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I have a question regarding differentiation when looking for extremas. I usualy set to find the stationary points (c for which f'(c)= 0) but if my function has absolute values.

I.e my function f(x) = x|x-2| for x in [-1,3] Since it has absolute values I separate my function into to functions

if x >= 2 $$ fx = x^2-2x $$ if x < 2 $$ fx = 2x-x^2$$

then I look for my stationary points and find only 1 that satisfies f'(-1) = 0 since my derivatives are 2x-2 and 2-2x and since f''(1) < 0 (-2) so at 1, I have a local MAX f(1)= 1 then I

  1. check my boundaries f(-1) = -3 -> absolute min and f(3)= 1 -> maximum?
  2. check my points where f'(x) does not exist but here I have none?

OR is there like a rule where you have to check for the value of x for which what's inside the absolute value = 0. (for instance x=2 here)? I think I have to check that to since it would mean that it is an undifferentiable value of my function but I do not understand why it is so? f(2) would equal 0 which would be

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    Check again. Your function's derivative does not exist at x=2 – 2017-01-12
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    Yes, I noticed anytime I have a function with absolute values, for the value of x that makes the inside of the absolute value=0 then the derivative of the function at that x will not exist. why is that so? – 2017-01-12
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    Because the absolute value function $f(x) = |x|$ is not differentiable at zero. Its derivative from the right is 1 and its derivative from the left is -1. – 2017-01-12
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    So it would be a good method to always look at the value for which |x| = |0| when looking for extremas which are either 1) stationary points (all x for which f'(x)=0 2)points where f'(x) does not exist ----> which would include any point y for which the absolute value part of my function would would equal to |0| 3) Bounderies if looking at a closed interval – 2017-01-12
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    By the way, "extrema" is already plural; "extremum" is the singular, and "extremas" isn't a word. – 2017-01-12
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    @SoHCahToha Yeah, that's a good guideline, I think. In general, just anywhere where the derivative DNE must be checked. You can see that formally here, by looking at the derivative function you wrote down... has a jump discontinuity from -2 to 2 at x = 2. (So the $x\ge 2$ you wrote next to the first equation should be a $x>2.$ The derivative DNE at 2). – 2017-01-13

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You have to check all points where the derivative fails to exist along with points where the derivative is zero and the ends of the interval. In your case the derivative fails to exist at $x=2$ To check this, check the derivative at $2^+$ and $2^-$. If they do not agree, the derivative fails to exist. Here they are $+2$ at $2^+$ and $-2$ at $2^-$. If you plot the graph (below) you can see the local minimum at $x=2$.
enter image description here