To express the function $h(x) = 2x - 6$ as $h(x) = f(g(x))$ using non-identity functions $f(x)$ and $g(x)$, we need to find two functions $f(x)$ and $g(x)$ such that:

$h(x) = f(g(x))$

A common strategy is to let $g(x)$ be a simpler function related to the argument of $h(x)$, and then find $f(x)$ accordingly. Here is one possible solution:

Let: $g(x) = x - 3$

Then: $f(x) = 2x$

Now, let's verify:

$h(x) = f(g(x)) = f(x - 3) = 2(x - 3) = 2x - 6$

This matches the original function $h(x) = 2x - 6$.

Thus, the functions $f(x)$ and $g(x)$ are:

$f(x) = 2x, \quad g(x) = x - 3$

Would you like more details or have any questions?

Here are 5 related questions to explore:

1. How can the composition of functions be verified?
2. What other pairs of functions could be used to express $h(x) = 2x - 6$ as $f(g(x))$?
3. How would you express a quadratic function as a composition of two functions?
4. What is the importance of using non-identity functions in composition?
5. Can the same approach be used to decompose trigonometric functions?

Tip: When decomposing functions into compositions, start by simplifying or isolating parts of the function that can be treated as $g(x)$, and then find the corresponding $f(x)$.