The closed interval method
To find the absolute maximum and minimum values of a continuous function f on a closed interval [a,b]:
1. Find the values of f at the critical numbers of f in (a,b).
2. Find the values of f at the endpoints of the interval.
3. The largest of the values from Steps 1 and 2 is the absolute maximum value; the smallest of these values is the absolute minimum value.
f(x)=6(x2−1)3,[−1,2]
1. Find the first derivative with respect to x
f′(x)=(6(x2−1)3)′=6(3)(x2−1)2(2x)=36x(x2−1)2 Find the critical number(s)
f′(x)=0=>36x(x2−1)2=0 x1=−1,x2=0,x3=1
f(−1)=6((−1)2−1)3=0
f(0)=6((0)2−1)3=−6
f(1)=6((1)2−1)3=0
2.
f(−1)=6((−1)2−1)3=0
f(2)=6((2)2−1)3=162
3. Absolute maxima
x=2 y=162
Absolute minima
x=0 y=−6
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