Abs. Max and Abs Min.

May 2010
8
0
Find the absolute maximum and absolute minimum values of f on the given interval.
f(x) = (x2 - 1)^3
[-1, 3]
 
Oct 2007
722
168
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 \(\displaystyle x=3\).