Question

If Rolle's theorem holds for the function $f\left(x\right)=\frac{2}{e^x+e^{-x}},x\in [-1,1]$ at the point $x=c$, then the value of $c$ is:

• A. $0$
• B. $2$
• C. $3$
• D. $5$

Answer 1

The correct option is A $0$

$f\left(x\right)=\frac{2}{e^x+e^{-x}}$ is continuous and differentiable for all real $x.$ As this is an even function, it is obvious that $f(-1)=f\left(1\right).$ Therefore Rolle's theorem is applicable to this function.
Find the derivative and equate it to zero to get $c.$
$f^{\prime }\left(x\right)={\left(\frac{2}{e^x+e^{-x}}\right)}^{\prime }=-\frac{2}{(e^x+e^{-x}{)}^2}\cdot (e^x+e^{-x}{)}^{\prime }=-\frac{2(e^x-e^{-x})}{(e^x+e^{-x}{)}^2}.$
$f^{\prime }\left(c\right)=0\Rightarrow -\frac{2(e^c-e^{-c})}{(e^c+e^{-c}{)}^2}=0,$
$\Rightarrow$ $\{\begin{array}{c}2(e^c-e^{-c})=0\\ (e^c+e^{-c}{)}^2\ne 0\end{array}$, $\Rightarrow e^c-e^{-c}=0,\Rightarrow e^{2c}=1,\Rightarrow c=0.$
$\therefore c=0$