Question

If $f\left(x\right)=(a{x}^2+b{)}^3$ then a function g(x) such that f(g(x)) = g (f(x)) is given by

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

Answer 1

The correct option is A $g\left(x\right)={\left(\frac{b-x^{1/3}}{a}\right)}^{1/2}$

Let $g\left(x\right)$ be $f^{-1}x$ as $f\left(f^{-1}\right(x\left)\right)=x$ and $f^{-1}\left(f\right(x\left)\right)=x$

$\sqrt[3]{3y}=a{x}^2+b\Rightarrow x^2=$ $\frac{\sqrt[3]{3y}-b}{a}$

$x=$${\left(\frac{\sqrt[3]{3y}-b}{a}\right)}^{\frac{1}{2}}$