How can you find the relative extrema of |1+\sqrt[3]{x}|?

[John O' Meara]

Use any method to find the relative extrema of |1+\sqrt[3]{x}|.
I get the following |1+\sqrt[3]{x}| = \left\{\begin{array}{cc} 1+\sqrt[3]{x}, & \mbox{if} x >-1 \\ -1-\sqrt[3]{x}, & \mbox{if} x<-1\end{array}\right
Using the first derivative test, I get f'(x) = \frac{1}{3x^{2/3}} \mbox{and} -\frac{1}{3x^{2/3}}. These suggest that there is a relative extremum at x=0 but not at x=-1, actually it is an inflection point at x=0. What is next? Maybe I am confused about absolute values! I think I have the function piece wise ok.Please help. Thanks.

 

[mathman]

At x=0, the derivative is infinite. I am confused about what you are confused about.

 

[John O' Meara]

Should not f'(x) =0 show a relative minimum at x=-1, i.e., if f(x) is differentiable at x=-1. According to my graphing calculator there is a relative minimum at x=-1. But I do not know to find the relative minimum at x=-1 by calculus? Thanks.

 

[arildno]

f(x) is not differentiable at x=-1.

That can be seen by the different values the right-hand derivative and left-hand derivative gets there.

Thus, x=-1 is a sharp corner, and it is, indeed, a relative extremum.