Relative extrema of absolute functions

[John O' Meara]

Find the relative extrema of $$f(x)=|3x-x^2|$$.
Normally you solve f'(x)=0 to find the critical points, in that case we have 3-2x=0 => x=3/2. However there are two other relative extrema that I cannot find by calculus as the absolute nature of f(x) has confused me. Indeed doing a sign analysis on the points x=0 and 3; shows that these points are not extrema points! I am studing on my own, Please help, Thanks.

 

[CRGreathouse]

Split the function into two parts, where neither have absolute value bars,using the definition of absolute value.

Then find the critical points of each, and adjoin the point where the two meet (probably another relative extremum).

 

[John O' Meara]

Ok,
|x| = x if x > 0, -x if x < 0. |3x-x^2|= 3x-x^2 if x>3, I do not know the rest?

 

[John O' Meara]

I think I have it now, |3x-x^2|= (x-3)^2 if x>3, 3x-x^2 if 0< x <3, -x^2 if x<0

 

[HallsofIvy]

Find the relative extrema of $$f(x)=|3x-x^2|$$.
Normally you solve f'(x)=0 to find the critical points

Your first statement is incorrect. "Critical points" are points where the derivative is 0 or where the derivative does not exist.

, in that case we have 3-2x=0 => x=3/2. However there are two other relative extrema that I cannot find by calculus as the absolute nature of f(x) has confused me. Indeed doing a sign analysis on the points x=0 and 3; shows that these points are not extrema points! I am studing on my own, Please help, Thanks.

The derivative does not exist where f(x)= 0. That is, at x=0 or at x= 3. Since the value of f is 0 there and the absolute value cannot be negative, they are obviously local minima.