Question

Which of the following functions is non-differentiable?

• A. $f\left(x\right)=(e^x-1)|e^{2x}-1|$ defined in $R$
• B. $f\left(x\right)=\frac{x-1}{x^2+1}$ defined in $R$
• C. $f\left(x\right)\{\begin{array}{cc}||x-3|-1|,& x<3\\ \frac{x}{3}\left[x\right]-2,& x\ge 3\end{array}$ where [.] represents the greatest integer function
• D. $f\left(x\right)=3(x-2{)}^{1/3}+3$ defined in $R$

Answer 1

The correct option is D $f\left(x\right)=3(x-2{)}^{1/3}+3$ defined in $R$

$f\left(x\right)=(e^x-1)|e^{2x}-1|$
$=(e^x-1)|e^x-1||e^x+1|$
$=(e^x+1)(e^x-1)|e^x-1|$
Now, both $e^x+1$ and $(e^x-1)|e^x-1|$ are differentiable, as $g\left(x\right)|g\left(x\right)|$ is differentiable when $g\left(x\right)=0$.
Hence, $f\left(x\right)$ is differentiable.
$f\left(x\right)=\frac{x-1}{x^2+1}$ is rational function in which denominator never becomes zero.
Hence, f(x) is differentiable.
$f\left(x\right)=\{\begin{array}{cc}||x-3|-1|,& x<3\\ \frac{x}{3}\left[x\right]-2,& x\ge 3\end{array}$
$=\{\begin{array}{cc}|3-x-1|,& x<3\\ \frac{x}{3}3-2,& 3\le x<4\end{array}$
$=\{\begin{array}{cc}|x-2|,& x<3\\ x-2,& 3\le x<4\end{array}$
$=x-2,x\in [2,4)$
Hence, f(x) is differentiable at $x=3$.
$f\left(x\right)=3(x-2{)}^{3/4}3$ or $f^{\prime }\left(x\right)=\frac{9}{4}(x-2{)}^{-1/4}$ which is non-differentiable at $x=2$.
Here,
$f\left(x\right)$ is continuous and the graph has vertical tangent at $x=2$;
however, the graph is smooth in the neighborhood of $x=2$.