# The set of point where the function

Question:

The set of point where the function $f(x)=|2 x-1|$ is differentiable, is_______________

Solution:

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 $\mathbf{R}-\left\{\frac{1}{2}\right\}$