The set of point where the function $f(x)=|2 x-1|$ is differentiable, is_______________
The given function is $f(x)=|2 x-1|$.
$f(x)=|2 x-1|= \begin{cases}2 x-1, & x \geq \frac{1}{2} \\ -(2 x-1), & x<\frac{1}{2}\end{cases}$
Now, $(2 x-1)$ and $-(2 x-1)$ are polynomial functions which are differentiable at each $x \in \mathrm{R} .$ So, $f(x)$ is differentiable for all $x>\frac{1}{2}$ and for all $x<\frac{1}{2}$.
So, we need to check the differentiability of $f(x)$ at $x=\frac{1}{2}$.
We have,
$L f^{\prime}\left(\frac{1}{2}\right)=\lim _{h \rightarrow 0} \frac{f\left(\frac{1}{2}-h\right)-f\left(\frac{1}{2}\right)}{-h}$
We have,
$L f^{\prime}\left(\frac{1}{2}\right)=\lim _{h \rightarrow 0} \frac{f\left(\frac{1}{2}-h\right)-f\left(\frac{1}{2}\right)}{-h}$
$\Rightarrow L f^{\prime}\left(\frac{1}{2}\right)=\lim _{h \rightarrow 0} \frac{1-2\left(\frac{1}{2}-h\right)-\left(2 \times \frac{1}{2}-1\right)}{-h}$
$\Rightarrow L f^{\prime}\left(\frac{1}{2}\right)=\lim _{h \rightarrow 0} \frac{2 h}{-h}$
$\Rightarrow L f^{\prime}\left(\frac{1}{2}\right)=-2$
And
$R f^{\prime}\left(\frac{1}{2}\right)=\lim _{h \rightarrow 0} \frac{f\left(\frac{1}{2}+h\right)-f\left(\frac{1}{2}\right)}{h}$
$\Rightarrow R f^{\prime}\left(\frac{1}{2}\right)=\lim _{h \rightarrow 0} \frac{2\left(\frac{1}{2}+h\right)-1-\left(2 \times \frac{1}{2}-1\right)}{h}$
$\Rightarrow R f^{\prime}\left(\frac{1}{2}\right)=\lim _{h \rightarrow 0} \frac{2 h}{h}$
$\Rightarrow R f^{\prime}\left(\frac{1}{2}\right)=2$
$\therefore L f^{\prime}\left(\frac{1}{2}\right) \neq R f^{\prime}\left(\frac{1}{2}\right)$
Thus, the set of points where the function $f(x)=|2 x-1|$ is differentiable is $\mathbf{R}-\left\{\frac{1}{2}\right\}$.
The set of point where the function $f(x)=|2 x-1|$ is differentiable, is $\mathbf{R}-\left\{\frac{1}{2}\right\}$