + 3. The derivative of this derivative is -6x + 4, which is negative for x > 4/6 = 2/3. Therefore, for x > 3, the derivative f’(x) is decreasing. Since at x = 3, it's -12, and it keeps decreasing as x increases. Therefore, for x > 3, the derivative is always negative, meaning the function is strictly decreasing on x > 3. So, on the entire domain, the function is decreasing on (-infty, 3], decreasing on (3, ∞), and continuous at x = 3. Therefore, the function is strictly decreasing overall. Therefore, it's one-to-one, so the inverse function exists and is well-defined. Therefore, each y-value has exactly one preimage. Therefore, f⁻¹(0) = 3 and f⁻¹(6) = -3. Hence their sum is 0. Therefore, the answer is 0.

But wait, let me verify with another example. Suppose someone thinks that maybe for f(x) = 6, there's another solution in the second piece. Let me check x = something else. For example, maybe there's a mistake in my factoring step?

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

Which is equivalent to -x³ + 2x² + 3x - 6 = 0. Let me try plugging x = 1: -1 + 2 + 3 -6 = -2 ≠ 0. x = 2: -8 + 8 + 6 -6 = 0. So yes, x=2 is a root. Then factoring as (x - 2)(quadratic). Then quadratic is x² - 3. So that's correct. Therefore, roots at 2, sqrt(3), -sqrt(3). All less than 3. Thus, no solution for x > 3.

Alternatively, maybe I can graph the function mentally. For x ≤ 3, it's a line decreasing from infinity to 0 at x = 3. For x > 3, it's a cubic decreasing from 0 to negative infinity. Therefore, the entire function is strictly decreasing. Therefore, 6 is achieved only once at x = -3, and 0 is achieved only once at x = 3. Therefore, their sum is 0.

Therefore, I think my answer is correct.

**Final Answer**
\boxed{0}

To solve the problem, we are given the 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) $, which means we must find all $ x $ such that $ f(x) = 0 $ and $ f(x) = 6 $, and then sum the corresponding values.

---

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

We search for values of $ x $ such that $ f(x) = 0 $.

**Case 1: $ x \leq 3 $**
$$
f(x) = 3 - x = 0 \Rightarrow x = 3
$$
This value lies in the domain $ x \leq 3 $, so it's valid.

**Case 2: $ x > 3 $**
$$
f(x) = -x^3 + 2x^2 + 3x = 0
$$
Factor the expression:
$$
x(-x^2 + 2x + 3) = 0 \Rightarrow x = 0 \text{ or } -x^2 + 2x + 3 = 0
$$
Solving the quadratic:
$$
-x^2 + 2x + 3 = 0 \Rightarrow x^2 - 2x - 3 = 0 \Rightarrow (x - 3)(x + 1) = 0
\Rightarrow x = 3, -1
$$
All these roots (0, 3, -1) are **not greater than 3**, so no solutions in this case.

**Conclusion:** The only solution to $ f(x) = 0 $ is $ x = 3 $, so:
$$
f^{-1}(0) = 3
$$

---

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

Now we find $ x $ such that $ f(x) = 6 $.

**Case 1: $ x \leq 3 $**
$$
f(x) = 3 - x = 6 \Rightarrow x = -3
$$
This value lies in the domain $ x \leq 3 $, so it's valid.

**Case 2: $ x > 3 $**
$$
f(x) = -x^3 + 2x^2 + 3x = 6 \Rightarrow -x^3 + 2x^2 + 3x - 6 = 0
$$
Multiply by $ -1 $:
$$
x^3 - 2x^2 - 3x + 6 = 0
$$
Use Rational Root Theorem: Try $ x = 2 $, plug in:
$$
2^3 - 2(2)^2 - 3(2) + 6 = 8 - 8 - 6 + 6 = 0 \Rightarrow \text{Valid root}
$$
So, $ x = 2 $ is a root. Factor:
$$
x^3 - 2x^2 - 3x + 6 = (x - 2)(x^2 - 3) = 0
\Rightarrow x = 2, \sqrt{3}, -\sqrt{3}
$$
None of these values are greater than 3, so no solutions in this case.

**Conclusion:** The only solution to $ f(x) = 6 $ is $ x = -3 $, so:
$$
f^{-1}(6) = -3
$$

---

### Final Step: Compute the Sum

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

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]