Question

If $y=g\left(x\right)$ is a curve which is obtained by the reflection of $f\left(x\right)=\frac{e^x-e^{-x}}{2}$ about the line $y=x,$ then which of the following is CORRECT?

• A. $g\left(x\right)$ has more than one tangent parallel to $x-$axis.
• B. $g\left(x\right)$ has more than one tangent parallel to $y-$axis.
• C. $y=-x$ is a tangent to $g\left(x\right)$ at $(0,0).$
• D. $g\left(x\right)$ has no extremum.

Answer 1

The correct option is D $g\left(x\right)$ has no extremum.

As $y=g\left(x\right)$ is a curve which is obtained by the reflection of $f\left(x\right)=\frac{e^x-e^{-x}}{2}$ about the line $y=x,$
$\therefore g\left(x\right)$ is inverse of $f\left(x\right).$
$f\left(x\right)=\frac{e^x-e^{-x}}{2}=y$
$\Rightarrow e^{2x}-2y\left(e^x\right)-1=0$
$\Rightarrow e^x=\frac{2y\pm \sqrt{4y^2+4}}{2}$
$\Rightarrow e^x=y+\sqrt{y^2+1}(\because e^x>0\forall x\in R)$
$\Rightarrow x=\ln (y+\sqrt{y^2+1})$
$\therefore g\left(x\right)=\ln (x+\sqrt{x^2+1})$

$g^{\prime }\left(x\right)=\frac{1+\frac{x}{\sqrt{x^2+1}}}{x+\sqrt{x^2+1}}$
$\Rightarrow g^{\prime }\left(x\right)=\frac{1}{\sqrt{x^2+1}}\ne 0\forall x\in R$
$\Rightarrow g\left(x\right)$ has no tangent parallel to $x-$axis.
$g\left(x\right)$ has no tangent parallel to $y-$axis as $g^{\prime }\left(x\right)=\frac{1}{\sqrt{x^2+1}}$ can never be undefined.

$g^{\prime }\left(0\right)=1$
Equation of tangent at $(0,0)$ is
$y-0=1(x-0)$
$\Rightarrow y=x$
Since $g^{\prime }\left(x\right)>0,$
$\Rightarrow g\left(x\right)$ has no extremum.