Solveeit Logo
Login

Question

Mathematics Question on Continuity and differentiability

Prove that the greatest integer function defined by f(x)=[x],0<x<3 is not differentiable at x=1 and x=2.

Answer

The given function is f(x)=[x],0<x<3
It is known that a function f is differentiable at a point x=c in its domain if both
limh→0−f(c+h)−f(c)h-f\frac{(c+h)-f(c)}{h} and limh→0+f(c+h)−f(c)hf\frac{(c+h)-f(c)}{h} are finite and equal
To check the differentiability of the given function at x=1,
consider the left hand limit of f at x=1
limh→0-f(1+h)−f(1)hf\frac{(1+h)-f(1)}{h} =limh→0-[1+h]−[1]h\frac{[1+h]-[1]}{h}
=limh→0-0−1h\frac{0-1}{h}=limh→0-[1+h]−[1]h\frac{[1+h]-[1]}{h}

Consider the right hand limit of f at x=1
limh→0+f(1+h)−f(1)hf\frac{(1+h)-f(1)}{h}=limh→0+f[1+h]−[1]hf\frac{[1+h]-[1]}{h}
=limh→0+1−1h\frac{1-1}{h}
=limh→0+0=0.

Since the left and right hand limits of f at x=1 are not equal,f is not differentiable at x=1

To check the differentiability of the given function at x=2,consider the left hand limit of f at x=2
limh→0-f(2+h)−f(2)h\frac{f(2+h)-f(2)}{h} =limh→0-[2+h]−[2]h\frac{[2+h]-[2]}{h}
=limh→0-1−2h\frac{1-2}{h}=limh→0--1/h =∞

Consider the right hand limit of f at x=1
limh→0+f(2+h)−f(2)hf\frac{(2+h)-f(2)}{h}=limh→0+f(2+h)−f(2)hf\frac{(2+h)-f(2)}{h}
=limh→0+2−2h\frac{2-2}{h}
=limh→0+0=0

Since the left and right hand limits of f at x=2 are not equal, f is not differentiable at x=2