To find two functions $f(x)$ and $g(x)$ such that $(f \circ g)(x) = H(x) = (5x - 4)^3$, we need to express $H(x)$ as a composition of two functions where neither function is the identity function. There can be more than one correct answer, but let's find one possible pair $f(x)$ and $g(x)$.

Step 1: Express $H(x) = (5x - 4)^3$ as $f(g(x))$.

Let's consider $g(x) = 5x - 4$.

Then, $f(x)$ should be the function that cubes its input, so: $f(x) = x^3$

Step 2: Verify the composition.

Now, check if $(f \circ g)(x)$ gives us $H(x)$: $(f \circ g)(x) = f(g(x)) = f(5x - 4) = (5x - 4)^3$

This matches $H(x)$, so the functions $f(x) = x^3$ and $g(x) = 5x - 4$ work correctly.

Final Answer:

• $f(x) = x^3$
• $g(x) = 5x - 4$

Alternatively, you could also express $H(x)$ differently, such as by choosing $g(x) = x - 4$ and $f(x) = (5x)^3$, but the above choice is straightforward.

Would you like to explore other possibilities or any more details?

Here are 5 related questions:

1. How can you determine if a function is a valid composition of two other functions?
2. What happens if you change the order of composition $f \circ g$ to $g \circ f$?
3. How do you find the derivative of a composed function?
4. Can $f(x)$ and $g(x)$ have domains that differ from $H(x)$?
5. What are other possible pairs of $f(x)$ and $g(x)$ that satisfy the same composition?

Tip: When decomposing a function into compositions, it helps to think about breaking down the operations step by step, starting from the inside out.