Let’s analyze the problem step by step based on the graph provided:

(a) Explain why $f$ has an inverse and find specific values.

1. Why $f$ has an inverse: The graph of $f$ passes the horizontal line test (i.e., no horizontal line intersects the graph more than once). This indicates $f$ is one-to-one and therefore has an inverse.

2. Find specific values using the graph:

   • $f^{-1}(2)$: Look at where $y = 2$ on the graph. The corresponding $x$-value is $8$. Thus, $f^{-1}(2) = 8$.
   • $f^{-1}(-1)$: Look at where $y = -1$ on the graph. The corresponding $x$-value is $-4$. Thus, $f^{-1}(-1) = -4$.
   • $f^{-1}(0)$: Look at where $y = 0$ on the graph. The corresponding $x$-value is $0$. Thus, $f^{-1}(0) = 0$.

(b) Find the domain and range of $f^{-1}$.

The domain of $f^{-1}$ is the range of $f$, and the range of $f^{-1}$ is the domain of $f$.

From the graph:

• The domain of $f^{-1}$ is $[-2, 2]$, because this is the range of $f$.
• The range of $f^{-1}$ is $[-8, 8]$, because this is the domain of $f$.

(c) Sketch the graph of $f^{-1}$.

To sketch $f^{-1}$:

• Reflect the graph of $f$ across the line $y = x$.
• Points such as $(-8, -2)$, $(-4, -1)$, $(0, 0)$, and $(8, 2)$ on $f$ become $(-2, -8)$, $(-1, -4)$, $(0, 0)$, and $(2, 8)$ on $f^{-1}$.

Would you like me to draw the graph of $f^{-1}$ or explain these reflections further?

Follow-up Questions:

1. Why is the horizontal line test essential for determining invertibility?
2. Can a function fail the horizontal line test but still have an inverse if restricted to a subset of its domain?
3. How does the reflection $y = x$ help construct the inverse function graphically?
4. What would happen to the domain and range of $f^{-1}$ if $f$'s domain were extended?
5. How would discontinuities in $f$ affect $f^{-1}$?

Tip: Always verify a function is one-to-one before finding its inverse to ensure it exists.