# Extrema of a closed interval??

• Oct 21st 2008, 05:03 PM
painterchica16
Extrema 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]

• Oct 21st 2008, 05:51 PM
Jhevon
Quote:

Originally Posted by painterchica16
How do I approach this problem?
Find the extrema of f(x)=x^(1/3) on the closed interval of [-1,1]

the same way you find extrema anywhere. find the derivative, set it equal to zero, solve for x. then just pick the x's that fall in that interval. then classify each as max, min, or inflection point
• Oct 21st 2008, 05:57 PM
painterchica16
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?
• Oct 21st 2008, 06:06 PM
Jhevon
Quote:

Originally Posted by painterchica16
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?

if you got x = 0, then x = 0 would be the extreme point in [-1,1]. however, x = 0 is wrong. $\frac 13 x^{-2/3} = \frac 1{3 \sqrt[3]{x^2}}$ 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
• Oct 21st 2008, 06:25 PM
painterchica16
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!
• Oct 21st 2008, 06:33 PM
Jhevon
Quote:

Originally Posted by painterchica16
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!

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
• Oct 21st 2008, 06:40 PM
Jhevon
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.
• Oct 21st 2008, 06:45 PM
painterchica16
Thank you so much! I really appreciate the help!