Question

The set of all points where the function $f\left(x\right)=\sqrt{1-e^{-x^2}}$ is differentiable is

• A. $(0,\infty )$
• B. $(-\infty ,\infty )$
• C. $(-\infty ,\infty )-\left\{0\right\}$
• D. $(-1,\infty )$

Answer 1

Answer: (C)

$(-\infty ,\infty )-\left\{0\right\}$

For $x\ne 0$, we have
$f^{\prime }\left(x\right)=\frac{1}{2}\frac{1}{\sqrt{1-e^{-x^2}}}[-(-2x)e^{-x^2}]=\frac{x{e}^{-x^2}}{\sqrt{1-e^{-x^2}}}$

Also, $f^{\prime }(0+)=\underset{h\to 0+}{lim}\frac{f\left(h\right)-f\left(0\right)}{h}=\underset{h\to 0+}{lim}\frac{\sqrt{1-e^{-h^2}}}{h}$

$=\underset{h\to 0+}{lim}{\left(\frac{e^{-h^2}-1}{-h^2}\right)}^{1/2}=1$

and $f^{\prime }(0-)=\underset{h\to 0-}{lim}-{\left(\frac{e^{-h^2}-1}{-h^2}\right)}^{1/2}=-1$

because, as h$\to$0-, h is a negative number, so that
$\frac{\sqrt{1-e^{-h^2}}}{h}=-\frac{\sqrt{1-e^{-h^2}}}{\left|h\right|}=-\sqrt{\frac{1-e^{-h^2}}{h^2}}=-{\left(\frac{e^{-h^2}-1}{-h^2}\right)}^{1/2}$

Hence f is not differentiable at x=0. Thus the points of differentiability are $(-\infty ,\infty )-\left\{0\right\}$.