Question

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

Answer 1

Given f(x) = [x] , (i) RHD of f(x) at x = 1;

Putting x = 1 + h as x $\to 1^{+}\Rightarrow h\to 0$

$R{f}^{\prime }\left(1\right)=\underset{h\to 0}{lim}\frac{[1+h]-\left[1\right]}{h}=\underset{h\to 0}{lim}\frac{1-1}{h}=0(\because R{f}^{\prime }(x)=\frac{f(x+h)-f\left(x\right)}{h})$

LHD of f(x) at x = 1; putting x = 1 - h as x $\to 1^{-}\Rightarrow h\to 0$

$L{f}^{\prime }\left(1\right)=\underset{h\to 0}{lim}\frac{[1-h]-\left[1\right]}{-h}=\underset{h\to 0}{lim}\frac{0-1}{-h}=\underset{h\to 0}{lim}\frac{1}{h}=\infty (\because L{f}^{\prime }(x)=\frac{f(x-h)-f\left(x\right)}{-h})$

$\therefore LHD\ne RHD.$Thus , f(x) is not diferentiable at x = 1.

(ii) At x = 2,

RHD = Rf '(2) =$\underset{h\to 0}{lim}\frac{f(2+h)-f\left(2\right)}{h}(\because R{f}^{\prime }(x)=\frac{f(x+h)-f\left(x\right)}{h})$

$=\underset{h\to 0}{lim}\frac{[2+h]-2}{h}=\underset{h\to 0}{lim}\frac{2-2}{h}=0$

LHD =Lf ' (2) = $\underset{h\to 0}{lim}\frac{f(2-h)-f\left(2\right)}{-h}(\therefore L{f}^{\prime }(x)=\frac{f(x-h)-f\left(x\right)}{-h})$

= $\underset{h\to 0}{lim}\frac{[2-h]-\left[2\right]}{-h}=\underset{h\to 0}{lim}\frac{1-2}{-h}=\infty$

$\therefore RHD\ne LHD$

Thus, f(x) is not differentiable at x = 2.

Hence proved.