Finding relative and absolute maxima of an absolute function

f(x) = |x^2 -16| + 2x

My first approach: Differentiating it w.r.t. x.
f'(x) = 2x (x^2-16) /|x^2 - 16| + 2
Set f'(x) = 0,
2x(x^2-16) = -2|x^2-16|
What can I do next? I figure I can only restrict the x value or square both sides. How do I get 4 and -4?

My second approach: Split the absolute value function to a piecewise function.
f(x) = {x^2+2x-16, x>=4 or x<=-4
_____{-x^2+2x+16, -4<=x<=4
Is this right? Both equations include equal signs?
Then I differentiate w.r.t. x.
f'(x) = {2x+2, x>=4 or x<=-4
______{-2x+2, -4<=x<=4
So the critical points are x = -1 or 1, which are wrong! f(-1) is neither a local max or local min. What's wrong?

How should I approach this question?

Final 2 questions:
Are end points never relative extrema?
Relative extrema include absolute extrema?

Thank you for your help!

f(x) = |x^2 -16| + 2x

My first approach: Differentiating it w.r.t. x.
f'(x) = 2x (x^2-16) /|x^2 - 16| + 2
Set f'(x) = 0,
2x(x^2-16) = -2|x^2-16|
What can I do next? I figure I can only restrict the x value or square both sides. How do I get 4 and -4?

My second approach: Split the absolute value function to a piecewise function.
f(x) = {x^2+2x-16, x>=4 or x<=-4
_____{-x^2+2x+16, -4<=x<=4
Is this right? Both equations include equal signs?
Then I differentiate w.r.t. x.
f'(x) = {2x+2, x>=4 or x<=-4
______{-2x+2, -4<=x<=4
So the critical points are x = -1 or 1, which are wrong! f(-1) is neither a local max or local min. What's wrong?

How should I approach this question?

Final 2 questions:
Are end points never relative extrema?
Relative extrema include absolute extrema?

Thank you for your help!

It's not sufficient to simply solve the piecewise equations, you have to make sure the solutions fall in the correct domain for where the piecewise equations are defined.

Solving 2x+2 = 0 gives x = -1, but, f'(x) = 2x+2 is only valid in the ranges x <=-4 or x>=4. So you have to rule out x = -1 here.

Solving -2x+2 = 0 gives x = 1; f'(x) = -2x+2 is valid for -4 <= x < = 4 and so x=1 is a valid solution here.

Any question involving the absolute value function you will have to check all the cases

now as you said:



Now clearly when $2-2x=0 \implies x=1$ so x=1 is a critical point (max by 2nd derivative test)
but what about $2+2x=0 \implies x=-1$ but $-4<-1<4$ so cannot use second part of function here since it only applies when $x\le-4, x \ge 4$. So no solution in this case.

Going back to the end points of the first part of function gives $x=4, x=-4$ as minimums since the derivative goes from decreasing to increasing at this points (as x is increasing) .

For such a simple graph (two quadratics) why bother to differentiate?

1. For fun.
To show off latex skills

why sad face?

I do not see what you are doing in part(a) but this would be my approach

if x^2>16, i.e. x<-4 or x>4, then f(x) = x^2+2x-16

if x^2<16 i.e. -4<x< 4, then f(x) =- x^2+2x+16

the equals I would include at both ends but Pure Mathematicians might disagree and they would probably be right.

see if this helps

[ps a graph will also help]

Any question involving the absolute value function you will have to check all the cases

now as you said:



Now clearly when $2-2x=0 \implies x=1$ so x=1 is a critical point (max by 2nd derivative test)
but what about $2+2x=0 \implies x=-1$ but $-4<-1<4$ so cannot use second part of function here since it only applies when $x\le-4, x \ge 4$. So no solution in this case.

Going back to the end points of the first part of function gives $x=4, x=-4$ as minimums since the derivative goes from decreasing to increasing at this points (as x is increasing) .

It's not sufficient to simply solve the piecewise equations, you have to make sure the solutions fall in the correct domain for where the piecewise equations are defined.

Solving 2x+2 = 0 gives x = -1, but, f'(x) = 2x+2 is only valid in the ranges x <=-4 or x>=4. So you have to rule out x = -1 here.

Solving -2x+2 = 0 gives x = 1; f'(x) = -2x+2 is valid for -4 <= x < = 4 and so x=1 is a valid solution here.

Thanks for your reply!

Can someone answer the following two questions?

Are end points never relative extrema?
Relative extrema include absolute extrema?

End points do not have to be relative extrema - easy to think of an example for this e.g.

Easy to check end points are not extrema.

End points can be relative extrema though:


Relative extrema include absolute extrema: i not really sure what you mean since obviously they do.

Maybe you are confused: neither relative nor absolute extrema mean the (first) derivative is 0. The derivative is only 0 for a stationary point which may or may not be one of the functions extrema (could be an inflection). A relative extrema is just the greatest/lowest value the function takes on a particular interval. Now if one choose the interval with the absolute extrema then that extrema are the relative extrema on that interval as well as the absolute extreme of the whole function.

Hope this clears things up

Thanks!!!!
I'm not sure about the 1st question because http://tutorial.math.lamar.edu/Classes/CalcI/MinMaxValues.aspx says that "in order for a point to be a relative extrema we must be able to look at function values on both sides of to see if it really is a maximum or minimum at that point. This means that relative extrema do not occur at the end points of a domain. They can only occur interior to the domain." I agree with this. Given an interval, we just look at the values within that interval... We ignore what happens outside that. So I thought end points can never be relative extrema. Now I'm really confused!!! Your 2nd example seems right...