# Application of derivative help

• Sep 10th 2009, 07:33 PM
gundu
Application of derivative help
How to find the absolute maximum and minimum for the function f(x)=x^3 between the limits [-2,2]?
• Sep 10th 2009, 07:41 PM
seld
for a function like f(x)=x^3 there is no absolute minimum or maximum. 3rd degree polynomials have relative mins and maxs.

Though your question is asking what's the smallest value and largest value of y=x^3, in the interval [-2,2].

well you know the derivative of f(x)=x^3 is:

\$\displaystyle f'(x)=3x^2\$

So you know the function is always increasing, so your maximum in that interval will be where x is at its largest value.

Since f(x) is always increasing I think you can also say the smallest value is when x is smallest. (i'm pretty sure though not 100% sure on this, it definitely makes sense though)

does that help?
• Sep 10th 2009, 09:31 PM
gundu
Ya thanks that helps
• Sep 11th 2009, 07:15 AM
HallsofIvy
Actually, any continuous function has absolute max and min on a closed and bounded interval like [-2, 2]. They must occur where the derivative is 0, or the derivative does not exist, or at the endpoints. Since the derivative of \$\displaystyle x^3\$, \$\displaystyle 3x^2\$ always exists and is equal to 0 only at x= 0, that means you have 3 points to check.