Discuss the continuity and differentiability of the function \[f(x)=\left| x \right|+\left| x-1 \right|\] in the interval (-1,2)

Hint: We know that a function is said to be continuous in an interval if it is continuous at every point in the interval and its left-hand limit is equal to the right-hand limit i.e.\[\underset{x\to {{c}^{-}}}{\mathop{\lim }}\,f(x)=\text{ }\underset{x\to {{c}^{+}}}{\mathop{\lim }}\,f(x)=f(c)\] . The function y = f (x) is said to be differentiable in the closed interval [a, b] if R f ′(a) and L f ′ (b) exist and f ′ (x) exists for every point of (a, b).

Complete step by step answer:
We have \[f(x)=\left| x \right|+\left| x-1 \right|\]
We can define the function at various interval
\[\begin{align}
  & \text{f}(\text{x})\text{=-x-(x-1)},\text{-1}<\text{x}<\text{0} \\
 & \text{=x-(x-1)},\text{0}<\text{x}<\text{1} \\
 & \text{=x+(x-1)},\text{1}\le \text{x}<\text{2} \\
\end{align}\]
On simplification, we get
\[\begin{align}
  & \text{f}(\text{x})\text{=-2x+1},\text{-1}<\text{x}<\text{0} \\
 & \text{=1},\text{0}<\text{x}<\text{1} \\
 & \text{=2x-1},\text{1}\le \text{x}<\text{2} \\
\end{align}\]
We will know find the continuity of the function in interval (-1,2)
As the value of function changes for different interval so we will find the continuity at a point where the value of function changes i.e. at x=0 and x=1
At x=0
Left hand limit = $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)$
The value of f(x) at x<0 is -2x+1
$\begin{align}
  & \Rightarrow \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,(-2x+1) \\
 & \Rightarrow -2\times 0+1=1 \\
 & \Rightarrow \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=1 \\
\end{align}$

Right Hand limit =$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)$
The value of f(x) at x>0 is 1
$\begin{align}
  & \Rightarrow \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,1 \\
 & \Rightarrow \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=1 \\
\end{align}$
As we know that the value of f(0) is 1
So, we can see that \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\text{ }\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=f(0)\]
Hence, the function f(x) is continuous at x=0

At x=1
Left hand limit = $\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)$
The value of f(x) at x<1 is 1
$\begin{align}
  & \Rightarrow \underset{x\to {{1}^{-}}}{\mathop{\lim }}\,(1) \\
 & \Rightarrow \underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)=1 \\
\end{align}$

Right Hand limit =$\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)$
The value of f(x) at x>0 is 2x+1
$\begin{align}
  & \Rightarrow \underset{x\to {{1}^{+}}}{\mathop{\lim }}\,(2x+1) \\
 & \Rightarrow 2\times 0+1 \\
 & \Rightarrow \underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)=1 \\
\end{align}$
As we know that the value of f (1)
$\begin{align}
  & \Rightarrow 2x-1 \\
 & \Rightarrow 2\times 1-1 \\
 & \Rightarrow f(1)=1 \\
\end{align}$
So, we can see that \[\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)=\text{ }\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)=f(1)\]
Hence, the function f(x) is continuous at x=1
 We can say that the function is continuous for all x in interval (-1,2)


We know that $\begin{align}
  & \text{f(x)= - 2, - 1 < x < 0} \\
 & \text{= 0, 0 < x < 1} \\
 & \text{= 2, 1 < x < 2}
\end{align}$
Differentiating f(x) w.r.t x, we get,
$\begin{align}
  & \text{f}(\text{x})\text{=-2},\text{-1}<\text{x}<\text{0} \\
 & \text{=0},\text{0}<\text{x}<\text{1} \\
 & \text{=2},\text{1}\le \text{x}<\text{2}
\end{align}$
We will check differentiability of function in interval (-1,2)
At x=0
Left hand derivative = $Lf'(0)$
$\Rightarrow Lf'(0)=-2$
Right hand derivative = $Rf'(0)$
$\Rightarrow Rf'(0)=0$
And $f'(0)=0$
As we can see that $Lf'(0)\ne f'(0)=Rf'(0)$ so the function is not differentiable at x=0

At x=1
Left hand derivative = $Lf'(1)$
$\Rightarrow Lf'(1)=0$
Right hand derivative = $Rf'(1)$
$\Rightarrow Rf'(1)=2$
And $f'(1)=2$
As we can see that $Lf'(0)\ne f'(0)=Rf'(0)$ so the function is not differentiable at x=1

So, the function f(x) is not differentiable in the interval (-1,2)

Note:
The points where the value of the function change are called as the branch points of the function. The differentiability and continuity of a function in an interval is only checked at its branch points. The students should keep in mind that the value of$\left| x \right|$ depends upon the value of x which is inside the modulus, if the value of x>0 then it will be positive and if x<0 it will be negative so don’t assume its value will always be positive. This should be kept in mind while defining the function of x for various intervals of x, at this place most of the mistakes are done.