Browse Questions

If $f: R\to R$ be given by $f(x)=(3-x^3)^\frac {1}{3}$, then $fof(x)$ is

$(A)\;\;x ^ \frac {1} {3}\qquad(B)\;\;x^3\qquad(C)\;\;x\qquad(D)\;\;(3-x^3)$

Toolbox:
• Given two functions $f:A \to B$ and $g:B \to C$, then composition of $f$ and $g$, $gof:A \to C$ by $gof (x)=g(f(x))\;for\; all \;x \in A$. Therefore it follows that $fof = f(f(x))$.
Given a function $f:R\to R$ given by $f(x)=(3-x^3)^{1/3}$
We know that $(fof)(x)=f(f(x))$
$\Rightarrow fof =f((3-x^3)^{1/3})$ $=[3-((3-x^3)^{1/3})^3]^{1/3}$
$\Rightarrow fof =[3-(3-x^3)]^{1/3}$ $=(x^3)^{1/3}=x$
Therefore $(fof)(x)= (C)\; x$ is the correct answer.
edited Mar 19, 2013