Question: The smallest distance between the origin and a point on the graph of $y=\frac{1}{\sqrt{2}}\left(x^2-3\right)$ can be expressed as $\sqrt{a}/b$, where $a$ and $b$ are positive integers such that $a$ is not divisible by the square of any integer greater than one.  Find $a+b$.

To find the smallest distance between the origin and a point on the graph of \( y = \frac{1}{\sqrt{2}}(x^2 - 3) \), we need to minimize the distance function \( d(x) \) from the origin \((0,0)\) to a point \((x, y)\) on the curve. The distance \( d(x) \) is given by:

\[
d(x) = \sqrt{x^2 + y^2}
\]

Given \( y = \frac{1}{\sqrt{2}}(x^2 - 3) \), substitute \( y \) into the distance formula:

\[
d(x) = \sqrt{x^2 + \left(\frac{1}{\sqrt{2}}(x^2 - 3)\right)^2}
\]

Simplify the expression inside the square root:

\[
d(x) = \sqrt{x^2 + \frac{1}{2}(x^2 - 3)^2}
\]

Expand \( (x^2 - 3)^2 \):

\[
(x^2 - 3)^2 = x^4 - 6x^2 + 9
\]

Thus,

\[
d(x) = \sqrt{x^2 + \frac{1}{2}(x^4 - 6x^2 + 9)}
\]

Distribute \( \frac{1}{2} \):

\[
d(x) = \sqrt{x^2 + \frac{1}{2}x^4 - 3x^2 + \frac{9}{2}}
\]

Combine like terms:

\[
d(x) = \sqrt{\frac{1}{2}x^4 - 2x^2 + \frac{9}{2}}
\]

To minimize \( d(x) \), it is equivalent to minimize \( d(x)^2 \):

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

Find the critical points by taking the derivative of \( f(x) \) and setting it to zero:

\[
f'(x) = \frac{d}{dx}\left(\frac{1}{2}x^4 - 2x^2 + \frac{9}{2}\right)
\]

Compute the derivative:

\[
f'(x) = 2x^3 - 4x
\]

Set the derivative equal to zero:

\[
2x^3 - 4x = 0
\]

Factor out \( 2x \):

\[
2x(x^2 - 2) = 0
\]

Solve for \( x \):

\[
x = 0 \quad \text{or} \quad x^2 = 2 \quad \Rightarrow \quad x = \pm \sqrt{2}
\]

Evaluate \( f(x) \) at these critical points:

1. For \( x = 0 \):

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

2. For \( x = \sqrt{2} \):

\[
f(\sqrt{2}) = \frac{1}{2}(\sqrt{2})^4 - 2(\sqrt{2})^2 + \frac{9}{2} = \frac{1}{2}(4) - 2(2) + \frac{9}{2} = 2 - 4 + \frac{9}{2} = -2 + \frac{9}{2} = \frac{-4 + 9}{2} = \frac{5}{2}
\]

3. For \( x = -\sqrt{2} \):

\[
f(-\sqrt{2}) = \frac{1}{2}(-\sqrt{2})^4 - 2(-\sqrt{2})^2 + \frac{9}{2} = \frac{1}{2}(4) - 2(2) + \frac{9}{2} = 2 - 4 + \frac{9}{2} = -2 + \frac{9}{2} = \frac{-4 + 9}{2} = \frac{5}{2}
\]

The minimum value of \( f(x) \) is \( \frac{5}{2} \), which occurs at \( x = \pm \sqrt{2} \).

Thus, the minimum distance \( d(x) \) is:

\[
d(x) = \sqrt{\frac{5}{2}} = \frac{\sqrt{10}}{2}
\]

Express this in the form \( \sqrt{a}/b \):

\[
\frac{\sqrt{10}}{2} = \sqrt{10}/2
\]

Here, \( a = 10 \) and \( b = 2 \). Therefore, \( a + b = 10 + 2 = 12 \).

\boxed{12}


Question:
Let $f$ be defined by  \[f(x) = \left\{
\begin{array}{cl}
3-x & \text{ if } x \leq 3, \\
-x^3+2x^2+3x & \text{ if } x>3.
\end{array}
\right.\]Calculate $f^{-1}(0)+f^{-1}(6)$.

