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 $x$. \[ f(x)=\sqrt[3]{4 x^{2}-2}+1 \] Answer \[ g(x)= \] \[ h(x)= \]
Solution
Step 1 ：Given the function \(f(x)=\sqrt[3]{4 x^{2}-2}+1\), we need to find functions \(g(x)\) and \(h(x)\) such that \(f(x)=g(h(x))\) and neither \(g(x)\) nor \(h(x)\) is solely \(x\).
Step 2 ：In the given function \(f(x)=\sqrt[3]{4 x^{2}-2}+1\), the most nested operation is the cube root, which is applied to the expression \(4 x^{2}-2\). Therefore, we can take \(h(x)=4 x^{2}-2\) as the inner function.
Step 3 ：The outer function is then \(g(x)=\sqrt[3]{x}+1\), which is applied to the result of \(h(x)\).
Step 4 ：By substituting \(h(x)\) into \(g(x)\), we can confirm that \(f(x)\) is indeed equal to \(g(h(x))\).
Step 5 ：Final Answer: The functions \(g(x)\) and \(h(x)\) such that \(f(x)=g(h(x))\) are \(\boxed{g(x)=\sqrt[3]{x}+1}\) and \(\boxed{h(x)=4 x^{2}-2}\).
