Suppose H(x)=(5x-4)^3. Find two functions f and g such that (fcirc g)(x)=H(x). Neither function can be the identity function. (There may be more than one correct answer.) f(x)=□ □ /□ sqrt(□ ) g(x)=□

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.

Learn how to find two functions f and g such that (f ◦ g)(x) = H(x) = (5x - 4)^3. Explore the step-by-step solution involving function composition and cubic functions. Suitable for advanced high school students.

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