To determine if $f(x)$ is the inverse of $g(x)$, we must check if the composition of the two functions, $f(g(x))$ and $g(f(x))$, results in $x$. In other words, if both $f(g(x)) = x$ and $g(f(x)) = x$, then $f(x)$ and $g(x)$ are inverses of each other.

Given:

$f(x) = -\left(\frac{1}{4} \cdot \sqrt[3]{x + 9}\right) - \frac{7}{4}$ $g(x) = (-4x - 7)^3 - 9$

Step 1: Find $f(g(x))$

Substitute $g(x) = (-4x - 7)^3 - 9$ into $f(x)$:

$f(g(x)) = -\left(\frac{1}{4} \cdot \sqrt[3]{(-4x - 7)^3 - 9 + 9}\right) - \frac{7}{4}$

Simplify: $f(g(x)) = -\left(\frac{1}{4} \cdot \sqrt[3]{(-4x - 7)^3}\right) - \frac{7}{4}$ $f(g(x)) = -\left(\frac{1}{4} \cdot (-4x - 7)\right) - \frac{7}{4}$ $f(g(x)) = -\left(\frac{-4x - 7}{4}\right) - \frac{7}{4}$ $f(g(x)) = x$

Step 2: Find $g(f(x))$

Substitute $f(x) = -\left(\frac{1}{4} \cdot \sqrt[3]{x + 9}\right) - \frac{7}{4}$ into $g(x)$:

$g(f(x)) = (-4 \left( -\left(\frac{1}{4} \cdot \sqrt[3]{x + 9}\right) - \frac{7}{4} \right) - 7)^3 - 9$

Simplify: $g(f(x)) = \left( 4 \cdot \frac{1}{4} \cdot \sqrt[3]{x + 9} + 7 - 7 \right)^3 - 9$ $g(f(x)) = \left( \sqrt[3]{x + 9} \right)^3 - 9$ $g(f(x)) = x + 9 - 9$ $g(f(x)) = x$

Conclusion:

Since both $f(g(x)) = x$ and $g(f(x)) = x$, $f(x)$ is the inverse of $g(x)$. The answer is "Yes".

Would you like more details or have any questions?

Here are 5 related questions to explore:

1. How can you algebraically derive the inverse of a function?
2. What happens when two functions are not inverses of each other?
3. How can you verify the inverse graphically?
4. What is the role of composition of functions in verifying inverses?
5. How does the domain and range affect the inverse of a function?

Tip: When dealing with cubic functions and cube roots, simplifying step-by-step is essential to avoid mistakes.