# Confused with finding the Absolute min of f(x)=x^(2/3)

• Jul 14th 2010, 12:07 PM
kensington
Confused with finding the Absolute min of f(x)=x^(2/3)
Find the absolute maximum and minimum values of f(x)=x^(2/3) on the interval [-2,3]

So i found f'(x) =(2/3)x^(-1/3)
Then when f'(x)=0 solved for x which i got x=0
Next i pluged in x=0,-2, 3 into f(x) and got
f(0)=0
f(-2)=-1.58
f(3)= 2.08

soo the absolute max is f(3)=2.08 and im having trouble finiding the absolute min.

I drew out f(x) on my graphing calculor and it came out looking like the absolute value of x so then f(0) would be the absolute min. But without looking at the calculator i would have assumed f(-2) is the absolute min.... So which is correct and why?
• Jul 14th 2010, 12:13 PM
Also sprach Zarathustra
abs. min, is f(-2)...

By some theorem.... If you have continuous increasing function on [a,b] then the abs. max is f(b) and abs. min is f(a).
• Jul 14th 2010, 12:27 PM
kensington
Quote:

Originally Posted by Also sprach Zarathustra
abs. min, is f(-2)...

By some theorem.... If you have continuous increasing function on [a,b] then the abs. max is f(b) and abs. min is f(a).

Thank you it just didnt seem to match the graph i was getting.. according to the graph the absolute min was 0 =/
• Jul 14th 2010, 12:39 PM
Ackbeet
The numbers you've gotten by plugging into the function are incorrect. Calculators are going to have trouble plotting this function, because of the algorithm they use to plot. I believe it's some sort of exponential or logarithmic algorithm. Obviously, your function is defined for all real numbers. You'll have to plot $(x^{2})^{1/3}$ to get a correct picture on some calculators. Plugging in -2 does not give me -1.58. Try again, with the formula I just gave.
• Jul 14th 2010, 12:48 PM
kensington
Quote:

Originally Posted by Ackbeet
The numbers you've gotten by plugging into the function are incorrect. Calculators are going to have trouble plotting this function, because of the algorithm they use to plot. I believe it's some sort of exponential or logarithmic algorithm. Obviously, your function is defined for all real numbers. You'll have to plot $(x^{2})^{1/3}$ to get a correct picture on some calculators. Plugging in -2 does not give me -1.58. Try again, with the formula I just gave.

OH ok soo i just put in (x^2)^1/3 into my graphing calculator and i got a parabola that is vertically stretched so once again f(0) is the min according to the graph. Then i relized u mentioned you got a different value for f(-2) and your correct. U didnt put the bracket around the -2 soo i ended up with -1.58 while its suppose to be 1.58 which then makes sence because now f(0) will be my absolute min.. Thanks sooooo much! :) haha stupid mistake