Question

given f(x), find g(x) and h(x) such that f(x) = g(h(x)) and neither g(x) nor h(x) is solely x.
f(x) = \sqrt3{-2x^2 - 1} + 1
g(x) =
h(x) =

Explanation:

Step1: Identify outer function $g(x)$

We take the outermost operation of $f(x)$: cube root plus 1. Let the input to $g$ be $u$, so $g(u) = \sqrt[3]{u} + 1$, which is $g(x) = \sqrt[3]{x} + 1$.

Step2: Identify inner function $h(x)$

The input to the outer function is the expression inside the cube root: $-2x^2 - 1$, so $h(x) = -2x^2 - 1$.

Step3: Verify composition

Check that $g(h(x)) = \sqrt[3]{h(x)} + 1 = \sqrt[3]{-2x^2 - 1} + 1 = f(x)$, which matches.

Answer:

$g(x) = \sqrt[3]{x} + 1$
$h(x) = -2x^2 - 1$