# Abs. Max and Abs Min.

#### sheva2291

Find the absolute maximum and absolute minimum values of f on the given interval.
f(x) = (x2 - 1)^3
[-1, 3]

Find the absolute maximum and absolute minimum values of f on the given interval.
f(x) = (x2 - 1)^3
[-1, 3]
You have to check three places. Critical points and both end points. These are the only places we can have a minimum or maximum on this interval
For the end points we have $$\displaystyle f(-1) = 0$$ and $$\displaystyle f(3) = 512$$.

Critical points.

$$\displaystyle f'(x) = 6x(x^2 - 1)^2$$

Set this equal to zero we find that $$\displaystyle x = 0$$, $$\displaystyle +1$$ or $$\displaystyle -1$$.

Put these value back into $$\displaystyle f(x)$$ to get...

$$\displaystyle f(0) = -1$$
$$\displaystyle f(1) = f(-1) = 0$$

Hence, from this you can conclude that the minimum is at $$\displaystyle x=0$$ and the maximum is at $$\displaystyle x=3$$.