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

Question

(A) The function is defined as $f(x) = |x|$ for $x \in [-2, 0) \cup (0, 2]$ and $f(0) = 1$.For any small neighborhood $(0 - h, 0 + h)$ around $x = 0$

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$A$ local maximum

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(A) The function is defined as $f(x) = |x|$ for $x \in [-2, 0) \cup (0, 2]$ and $f(0) = 1$.For any small neighborhood $(0 - h, 0 + h)$ around $x = 0$ (where $h > 0$ is very small),we have:$f(0) = 1$For $x 
eq 0$ in this neighborhood,$f(x) = |x|$. Since $x$ is very close to $0$,$|x| Thus,$f(x) By the definition of a local maximum,if $f(x) \le f(a)$ for all $x$ in some neighborhood of $a$,then $f$ has a local maximum at $x = a$.Since $f(x) < f(0)$ for all $x \in (-h, h) \setminus \{0\}$,the function $f$ has a local maximum at $x = 0$.

Let $f(x) = \begin{cases} |x|, & 0 < |x| \le 2 \\ 1, & x = 0 \end{cases}$,then at $x = 0$,$f$ has

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