Let f(x) = begin{cases} |x| & text{for } 0

Question

(A) Given the function $f(x) = \begin{cases} |x| & 0  0$. Sinc

Vedclass · Accepted Answer

Local maximum.

Vedclass · Answer

(A) Given the function $f(x) = \begin{cases} |x| & 0 For $x$ close to $0$ (but $x 
eq 0$),$|x| > 0$. Since $f(x) = |x|$ for $x 
eq 0$,we have $f(x) > 0$ for all $x$ in the neighborhood of $0$ except at $x=0$.At $x = 0$,$f(0) = 1$.For any $x$ in the interval $(-\delta, \delta)$ where $\delta$ is small (e.g.,$\delta = 0.5$),we have $f(x) = |x|$.Since $|x| Specifically,for $x \in (-0.5, 0.5)$ and $x 
eq 0$,$f(x) = |x| Since $f(x) < f(0)$ for all $x$ in the neighborhood of $0$,$x = 0$ is a point of local maximum.

Let $f(x) = \begin{cases} |x| & \text{for } 0 < |x| \leqslant 2 \\ 1 & \text{for } x = 0 \end{cases}$. What is the nature of $f$ at $x = 0$?

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