Question

If $f(x)=(ax^2+b)^3$, then the function $g$ such that $f(g(x))=g(f(x))$ is given by

• A. $g\left(x\right)=\left(\frac{b-x^{1/3}}{a}\right)$
• B. $g\left(x\right)=\frac{1}{\left(ax^{2}+b\right)^{3}}$
• C. $g(x) = (ax^2 + b)^{1/3}$
• D. $g\left(x\right)=\left(\frac{x^{1/3}-b}{a}\right)^{1/2}$

Answer 1

Answer: (D)

$y=f \left(x\right)=\left(ax^{2}+b\right)^{3}$
$\Rightarrow ax^{2}+b=y^{1/3}$
$\Rightarrow x^{2}=\frac{y^{1/3}-b}{a}$
$\Rightarrow x=\left(\frac{y^{1/3}-b}{a}\right)^{1/2}$
$=f^{-1}\left(y\right)$
Since, $f(g(x)) = g(f(x))$
$\therefore $ We have, $g\left(x\right)=\left(\frac{x^{1/3}-b}{a}\right)^{1/2}$.