Question

Let $f:R\to R,g:R\to R$ be the two functions given by $f\left(x\right)=2x-3,g\left(x\right)=x^3+5$. Then, ${(f\circ g)}^{-1}\left(x\right)$ is equal to

• A. ${\left(\frac{x+7}{2}\right)}^{\frac{1}{3}}$
• B. ${(x-\frac{7}{2})}^{\frac{1}{3}}$
• C. ${\left(\frac{x-2}{7}\right)}^{\frac{1}{3}}$
• D. ${\left(\frac{x-7}{2}\right)}^{\frac{1}{3}}$

Answer 1

The correct option is D ${\left(\frac{x-7}{2}\right)}^{\frac{1}{3}}$

Since, $f:R\to R$ and $g:R\to R$, given by
$f\left(x\right)=2x-3$ and $g\left(x\right)=x^3+5$ respectively, are bijections
Therefore, $f^{-1}$ and $g^{-1}$ exist
We have
$f\left(x\right)=2x-3$
$\therefore$ $f\left(x\right)=y$
$\Rightarrow 2x-3=y$
$\Rightarrow x=\frac{y+3}{2}$
$\Rightarrow f^{-1}\left(y\right)=\frac{y+3}{2}$
thus, $f^{-1}$ is given by $f^{-1}\left(x\right)=\frac{x+3}{2}$ for all $x\in R$
Similarly
$g^{-1}\left(x\right)={(x-5)}^{1/3}$, for all $x\in R$
Now, ${(f\circ g)}^{-1}=g^{-1}\circ f^{-1}=g^{-1}\left(f^{-1}\right(x\left)\right)=g^{-1}\left(\frac{x+3}{2}\right)={(\frac{x+3}{2}-5)}^{1/3}={\left(\frac{x-7}{2}\right)}^{1/3}$