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!

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- October 21st 2008, 06:03 PMpainterchica16Extrema of a closed interval??
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! - October 21st 2008, 06:51 PMJhevon
- October 21st 2008, 06:57 PMpainterchica16
I tried that, but when I set the derivative (1/3x^-2/3) equal to 0, I get x=0, and then I plug in -1 and 1 in the derivative and get one for both, which doesnt seem right. What am I doing wrong?

- October 21st 2008, 07:06 PMJhevon
if you got x = 0, then x = 0 would be the extreme point in [-1,1]. however, x = 0 is wrong. 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 - October 21st 2008, 07:25 PMpainterchica16
I'm trying this over and over again, but I still get x=0. I know you've already helped me so much, but could take it back from the derivative and go from there? This topic kinda confuses me more than it should and I'm still lost. Thank you so much!

- October 21st 2008, 07:33 PMJhevon
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 - October 21st 2008, 07:40 PMJhevon
be sure to familiarize yourself with what extreme points are and what absolute extreme points are. i think that's one of your problems here.

- October 21st 2008, 07:45 PMpainterchica16
Thank you so much! I really appreciate the help!