Question

If $f\left(x\right)=2x+1;x\le 1$
$=x^2+2;1<x\le 2$
$=4x^2+2;x>2$
then number of points where $f\left(x\right)$ is not differentiable is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is B 1

If $f\left(x\right)=2x+1;x\le 1$ Find no of pts where $f\left(x\right)$ is not differentiable.

$=x^2+2;2x\le 2$

$=4x^2+2;x>2$

At $x=1$

$f^{\prime }\left(x\right){|}_{x=1^{-}}=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$

$=\underset{h\to 0}{lim}{\frac{(2(x+h)+1)-(2x+1)}{h}|}_{x=1}=2$

$f^{\prime }\left(x\right){|}_{x=1^{+}}=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$

$=\underset{h\to 0}{lim}\frac{(x+h{)}^2+2-(x^2+2)}{h}=\underset{h\to 0}{lim}{\frac{h^2+2xh}{h}|}_{x=1}$

$=2$

Hence differentiable at $x=1$

At $x=2$

$f^{\prime }\left(x\right){|}_{x=2^{-}}=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$

$=\underset{h\to 0}{lim}{\frac{({(x+h)}^2+2)-(x^2+2)}{h}=\underset{h\to 0}{lim}\frac{h^2+2xh}{h}|}_{x=2}=4$

$f^{\prime }\left(x\right){|}_{x=2^{+}}=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$

$=\underset{h\to 0}{lim}{\frac{(4{(x+h)}^2+2)-(4x^2+2)}{h}=\underset{h\to 0}{lim}\frac{4h^2+8xh}{h}|}_{x=2}$

$=16$

Hence Not differentiable at $x=2$

$\therefore$ There is a $1$ ptnd differentiability.