Question

Write the number of points where f(x) = |x + 2| + |x – 3| is not differentiable.​

Answer 1

Given,

$\longrightarrow f(x)=|x+2|+|x-3|$

Case 1:-

Let $x\leq-2.$

This implies $x+2\leq0$ and $x-3\leq0.$

Then,

$\longrightarrow f(x)=-(x+2)-(x-3)$

$\longrightarrow f(x)=-2x+1$

Case 2:-

Let $-2\leq x\leq3.$

This implies $x+2\geq0$ and $x-3\leq0.$

Then,

$\longrightarrow f(x)=(x+2)-(x-3)$

$\longrightarrow f(x)=5$

Case 3:-

Let $x\geq3.$

This implies $x+2\geq0$ and $x-3\geq0.$

Then,

$\longrightarrow f(x)=(x+2)+(x-3)$

$\longrightarrow f(x)=2x-1$

This makes new definition for $f(x).$

$\longrightarrow f(x)=\left\{\begin{array}{lr}-2x+1,&x\leq-2\\5,&-2\leq x\leq3\\2x-1,&x\geq3\end{array}\right.$

So the derivative of $f(x)$ wrt $x$ will be,

$\longrightarrow f'(x)=\left\{\begin{array}{rr}-2,&x\leq-2\\0,&-2\leq x\leq3\\2,&x\geq3\end{array}\right.$

Here $f'(x)$ is not continuous at $x\in\{-2,\ 3\}.$

This implies $f(x)$ is not differentiable at $x\in\{-2,\ 3\}.$

Hence there are two points.

Answer 2

Answer:

two points where it’s not Differentiable