# The function f(x) = |x + 1| is not differentiable

Question:

The function $f(x)=|x+1|$ is not differentiable at $x=$___________

Solution:

The given function is $f(x)=|x+1|$.

$f(x)=|x+1|= \begin{cases}x+1, & x \geq-1 \\ -(x+1), & x<-1\end{cases}$

Now, $(x+1)$ and $-(x+1)$ are polynomial functions which are differentiable at each $x \in \mathrm{R} .$ So, $f(x)$ is differentiable for all $x>-1$ and for all $x<-1$.

So, we need to check the differentiability of $f(x)$ at $x=-1$.

We have,

$L f^{\prime}(-1)=\lim _{h \rightarrow 0} \frac{f(-1-h)-f(-1)}{-h}$

$\Rightarrow L f^{\prime}(-1)=\lim _{h \rightarrow 0} \frac{-(-1-h+1)-(-1+1)}{-h}$

$\Rightarrow L f^{\prime}(-1)=\lim _{h \rightarrow 0} \frac{h}{-h}$

$\Rightarrow L f^{\prime}(-1)=-1$

And

$R f^{\prime}(-1)=\lim _{h \rightarrow 0} \frac{f(-1+h)-f(-1)}{h}$

$\Rightarrow R f^{\prime}(-1)=\lim _{h \rightarrow 0} \frac{(-1+h+1)-(-1+1)}{h}$

$\Rightarrow R f^{\prime}(-1)=\lim _{h \rightarrow 0} \frac{h}{h}$

$\Rightarrow R f^{\prime}(-1)=1$

$\therefore L f^{\prime}(-1) \neq R f^{\prime}(-1)$

So, $f(x)$ is not differentiable at $x=-1$.

Thus, the function $f(x)=|x+1|$ is not differentiable at $x=-1$.

The function $f(x)=|x+1|$ is not differentiable at $x=$ ___−1___.