Question

Let $f:R\to R,g:R\to R$ be two functions given by $f\left(x\right)=5x-4$ and $g\left(x\right)=x^3+7$ then $\left(fog{)}^{-1}\right(x)$ equals

• A. ${\left(\frac{x+31}{5}\right)}^{1/3}$
• B. ${\left(\frac{x-31}{5}\right)}^{1/3}$
• C. ${\left(\frac{x-5}{7}\right)}^{1/3}$
• D. ${(x-\frac{31}{5})}^{1/3}$

Answer 1

Answer: (B)

${\left(\frac{x-31}{5}\right)}^{1/3}$

$f\left(x\right)=5x-4$ and $g\left(x\right)=x^3+7$ are one-one onto i.e. bijections, so $f^{-1}$ and $g^{-1}$ both exist.
$f\left(x\right)=5x-4$
$\therefore f\left(x\right)=y=5x-4$
$\therefore x=\frac{y+4}{5}i.e.f^{-1}\left(y\right)=\frac{y+4}{5}$
$f^{-1}\left(x\right)=\frac{x+4}{5}\forall xϵR$
Also $g\left(x\right)=x^3+7=y$
$\therefore x^3=y-7,x={(y-7)}^{1/3}$
$\therefore g^{-1}\left(y\right)={(y-7)}^{1/3}$
$\therefore g^{-1}\left(x\right)={(x-7)}^{1/3}\forall xϵR$
now ${\left(fog\right)}^{-1}\left(x\right)=\left(g^{-1}o{f}^{-1}\right)\left(x\right)=g^{-1}\left(f^{-1}\left(x\right)\right)$
$\therefore g^{-1}\left(\frac{x+4}{5}\right)={(\frac{x+4}{5}-7)}^{1/3}={\left(\frac{x-31}{5}\right)}^{1/3}$