find total number of point at which f attains extremum is , $F = | x^{2} - 25 | $ To find total numberof point at which F attains extremum is 
$F = | x^{2} - 25 | $
I have made its graph and and from it i have got answer , but how do i use calculus to see ? I am confused with absolute value sign . Thanks for help 
 A: We know that $x^2-25$ is positive for $x<-5$ and $x>5$, and negative for $-5<x<5$. This divides your domain into three intervals. Your function is $x^2-25$ in the outside intervals, and $-(x^2-25)$ in the middle one.
You can use calculus to look for extreme values in the interior of each interval, and then check the critical points $x=-5$ and $x=5$ where the derivative of $F$ is undefined.
A: the graph of $y = x^2 - 25$ is the graph of $y = x^2$ translated $25$ units vertically down. the $x$-intercepts are $x = \pm 5,$ the solutions of $x^2 - 25 = 0$
observe that the graph of $y$ between $-5 < x < 5$ is negative. to draw the graph of $y = |x^2 - 25|,$  flip the part of the graph $y = x^2 - 25$ between $-5$ and $5$ so as to make the $y$ nonnegative everywhere. 
can you see that the graph of $y = |x^2 - 25|$ has two local minima at $x = \pm 5$ and a local maximum at $x = 0?$
now to show that the critical numbers of $|x^2 - 25|$ are $x = 0, -5, 5.$ look at the tangent at the point $x = -5$ of the graph $y = x^2 - 25.$  it is $\dfrac{d}{dx}(x^2 - 25) = 2x$ evaluated at $x = -5$ which is $-10.$ when we reflected the graph to get $y = |x^2 - 25|,$ this tangent got broken in to two half lines with slopes $-25$ and $25.$  that made the point $(-5,0)$ on the graph $y = |x^2 - 25|$ a cusp; that is the function $|x^2 - 25|$ does not have a derivative and the graph does not have a unbroken tangent. 
you can do the same at $x = 5.$ at $x = 0,$ both graphs have a horizontal tangent, so that makes $x = 0$ a critical point for both graphs.
A: The function is not differentiable at $-5$ and $5$. For $x\ne-5$ and $x\ne5$,
$$
f'(x)=\frac{x^2-25}{|x^2-25|}\cdot 2x=
\begin{cases}
-2x & \text{if $x<-5$ or $x>5$}\\
2x & \text{if $-5<x<5$}
\end{cases}
$$
Thus it's easy to see that $f$ is decreasing in $(-\infty,-5]$, increasing in $[-5,0]$, decreasing in $[0,5]$ and increasing in $[5,\infty)$.
