To find functions $g(x)$ and $h(x)$ such that $f(x) = g(h(x))$ and neither $g(x)$ nor $h(x)$ is solely $x$, let's consider the given function:

$f(x) = -\frac{4}{3x - 2}$

Step-by-Step Solution

We want to express $f(x)$ as $g(h(x))$, where neither $g(x)$ nor $h(x)$ is just $x$.

Step 1: Choose a form for $h(x)$

One approach is to look at the inner expression $3x - 2$ and consider this as our $h(x)$. Let’s set:

$h(x) = 3x - 2$

This choice ensures that $h(x)$ is not just $x$, but a linear function.

Step 2: Determine $g(x)$

Given that $h(x) = 3x - 2$, we want to find $g(x)$ such that:

$f(x) = g(h(x)) = g(3x - 2)$

Substituting $h(x)$ into $f(x)$, we get:

$f(x) = -\frac{4}{h(x)}$

So we can define $g(x)$ based on this relationship:

$g(x) = -\frac{4}{x}$

Step 3: Verify the decomposition

Now we have:

$g(x) = -\frac{4}{x}, \quad h(x) = 3x - 2$

Let's verify that $f(x) = g(h(x))$:

$g(h(x)) = g(3x - 2) = -\frac{4}{3x - 2}$

Which is exactly $f(x)$.

Conclusion

Thus, one possible pair of functions $g(x)$ and $h(x)$ such that $f(x) = g(h(x))$ are:

$\boxed{g(x) = -\frac{4}{x}, \quad h(x) = 3x - 2}$

Neither $g(x)$ nor $h(x)$ is solely $x$.

Would you like more details or have any questions?

Here are 5 related questions to explore further:

1. Can you find another pair of functions $g(x)$ and $h(x)$ that would work for the same $f(x)$?
2. How would the decomposition change if $f(x)$ had a more complicated numerator or denominator?
3. What are some examples of decomposing quadratic functions into two non-quadratic functions?
4. How does function composition affect the domain and range of the resulting function?
5. What strategies can be used to decompose a rational function like $f(x) = \frac{ax + b}{cx + d}$?

Tip: When decomposing a function into two others, try to identify a part of the expression that could be isolated as one function, making the rest easier to express as a composition.