Question

Show that the function $f\left( x \right)=|x-1|+|x+1|$ , for all x $\in $ R, is not differentiable at the point x = -1 and x = 1.

Answer 1

Hint: Here, we will try to form a piecewise function for f(x) and then we will define its value in different intervals. After that we can find the left hand limit and right hand limit at 1 and -1. After this we will check whether the left hand limit is equal to the right hand limit or not.

Complete step-by-step answer:
$f\left( x \right)=\left\{ \begin{align}
  & x-1+x+1;\text{ x}\ge \text{1} \\
 & \text{-}\left( x-1 \right)+x+1;\text{ x}\in \left( -1,1 \right) \\
 & -\left( x-1 \right)-\left( x+1 \right);\text{ x}\le \text{-1} \\
\end{align} \right.$ A piecewise function is a function defined by multiple sub-functions, each such function applying to a certain interval of the main function’s domain, a sub-domain.
Piecewise is actually a way of expressing the function rather than a characteristic of the function itself, but with additional qualification it can describe the nature of the function.
While finding the left hand limit, we are interested in what happens to the function to the right of the cluster point.
The limit of a function exists at a point if the left hand limit and right hand limit are equal at the point. For a function to be differentiable at a point limit must exist at that point. The left hand derivative should be equal to the right hand derivative.
Since, the function given to us is:
$f\left( x \right)=|x-1|+|x+1|$ for all x $\in $ R
On forming piecewise function, we get:

We can also write as:
\[f\left( x \right)=\left\{ \begin{align}
  & 2x\text{ };\text{ x}\ge \text{1} \\
 & 2\text{ };\text{ x}\in \left( -1,1 \right) \\
 & -2x\text{ };\text{ x}\le \text{-1} \\
\end{align} \right.\]
The left hand derivative at -1 is = $\dfrac{d\left( -2x \right)}{dx}=-2$
Right hand derivative at-1 is = $\dfrac{d\left( 2 \right)}{dx}=0$
Similarly,
Left hand derivative at 1 is = $\dfrac{d\left( 2 \right)}{dx}=0$
Right hand derivative at 1 is =$\dfrac{d\left( 2x \right)}{dx}=2$
Since, at both the points 1 and -1, the left hand derivative is not equal to the right hand derivative.
Hence, the function is not differentiable at 1 and -1.

Note: Students should note here that a function is differentiable at a point only when the left hand derivative is equal to the right hand derivative at that point. Also, the existence of the limit is necessary for a function to be differentiable.