Let's work through this question step by step:

The function $f(x) = x^3 - 5$ is one-to-one.

(a) Finding the inverse of $f(x)$

1. Start with $y = x^3 - 5$.
2. Solve for $x$ in terms of $y$: $y + 5 = x^3 \quad \implies \quad x = \sqrt[3]{y + 5}.$
3. Swap $x$ and $y$ to find the inverse function: $f^{-1}(x) = \sqrt[3]{x + 5}.$

So, the inverse function is $f^{-1}(x) = \sqrt[3]{x + 5}$.

(b) Domain and range of $f(x)$ and $f^{-1}(x)$

• Domain of $f(x)$: Since $x^3 - 5$ is a cubic function, it is defined for all real numbers $x$. Therefore: $\text{Domain of } f(x) = (-\infty, \infty).$

• Range of $f(x)$: A cubic function is surjective (covers all real numbers). Therefore: $\text{Range of } f(x) = (-\infty, \infty).$

• Domain of $f^{-1}(x)$: The domain of $f^{-1}(x)$ is the range of $f(x)$, which is: $\text{Domain of } f^{-1}(x) = (-\infty, \infty).$

• Range of $f^{-1}(x)$: The range of $f^{-1}(x)$ is the domain of $f(x)$, which is: $\text{Range of } f^{-1}(x) = (-\infty, \infty).$

(c) Graph $f(x)$, $f^{-1}(x)$, and $y = x$ on the same coordinate axes

To graph these:

1. $f(x) = x^3 - 5$ is a cubic function shifted down by 5.
2. $f^{-1}(x) = \sqrt[3]{x + 5}$ is the reflection of $f(x)$ over $y = x$.
3. $y = x$ is the line of symmetry.

Would you like me to sketch this graph for you?

Questions to Expand Learning:

1. What are the key properties of one-to-one functions, and why are they necessary for finding an inverse?
2. How would the inverse function change if $f(x) = x^3 + c$ instead of $x^3 - 5$?
3. How can we verify if two functions are inverses of each other?
4. What happens to the domain and range when you apply transformations to a cubic function, such as shifting or reflecting?
5. Why is $y = x$ the line of symmetry for inverse functions?

Tip:

When solving for an inverse function, always verify your solution by checking if $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$.