The total number of points of $\mathbb{R}$ at which $f$ attains a local extremum Let $f(x) = \vert x^2-25 \vert$ for all $x \in \mathbb{R}$. The total number of points of $\mathbb{R}$ at which $f$ attains a local extremum is 
$A$. $1$
$B$. $2$
$C$. $3$
$D$. $4$
What I was thinking where $f'(x)=0$. but this only gives you $x=0$. Now here I'm stuck. Help me!
 A: Recall that

a function can have a relative maximum or relative minimum only at those numbers in its domain at which the derivative is undefined or is zero (these numbers are called critical points).

Notice that $f'(x) = \begin{cases} -2x & x \in (-5, 5) \\ 2x & x \in (-\infty, -5) \cup (5, \infty) \\ \text{DNE} & x = -5, 5 \end{cases}$
Notice that if we consider $f'$ on the domain in which it is defined, $f'(x) = 0$ means $x = 0$.
So our critical points are $x = -5, 0, 5$. Hence your answer (d) is out of the question since there are at most 3 local extremum.
Now we check each point one by one:


*

*($x = 5$): $f(5) = 0$ and $f(5) \leq f(x)$ for all $x \in [4, 6]$.

*($x = 0$): $f(0) = 25$ and $f(0) \geq f(x)$ for all $x \in [-1, 1]$.

*($x = -5$): $f(-5) = 0$ and $f(-5) \leq f(x)$ for all $x \in [-6, -4]$. Here you could notice that $f$ is symmetric with respect to the $y$-axis and refer to the case where $x = 5$.


Hence we have 3 local extrema. This should have been obvious from looking at the graph, which is simple enough to draw pretty quickly!
