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

The correct option is C

$(–\infty ,\infty )-\left\{0\right\}$

Explanation for the correct option:

Step 1: Solve the right-hand limit for $f\left(x\right)=\sqrt{1-e^{-x^2}}$

For $\begin{array}{rcl}x& \ne & 0\end{array}$,

$\begin{array}{rcl}f\text{'}\left(x\right)& =& \frac{1}{2}\times \frac{(-e^{-x^2})(-2x)}{\sqrt{1-e^{-x^2}}}\\ & =& \frac{x{e}^{-x^2}}{\sqrt{1-e^{-x^2}}}\end{array}$

$f\text{'}\left(0^{+}\right)=\underset{h\to 0^{+}}{\lim }\frac{f\left(h\right)–f\left(0\right)}{h}=\underset{h\to 0^{+}}{\lim }\frac{\sqrt{1-e^{-h^2}}}{h}$

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

Step 2: Solve the left-hand limit for $f\left(x\right)=\sqrt{1-e^{-x^2}}$

$f\text{'}\left(0^{-}\right)=\underset{h\to 0^{-}}{\lim }{\left(-\frac{e^{-h^2}-1}{-h^2}\right)}^{\frac{1}{2}}=–1$

As $h\to 0^{-}$ is a negative number,

$\begin{array}{rcl}\frac{\sqrt{1-e^{-h^2}}}{h}& =& –\frac{\sqrt{1-e^{-h^2}}}{\left|h\right|}\\ & =& –\frac{\sqrt{1-e^{-h^2}}}{h^2}\\ & =& –{\left(\frac{e^{-h^2}-1}{-h^2}\right)}^{\frac{1}{2}}\end{array}$

$f$ is not differentiable at$x=0.$

The differentiable points are $(–\infty ,\infty )-\left\{0\right\}.$

Hence, option C is the correct option.