Question

If $f\left(x\right)=\{\begin{array}{cc}\frac{x}{1+e^{1/x}}& x\ne 0\\ 0& x=0\end{array}$, then the function $f\left(x\right)$ is differentiable for:

• A. $x\in R^{+}$
• B. $x\in R$
• C. $x\in R-\left\{0\right\}$
• D. $x\in R-\{0,1\}$

Answer 1

The correct option is C $x\in R-\left\{0\right\}$

$f\left(x\right)=\{\begin{array}{cc}\frac{x}{1+e^{1/x}}& x\ne 0\\ 0& x=0\end{array}$
We need to check at $x=0$
$f{(}^{\prime }x)=\{\begin{array}{cc}\frac{1(1+e^{1/x})+x\left(e^{1/x}\right)\frac{1}{x^2}}{{(1+e^{1/x})}^2}& x\ne 0\\ 0& x=0\end{array}=\{\begin{array}{cc}\frac{1+e^{1/x}+\frac{e^{1/x}}{x}}{{(1+e^{1/x})}^2}& x\ne 0\\ 0& x=0\end{array}{lim}_{x\to 0}{f}^{\prime }\left(x\right)=\{\begin{array}{cc}\ne 0& x\ne 0\\ 0& x=0\end{array}$
$\therefore f\left(x\right)$ is not differentiable at $x=0$
$x\in R-\left\{0\right\}$