Question

Let $f:R\to R$ be defined as
$f\left(x\right)=x^3+x-5$
If $g\left(x\right)$ is a function such that $f\left(g\right(x\left)\right)=x,\forall x\in R$, then $g^{\prime }\left(63\right)$ is equal to

Answer 1

$f^{\prime }\left(x\right)=3x^2+1$
$f\left(x\right)$ is bijective function
and $f\left(g\right(x\left)\right)=x\Rightarrow g\left(x\right)$ is inverse of $f\left(x\right)$
$g\left(f\right(x\left)\right)=x$
$g^{\prime }\left(f\right(x\left)\right)\cdot f^{\prime }\left(x\right)=1$
$g^{\prime }\left(f\right(x\left)\right)=\frac{1}{3x^2+1}....\left(i\right)$
$\because f\left(x\right)=x^3+x-5=63$
$\Rightarrow x=4$
Put $x=4$ in $\left(i\right)$ we get
$g^{\prime }\left(63\right)=\frac{1}{49}$