Question

If the function $f(x)=x^3+e^{x/2}$ and $g(x)=f^{-1}$, then the value of $g^{\text{'}}(1)$ is

Answer 1

Correct Answer: 2

Explanation:
$g(x)$ is the inverse of $f(x)$
So,$g(f(x))=x$
$\text{'}{g}^{\text{'}}(f(x))\text{'}{f}^{\text{'}}(x)=1$
$\text{'}{g}^{\text{'}}(f(x))=\frac{1}{f^{\text{'}}(x)}$
We need to find out when $f(x)=1$
Clearly this happens for $x=0$
So ${\mathrm{g}}^{\text{'}}(1)=\frac{1}{f^{\text{'}}(0)}$
$f^{\text{'}}(x)=3x^2+\frac{1}{2}{e}^{\frac{l}{2}}$
$f^{\text{'}}(0)=0+\frac{1}{2}=\frac{1}{2}$
So $g^{\text{'}}(1)=2$
Hence, the correct answer is (2).