Question

Consider the function $y=|x-1|+|x-2|$ in the interval $[0,3]$ and discuss the continuity and differentiability of the function in this interval.

• A. continuous everywhere
• B. differentiable everywhere except at $x=1$ and $x=2$
• C. differentiable everywhere
• D. continuous everywhere except at $x=1$ and $x=2$

Answer 1

Answer: (A), (B)

The correct options are
A continuous everywhere
B differentiable everywhere except at $x=1$ and $x=2$

We know that $\left|x\right|=x$ when x is +ive and $\left|x\right|=-x$ when x is negative.
From the definition of function,we have $y=-(x-1)-(x-2)=3-2x$ when $x\le 1$
$y=(x-1)-(x-2)=1$ when $1\le x\le 2$
$f\left(x\right)=(x-1)+(x-2)=2x-3$ when $x>2.$
Thus the function has been defined for $[0,1],[1,2]and[2,3]$ respectively.
Hence the graph drawn for the interval $[0,3]$ is as shown.
From the graph it is clear that the function is continuous throughout the interval but is not differentiable at $x=1,2$ since the slopes at these points are different on the left hand and right hand sides.