Calculus 1: Finding the maxima and minima of the function 1-x^(2/3)

[dez_cole]

Homework Statement

I am given the problem "A function is defined by f(x) = 1-x^(2/3).

a. find the minima and maxima of the function. State whether they are relative or absolute.

b. graph the function."

Homework Equations

I found the derivative and set it equal to zero. y' = -(2/3)x^(-1/3), -(2/3)x^(-1/3) = 0

The Attempt at a Solution

Now I really have little idea of where to go from here because I am left with a few questions. If i set the derivative equal to zero and set x to be zero would the statement "-(2/3)0^(-1/3) = 0" be true or would I essentially be dividing by zero because x^(-1/3) = 1/ \sqrt[3]{x}? Now if this is a spot where the velocity is zero would you consider it a max, a min, or would it be irrelevant to my answer because you can't quite classify it either way. I think it is simply an interesting value on the graph, but is neither a maximum or minimum relative or absolute, and this graph has no maxes or mins.

Thanks for your help. I really appreciate it.

 

[Dick]

Sketch a graph. You are correct that the derivative at x=0 doesn't exist. It still might be a max or a min. It's an "interesting point" as you put it.

 

[dez_cole]

Thanks again.