Question

If $f\left(x\right)=\left\{\begin{array}{ll}\left|x\right|,& 0<\left|x\right|\le 2\\ 1,& x=0\end{array}\right.$, then at $x=0$, $f$ has

• A. a local maximum.
• B. a local minimum.
• C. no local extremum.
• D. no local maximum.

Answer 1

Answer: (A), (B)

a local minimum.

Explanation for the correct option

Given that, $f\left(x\right)=\left\{\begin{array}{ll}\left|x\right|,& 0<\left|x\right|\le 2\\ 1,& x=0\end{array}\right.$

It can be also represented as, $f\left(x\right)=\left\{\begin{array}{ll}-x,& -2\le x<0\\ 1,& x=0\\ x,& 0<x\le 2\end{array}\right.$ $\left[\because f\left(x\right)=\left|x\right|\Rightarrow f\left(x\right)=\left\{\begin{array}{l}+x;\forall x>0\\ -x;\forall x<0\end{array}\right.\right]$.

Thus, the left side of $x=0$ is $f\left(x\right)=-x$.

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

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

Therefore, while approaching $x=0$ from the left side, the function tends to $0$.

Thus, the right side of $x=0$ is $f\left(x\right)=x$.

So, $\underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }\left(x\right)$.

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

Therefore, while approaching $x=0$ from the right side, the function tends to $0$.

But, at $x=0$, $f\left(x\right)=1$, which is greater than $0$.

So, the given function has a local maximum at $x=0$.

Hence, option(A) i.e. a local maximum is the correct option.