Question

If $f\left(x\right)=\{\begin{array}{c}{x}^3-3x+2x<2\\ x^3-6x^2+9x+2,x\ge 2\end{array}$
then

• A. $\underset{x\to 2}{lim}f\left(x\right)$
• B. $f$ is not continuous at $x=2$
• C. $f$ is continuous but not differentiable at $x=2$
• D. $f$ is continuous but differentiable at $x=2$

Answer 1

Answer: (C)

$f$ is continuous but not differentiable at $x=2$

Given

$f\left(x\right)=\{\begin{array}{c}{x}^3-3x+2x<2\\ x^3-6x^2+9x+2,x\ge 2\end{array}$

LHL $=f(2-0)=\underset{h\to 0}{lim}{(2-h)}^3-3(2-h)+2$
$=2^3-6+2=8-6+2=4$

RHL $=f(2+0)=\underset{h\to 0}{lim}{(2-h)}^3-6{(2+h)}^2+9(2+h)+2$
$2^3-6{\left(2\right)}^2+9\left(2\right)+2=8-24+18+2=4$
$\because$ LHL $=$ RHL
$\therefore$ $\underset{x\to 2}{lim}f\left(x\right)$ exist
and

$f\left(2\right)={\left(2\right)}^3-6{\left(2\right)}^2+9\left(2\right)+2$
$=8-24+18+2=4$

$\therefore$ LHL $=$ RHL $=f\left(2\right)$

So, $f\left(x\right)$ is continuous at $x=2$

Now

$f^{\prime }\left(x\right)=\{\begin{array}{c}3x^2-3,x<2\\ 3x^2-12x+9,x\ge 2\end{array}$
$\therefore$ $LHD:f^{\prime }\left(2\right)=3{\left(2\right)}^2-3=12-3=9$
and

$RHD:f^{\prime }\left(2\right)=3{\left(2\right)}^2-12\left(2\right)+9$
$=12-24+9=-3$

$\because$ $L{f}^{\prime }\left(2\right)\ne R{f}^{\prime }\left(2\right)$

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

Hence $f$ is continuous but not differentiable at $x=2$