Question

The exhaustive set of values of $x$ for which $f(x)=\sqrt[5]{x^2|x{|}^3}-\sqrt[3]{x^2|x|}-1$ is not differentiable is

• A. $\{0,1\}$
• B. $\{0,1,-1\}$
• C. $\{0,-1\}$
• D. none of these

Answer 1

Answer: (D)

$f(x)={\sqrt[5]{x}}^2|x{|}^3-\sqrt[3]{x^2|x|-1}$
$\Rightarrow f(x)=\{\begin{array}{ll}{\left(x^5\right)}^{\frac{1}{5}}-{\left(x^3\right)}^{\frac{1}{3}}-1,& \text{ x≥0 }\\ {\left(-x^5\right)}^{\frac{1}{5}}-{\left(-x^3\right)}^{\frac{1}{3}}-1,& \text{x<0 }\end{array}$
$=\{\begin{array}{ll}x-x-1,& \text{if x≥0}\\ -x+x-1,& \text{if x<0 }\end{array}$
$=-1$ for all x
Thus, f(x) is differentiable everywhere