Let's work through both parts of this problem:

a) $f(x) = x^2 - 4$

1. Finding the inverse of $f(x)$:

   • To find the inverse, replace $f(x)$ with $y$: $y = x^2 - 4$

   • Swap $x$ and $y$ and solve for $y$: $x = y^2 - 4$ $y^2 = x + 4$ $y = \pm \sqrt{x + 4}$ Thus, the inverse relation of $f(x)$ is $f^{-1}(x) = \pm \sqrt{x + 4}$.

   • Graphing:

     • The original function $f(x) = x^2 - 4$ is a parabola opening upwards, with a vertex at $(0, -4)$.
     • The inverse relation $f^{-1}(x) = \pm \sqrt{x + 4}$ is not a function, since it gives two values (positive and negative square roots) for each $x$, except when $x = -4$.

   • Is the inverse a function? No, the inverse is not a function because it does not pass the vertical line test (it provides two outputs for one input).

b) $g(x) = 2^x$

1. Finding the inverse of $g(x)$:

   • Replace $g(x)$ with $y$: $y = 2^x$

   • Swap $x$ and $y$, and solve for $y$: $x = 2^y$ $y = \log_2(x)$ Thus, the inverse function is $g^{-1}(x) = \log_2(x)$.

   • Graphing:

     • The original function $g(x) = 2^x$ is an exponential curve that increases rapidly for positive $x$ and approaches zero as $x$ becomes more negative.
     • The inverse function $g^{-1}(x) = \log_2(x)$ is a logarithmic curve that grows slowly for larger values of $x$ and is undefined for $x \leq 0$.

   • Is the inverse a function? Yes, the inverse of $g(x)$ is a function because it passes the vertical line test.

Do you need help with a graphical representation, or do you have any specific questions about the solutions?

Here are 5 questions to further your understanding:

1. What does the vertical line test indicate about a relation being a function?
2. How do we algebraically find the inverse of a quadratic function?
3. Why is the inverse of an exponential function always a logarithmic function?
4. What restrictions exist on the domain of $f(x) = x^2 - 4$ and its inverse?
5. Can every function have an inverse that is also a function? Why or why not?

Tip: Always check if the inverse relation is a function by applying the vertical line test to its graph.