Question

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

Answer 1

$f\left(x\right)=\left[x\right],0<x<3$

At $x=1$

f(x) is differentiable at $x=1y$ if

${lim}_{h\to 0}\frac{f\left(x\right)-f(x-h)}{h}={lim}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$

${lim}_{h\to 0}\frac{f\left(1\right)-f(1-h)}{h}={lim}_{h\to 0}\frac{f(1+h)-f\left(1\right)}{h}$

${lim}_{h\to 0}\frac{\left(1\right)-(1-h)}{h}={lim}_{h\to 0}\frac{(1+h)-\left(1\right)}{h}$

${lim}_{h\to 0}\frac{1-0}{h}={lim}_{h\to 0}\frac{1-1}{h}$

${lim}_{h\to 0}\frac{1}{h}={lim}_{h\to 0}\frac{0}{h}$

$\infty \ne 0$

$\therefore f\left(x\right)$ is not differentaible at $x=1$

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For $x=2$

${lim}_{h\to 0}\frac{f\left(x\right)-f(x-h)}{h}={lim}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$

${lim}_{h\to 0}\frac{f\left(2\right)-f(2-h)}{h}={lim}_{h\to 0}\frac{f(2+h)-f\left(2\right)}{h}$

${lim}_{h\to 0}\frac{\left(2\right)-(2-h)}{h}={lim}_{h\to 0}\frac{(2+h)-\left(2\right)}{h}$

${lim}_{h\to 0}\frac{2-0}{h}={lim}_{h\to 0}\frac{2-2}{h}$

${lim}_{h\to 0}\frac{2}{h}={lim}_{h\to 0}\frac{0}{h}$

$\infty \ne 0$

$\therefore f\left(x\right)$ is not differentaible at $x=2$