Question

If $f\left(x\right)=\{\begin{array}{cc}2x^2+12x+16& ,-4\le x\le -2\\ 2-|x|& ,-2<x\le 1\\ 4x-x^2-2& ,1<x\le 3\end{array}$. Then $f\left(x\right)$ is

• A. Continuous everywhere and not differentiable at 3 point
• B. Discontinuous at 3 points
• C. Not differentiable at exactly 2 points
• D. Not continous at 3 points

Answer 1

Answer: (A)

Continuous everywhere and not differentiable at 3 point

$f\left(x\right)=\{\begin{array}{c}2x^2+12x+16;-4\le x\le -2\\ 2+x;-2<x\le 0\\ 2-x;0<x\le 1\\ 4x-x^2-2;1<x\le 3\end{array}$

at $x=-2$

$\underset{x\to 2^{-}}{lim}f\left(x\right)=\underset{x\to 2^{-}}{lim}(2x^2+12x+16)$

$=2{(-2)}^2+12(-2)+16$

$=0$

$\underset{x+-2^{+}}{lim}f\left(x\right)=\underset{x+-2^{+}}{lim}(2+x)=2-2=0$

LHL$=$RHL $\Rightarrow$continuous

at $x=0$

$\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{-}}{lim}(2-x)=2$

$\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0^{-}}{lim}(2-x)=2$

At $x=1$ LHL$=$RHL $\Rightarrow$continuous

$\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{-}}{lim}(2-x)=2-1=1$

$\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}(4x-x^2-2)=4-1-2=1$

LHL$=$RHL $\Rightarrow$continuous

$f^{\prime }\left(x\right)=\{\begin{array}{c}4x+12;-4\le x\le -2\\ 1;-2<x\le 0\\ -1;0<x\le 1\\ 4-2x;1<x\le 3\end{array}$

at $x=-2$

$\underset{x\to -2^{-}}{lim}{f}^{\prime }\left(x\right)=\underset{x\to -2^{-}}{lim}(4x+12)=4$

$\underset{x+-2^{+}}{lim}{f}^{\prime }\left(x\right)=\underset{x+-2^{+}}{lim}\left(1\right)=1$

LHL$\ne$RHL $\Rightarrow$non differentiable

at $x=0$

$\underset{x\to 0^{-}}{lim}{f}^{\prime }\left(x\right)=1$

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

LHL$\ne$RHL $\to$ differentiable

at $x=1$

$\underset{x\to 1^{-}}{lim}{f}^{\prime }\left(x\right)=\underset{x\to 1^{-}}{lim}(-1)=-1$

$\underset{x\to 1^{+}}{lim}{f}^{\prime }\left(x\right)=\underset{x\to 1^{+}}{lim}(4-2x)=4-2=2$

LHL$\ne$RHL $\Rightarrow$not differentiable