How to find the absolute maximum and minimum for the function f(x)=x^3 between the limits [-2,2]?
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?
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.