How do I approach this problem?
Find the extrema of f(x)=x^(1/3) on the closed interval of [-1,1]
Help! Please and thank you!
if you got x = 0, then x = 0 would be the extreme point in [-1,1]. however, x = 0 is wrong. $\displaystyle \frac 13 x^{-2/3} = \frac 1{3 \sqrt[3]{x^2}}$ is never zero.
if this is a problem that asks you about the absolute extrema, then it remains for you to test the end points. the one that gives the largest value is your absolute max, the one that gives the smallest value is the absolute min
and to get the points, you do not plug into the derivative, you plug into the original function. you want the value of the function at that point
your derivative is correct. if you set it equal to zero, there is no solution. you have x in the denominator, it cannot be zero!
now, test the end points. that is, find f(-1) and f(1). the larger one of those values is you absolute max, and the smaller is your absolute min