To solve for \( f^{-1}(0) + f^{-1}(6) \), we need to find the values of \( x \) such that \( f(x) = 0 \) and \( f(x) = 6 \), and then sum these values.

First, let's find \( f^{-1}(0) \). This means we need to solve \( f(x) = 0 \).

1. For \( x \leq 3 \), \( f(x) = 3 - x \). Set \( 3 - x = 0 \):
   \[
   3 - x = 0 \implies x = 3.
   \]
   Since \( x = 3 \) is in the domain \( x \leq 3 \), \( x = 3 \) is a solution.

2. For \( x > 3 \), \( f(x) = -x^3 + 2x^2 + 3x \). Set \( -x^3 + 2x^2 + 3x = 0 \):
   \[
   -x(x^2 - 2x - 3) = 0 \implies -x(x - 3)(x + 1) = 0.
   \]
   The solutions are \( x = 0 \), \( x = 3 \), and \( x = -1 \). However, since we are considering \( x > 3 \), none of these solutions are valid in this domain. Therefore, there are no solutions in this domain.

Thus, the only solution to \( f(x) = 0 \) is \( x = 3 \). Therefore, \( f^{-1}(0) = 3 \).

Next, let's find \( f^{-1}(6) \). This means we need to solve \( f(x) = 6 \).

1. For \( x \leq 3 \), \( f(x) = 3 - x \). Set \( 3 - x = 6 \):
   \[
   3 - x = 6 \implies x = -3.
   \]
   Since \( x = -3 \) is in the domain \( x \leq 3 \), \( x = -3 \) is a solution.

2. For \( x > 3 \), \( f(x) = -x^3 + 2x^2 + 3x \). Set \( -x^3 + 2x^2 + 3x = 6 \):
   \[
   -x^3 + 2x^2 + 3x - 6 = 0.
   \]
   We can try to find rational roots using the Rational Root Theorem. The possible rational roots are \( \pm 1, \pm 2, \pm 3, \pm 6 \). Testing \( x = 2 \):
   \[
   -2^3 + 2 \cdot 2^2 + 3 \cdot 2 - 6 = -8 + 8 + 6 - 6 = 0.
   \]
   So, \( x = 2 \) is a root. However, since we are considering \( x > 3 \), \( x = 2 \) is not in the domain. Testing \( x = 3 \):
   \[
   -3^3 + 2 \cdot 3^2 + 3 \cdot 3 - 6 = -27 + 18 + 9 - 6 = -6 \neq 0.
   \]
   Testing \( x = 4 \):
   \[
   -4^3 + 2 \cdot 4^2 + 3 \cdot 4 - 6 = -64 + 32 + 12 - 6 = -26 \neq 0.
   \]
   Testing \( x = 6 \):
   \[
   -6^3 + 2 \cdot 6^2 + 3 \cdot 6 - 6 = -216 + 72 + 18 - 6 = -132 \neq 0.
   \]
   It seems there are no rational roots in the domain \( x > 3 \). To confirm, we can check the derivative to see if there are any real roots. The derivative of \( -x^3 + 2x^2 + 3x \) is \( -3x^2 + 4x + 3 \). The discriminant of the quadratic \( -3x^2 + 4x + 3 \) is \( 4^2 - 4(-3)(3) = 16 + 36 = 52 \), which is positive, so the derivative has two real roots. However, we need to check the behavior of the function. The function \( -x^3 + 2x^2 + 3x \) is a cubic polynomial with a negative leading coefficient, so it decreases to \(-\infty\) as \( x \to \infty \). At \( x = 3 \), the function value is \( -27 + 18 + 9 = 0 \). Since the function is decreasing for \( x > 3 \) (as the derivative is negative for \( x > 3 \)), the function cannot reach 6 for \( x > 3 \). Therefore, there are no solutions in this domain.

Thus, the only solution to \( f(x) = 6 \) is \( x = -3 \). Therefore, \( f^{-1}(6) = -3 \).

Finally, we sum the two values:
\[
f^{-1}(0) + f^{-1}(6) = 3 + (-3) = 0.
\]

Therefore, the answer is \(\boxed{0}\).