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)=−3x+4⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√3+5

Answer 1

Answer:

$h(x)=-2x+4$

$g(x)=\sqrt[3]{x}-5$

Step-by-step explanation:

Consider the given function is

$f(x)=\sqrt[3]{-2x+4}-5$

It is given that $f(x)=g(h(x))$ and neither g(x) nor h(x) is solely x.

$f(x)=\sqrt[3]{(-2x+4)}-5$

Let $h(x)=-2x+4$, then we get

$f(x)=g(h(x))=\sqrt[3]{h(x)}-5$

Substitute h(x)=x in the above function.

$g(x)=\sqrt[3]{x}-5$

Therefore, the required functions are $h(x)=-2x+4$ and $g(x)=\sqrt[3]{x}-5$.

Check the solutions.

$g(h(x))=g(-2x+4)$             $[\because h(x)=-2x+4]$

$g(h(x))=\sqrt[3]{-2x+4}-5$            $[\because g(x)=\sqrt[3]{x}-5]$

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

Therefore, our solution is correct.