Question

If $f(x)=x^{3}+e^{\frac{x}{2}}$ and $g(x)$ is the inverse of $f(x)$, then find $g'(1)$.

Answer 1

$g(x)$ is the inverse of $f(x)$.
$\Rightarrow g(x)=f^{-1}(x)$ ...(i)
$\Rightarrow f[g(x)]=x$
Differentiating with respect to $x$, we get
$f'[g(x)] g'(x)=1$
$\Rightarrow f'[g(1)] g'(1)=1$
$\Rightarrow g'(1)=\frac{1}{f'[g(1)]}$ ... (ii)
$f(x)=x^{3}+e^{\frac{x}{2}}$
$\Rightarrow f(0)=0+e^{0}=1$
$\Rightarrow 0=f^{-1}(1)$
$\Rightarrow g(1)=0$ ...[From (i)]]
$\Rightarrow g'(1)=\frac{1}{f'(0)}$ ...(iii) [From (ii)]
$f(x)=x^{3}+e^{\frac{x}{2}}$
$\Rightarrow f'(x)=3 x^{2}+\frac{e^{\frac{x}{2}}}{2}$
$\Rightarrow f'(0)=0+\frac{e^{0}}{2}=\frac{1}{2}$ ...(iv)
$\Rightarrow g'(1)=\frac{1}{f'(0)}=2$ ....[From (iii) and (iv)]