Question

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

Answer 1

Answer: 2

$g(x)$ 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
$f^{\prime }[g(x)]g^{\prime }(x)=1$
$\Rightarrow f^{\prime }[g(1)]g^{\prime }(1)=1$
$\Rightarrow g^{\prime }(1)=\frac{1}{f^{\prime }[g(1)]}$ ... (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)]]
$\Rightarrow g^{\prime }(1)=\frac{1}{f^{\prime }(0)}$ ...(iii) [From (ii)]
$f(x)=x^3+e^{\frac{x}{2}}$
$\Rightarrow f^{\prime }(x)=3x^2+\frac{e^{\frac{x}{2}}}{2}$
$\Rightarrow f^{\prime }(0)=0+\frac{e^0}{2}=\frac{1}{2}$ ...(iv)
$\Rightarrow g^{\prime }(1)=\frac{1}{f^{\prime }(0)}=2$ ....[From (iii) and (iv)]