Question

Let a function $f:(0,\infty )\to (1,\infty )$ is defined as $f\left(x\right)=x+e^{-x},$ then which of the following is correct

• A. $f^{-1}\left(x\right)$ does not exist
• B. $f^{-1}\left(x\right)$ is decreasing with concave upwards
• C. $f^{-1}\left(x\right)$ is increasing with concave upwards
• D. $f^{-1}\left(x\right)$ is increasing with concave downwards

Answer 1

The correct option is D $f^{-1}\left(x\right)$ is increasing with concave downwards

Given : $f\left(x\right)=x+e^{-x}$
$\Rightarrow f^{\prime }\left(x\right)=1-e^{-x}>0\forall x>0$
$\Rightarrow f\left(x\right)$ is increasing (one to one and onto both)
$\therefore f^{-1}\left(x\right)$ exist.

Let inverse of $f\left(x\right)$ is $g\left(x\right).$
$\Rightarrow f\left(g\right(x\left)\right)=x$
$\Rightarrow f^{\prime }\left(g\right(x\left)\right)\cdot g^{\prime }\left(x\right)=1$
$\Rightarrow g^{\prime }\left(x\right)=\frac{1}{f^{\prime }\left(g\right(x\left)\right)}$
$\Rightarrow g^{\prime \prime }\left(x\right)=-\frac{f^{\prime \prime }\left(g\right(x\left)\right)\cdot g^{\prime }\left(x\right)}{\left[f^{\prime }\right(g\left(x\right)){]}^2}\cdots \left(i\right)$
As, $f\left(x\right)$ is increasing, $g\left(x\right)$ will also be increasing.
$\Rightarrow g^{\prime }\left(x\right)>0$ and $f^{\prime \prime }\left(x\right)=e^{-x}>0$
From $\left(i\right),g^{\prime \prime }\left(x\right)<0$
$\therefore g\left(x\right)$ has downward concavity.