Question

Discuss the continuity and differentiability of the function f(x) = |x|+|x-1| in the interval (-1,2).

Answer 1

Method 1: Since the continuity and differentiability of modulus function is doubtful at the corner points. So well shall check continuity and differentiability at the critical points x=0, 1 $ϵ$ (-1,2).

Therefore, f(x) = |x|+|x-1|= $$\begin{array}{cc}-x-(x-1)=-2x+1,& \text{if}x<0\\ x-(x-1)=1,& \text{if}0\le x<1\\ x+x-1=2x-1,& \text{if}x\ge 1\end{array}$$

Continuity at x=0 : We have f(0)=1.

LHL (at x=0): $\stackrel{lim}{x\to 0^{-}}(-2x+1)=-2\times 0+1=1$

RHL (at x=1): $\stackrel{lim}{x\to 0^{+}}1=1$

Since $\stackrel{lim}{x\to 1}$ f(x) = f(1) so f(x) is continuous at x=0.

Continuity at x=1: We have f(1)= 2(1)-1=1.

LHL (at x=1) :$\stackrel{lim}{x\to 1^{-}}=1$
RHL (at x=1) : $\stackrel{lim}{x\to 1^{+}}2x-1=2\times 1-1=1$

Since $\stackrel{lim}{x\to 1}$ f(x)= f(1) so f(x) is continuous at x=1.

Differentiability at x=0:
LHD (at x=1) : $\stackrel{lim}{x\to 0^{-}}\frac{|x|+|x-1|-1}{x-0}=\stackrel{lim}{x\to 0^{-}}\frac{-2x+1-1}{x}=\stackrel{lim}{x\to 0^{-}}\frac{-2x}{x}=-2$

RHD (at x=0): $\stackrel{lim}{x\to 0^{+}}\frac{1-1}{x-0}=\stackrel{lim}{x\to 0^{+}}\frac{0}{x}=\stackrel{lim}{x\to 0^{+}}0=0\ne$ LHD (at x=0)

$\therefore$ f(x) is not differentable at x=0.

Differentibility at x=1:

LHD (at x=1) : $\stackrel{lim}{x\to 1^{-}}\frac{1-1}{x-1}=\stackrel{lim}{x\to 1^{-}}\frac{0}{x-1}=\stackrel{lim}{x\to 1^{-}}0=0$

RHD (at x=1): $\stackrel{lim}{x\to 1^{+}}\frac{x+x-1-1}{x-1}=\stackrel{lim}{x\to 1^{+}}\frac{2(x-1)}{x-1}=2\ne$ LHD (at x=1)

$\therefore$ f(x) is not differentiable at x=1.

Method 2: Since the continuity and differentiability of modulus function is doubtful at the corner points. So well shall check continuity and differentiability at the critical points x =0, 1 $ϵ$ (-1,2).

Continuity at x=0 : We have f(0)= |0|+|0-1|=1.

LHL (at x=0): $\stackrel{lim}{x\to 0^{-}}|x|+|x-1|=|0|+|0-1|=1$

RHL (at x=0): $\stackrel{lim}{x\to 0^{+}}|x|+|x-1|=|0|+|0-1|=1$

Since $\stackrel{lim}{x\to 0}$ f(x) = f(0) so f(x)is continous at x=0.

Continuity at x=1: We have f(1)= |1|+|1-1|=1. LHL (at x=1) : $\stackrel{lim}{x\to 1^{-}}|x|+|x-1|=|1|+|1-1|=1$

RHL (at x =1): $\underset{x\to 1^1}{lim}$|x|+|x-1|=|1|+|1-1|=1

Since $\underset{x\to 1}{lim}$ f(x)= f(1) so f(x) is continuous at x=1.

Differentiability at x=0 :

LHD (at x=0): $\underset{x\to 0^{-}}{lim}\frac{|x|+|x-1|-1}{x-0}=\underset{x\to 0^{-}}{lim}\frac{-x-x+1-1}{x}=\underset{x\to 0^{-}}{lim}\frac{-2x}{x}=-2$

RHD (at x=0): $\underset{x\to 0^{+}}{lim}\frac{|x|+|x-1|-1}{x-0}=\underset{x\to 0^{+}}{lim}\frac{x-x+1-1}{x-1}=\underset{x\to 0^{+}}{lim}\frac{0}{x}=0\ne$ LHD (at x=0)

$\therefore$ f(x) is not differentiable at x=0.

Differentiability at x=1 :

LHD (at x=1):$\underset{x\to 1^{-}}{lim}\frac{|x|+|x-1|-1}{x-1}=\underset{x\to 1^{-}}{lim}\frac{x-x+1-1}{x-1}=\underset{x\to 1^{-}}{lim}\frac{0}{x-1}=0$

RHD (at x=1): $\underset{x\to 1^{+}}{lim}\frac{|x|+|x-1|-1}{x-1}=\underset{x\to 1^{+}}{lim}\frac{x+x-1-1}{x-1}=\underset{x\to 1^{+}}{lim}\frac{2(x-1)}{x-1}=2\ne$ LHD (at x=0)

$\therefore$ f(x)is not differentiable at x=1.