The points of non differentiability of $f(x) =|x-2|+|x-3|$
A. 1
B. 2
C. 3
D. 4
E. 5
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → B and C only
$f(x) =|x-2|+|x-3|=\left\{\begin{matrix}2x-5,&x>3\\1,&2≤x≤3\\-2x+5,&x<2\end{matrix}\right.$
so $f'(x)=\left\{\begin{matrix}2,&x>3\\0,&2<x<3\\-2,&x<2\end{matrix}\right.$
so $\lim\limits_{x→2^+}f'(x)=2≠0=\lim\limits_{x→3^-}f'(x)$
$\lim\limits_{x→2^+}f'(x)=0≠-2=\lim\limits_{x→2^-}f'(x)$
point of non differentiability → 2, 3