Search any question & find its solution
Question:
Answered & Verified by Expert
The set of all points where the function f (x) = 2x| x|is differentiable is
Options:
Solution:
2681 Upvotes
Verified Answer
The correct answer is:
$(-\infty, \infty)$
f(x) = 2x | x |
$f(x)=\left\{\begin{array}{cc}2 x^2 & , \quad x \geq 0 \\ -2 x^2 & , \quad x < 0\end{array}\right.$
Since, f(x) is a polynomial function, hence it will be differentiable everywhere. We need to check only at x = 0.
LHD $f^{\prime}\left(0^{-}\right)=\lim _{h \rightarrow 0^{-}} \frac{f(0+h)-f(0)}{h}$
$=\lim _{h \rightarrow 0^{-}} \frac{-2\left(h^2\right)-0}{h}=0$
RHD $f^{\prime}\left(0^{+}\right)=\lim _{h \rightarrow 0^{+}} \frac{f(0+h)-f(0)}{h}$
$=\lim _{h \rightarrow 0^{+}} \frac{2 h^2-0}{h}=\lim _{h \rightarrow 0^{+}} 2 h=0$
$\therefore f(x)$ is differentiable in $(-\infty, \infty)$
$f(x)=\left\{\begin{array}{cc}2 x^2 & , \quad x \geq 0 \\ -2 x^2 & , \quad x < 0\end{array}\right.$
Since, f(x) is a polynomial function, hence it will be differentiable everywhere. We need to check only at x = 0.
LHD $f^{\prime}\left(0^{-}\right)=\lim _{h \rightarrow 0^{-}} \frac{f(0+h)-f(0)}{h}$
$=\lim _{h \rightarrow 0^{-}} \frac{-2\left(h^2\right)-0}{h}=0$
RHD $f^{\prime}\left(0^{+}\right)=\lim _{h \rightarrow 0^{+}} \frac{f(0+h)-f(0)}{h}$
$=\lim _{h \rightarrow 0^{+}} \frac{2 h^2-0}{h}=\lim _{h \rightarrow 0^{+}} 2 h=0$
$\therefore f(x)$ is differentiable in $(-\infty, \infty)$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.