This problem involves the function $f(x) = x^2 + 1$ for $0 \leq x \leq 3$. The graph of this function is shown, and you're tasked with finding information about its inverse.

Let's address each part:

(a) Draw the graph of $f^{-1}$

To sketch the inverse function $f^{-1}$, recall that the graph of an inverse function is a reflection of the original graph across the line $y = x$. For $f(x) = x^2 + 1$, the inverse will involve reflecting the curve over this line.

Steps to follow:

• Identify key points on the original graph (such as $(0, 1)$, $(1, 2)$, $(2, 5)$, and $(3, 10)$).
• Swap the coordinates of these points to get the points on the graph of $f^{-1}$.
  • $(0, 1) \to (1, 0)$
  • $(1, 2) \to (2, 1)$
  • $(2, 5) \to (5, 2)$
  • $(3, 10) \to (10, 3)$
• Plot these points and sketch a smooth curve.

(b) Write the range of $f^{-1}$

The range of the inverse function $f^{-1}$ is the domain of the original function $f$, which is $[0, 3]$. Thus, the range of $f^{-1}$ is: $[0, 3]$

(c) Find $f^{-1}(x)$

To find the inverse function, solve $f(x) = y$ for $x$: $y = x^2 + 1 \implies x^2 = y - 1 \implies x = \sqrt{y - 1}$ However, since the domain of $f$ is restricted to $[0, 3]$, we only take the positive root: $f^{-1}(x) = \sqrt{x - 1}, \quad \text{for } x \geq 1$

Summary:

• (a) The graph of $f^{-1}$ is the reflection of $f(x) = x^2 + 1$ over $y = x$, using key points.
• (b) The range of $f^{-1}$ is $[0, 3]$.
• (c) The inverse function is $f^{-1}(x) = \sqrt{x - 1}$, valid for $x \geq 1$.

Let me know if you'd like further clarification or details on any part of the solution. Here are some related questions:

1. How do we determine if a function has an inverse?
2. What conditions must be met for a function to be invertible?
3. How does restricting the domain affect the invertibility of a function?
4. Can all quadratic functions have inverses over their entire domain?
5. What happens graphically when we reflect a function over $y = x$?

Tip: To verify your inverse, remember that applying $f$ and then $f^{-1}$ (or vice versa) should return the original input.