To find functions $g(x)$ and $h(x)$ such that $f(x) = g(h(x))$ and neither $g(x)$ nor $h(x)$ is solely $x$, we need to decompose the function $f(x) = -\frac{4}{3}x - 2$ into a composition of two functions.

Step-by-Step Solution

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

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

Step 1: Identify potential structures for $h(x)$ and $g(x)$

Let's choose a function $h(x)$ that is neither too simple nor too complex, and then find $g(x)$ accordingly.

A good strategy is to let $h(x)$ be a linear function, because the given function $f(x)$ is also linear. We could choose:

$h(x) = ax + b$

And then $g(x)$ will be:

$g(x) = c \cdot x + d$

Step 2: Choose specific forms for $h(x)$ and $g(x)$

Let's try $h(x) = 2x + 3$ and find $g(x)$ such that $g(h(x)) = f(x)$.

Substitute $h(x)$ into $g(x)$:

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

Now, let’s find $g(x)$.

We need: $g(2x + 3) = -\frac{4}{3}x - 2$

To match this, set up a function $g(x)$:

If $g(x) = -\frac{2}{3}x + c$, substituting $x = 2x + 3$ gives:

$g(2x + 3) = -\frac{2}{3}(2x + 3) + c$

Simplify: $g(2x + 3) = -\frac{4}{3}x - 2 + c$

To match $f(x) = -\frac{4}{3}x - 2$, set $c = 0$.

So, $g(x) = -\frac{2}{3}x$.

Step 3: Verify the decomposition

Now we have:

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

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

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:

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

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. How can you find a different pair of functions $g(x)$ and $h(x)$ for the same $f(x)$?
2. What is the composition of two quadratic functions to form a cubic function?
3. Can you decompose a nonlinear function into two linear functions?
4. How does the chain rule in calculus relate to function composition?
5. What are the properties of invertible functions in function decomposition?

Tip: When decomposing functions, start by looking for simpler component functions (like linear or quadratic ones) that, when combined, match the original function.