Find the absolute maximum and minimum values of the
function f(x) = x4 – 2x2 + 2, [–2, 4].
Sol:
Since f is continuous on [–2,
4], we can use the Closed interval method.
f(x)
=
x4 – 2x2 + 2
f '(x)
=
4x3 – 4x
=
4x(x2 – 1)
=
4x(x – 1)(x + 1)
Since f '(x) exists for all 'x', the only critical numbers of f
occur when f '(x) = 0, that is, x = 0
or x = 1 or x = –1. Notice that each of these critical numbers lies in the interval (–2, 4).
The values
of f at these critical numbers are: f(0) = 2, f(1) = 1, and
f(–1) = 1.
The values of f at the endpoints of the interval are: f(–2) =
10 and f(4) = 226.
Comparing these five numbers, we see that the absolute maximum value is f(4) = 226 and the absolute
minimum value is f(±1) = 1.