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3.1 Extrema on an Interval
Don’t get behind the
Do your homework meticulously!!!
Definition of Extrema
1. f(c) is the minimum of f on I if f(c) £ f(x) " x in I.
2. f(c) is the maximum of f on I if f(c) ! f(x) " x in I.
If f is continuous on a closed interval [a,b], then f has
both a minimum and a maximum on the interval.
If f is defined at c, then c is called a critical number
of f if f’(c) = 0 or if f’(c) is undefined at c.
Relative Extrema occur only at critical numbers.
If f has a relative min or relative max at x = c, then
c is a critical number of f.
Ex. Find the absolute extrema of f(x) = 3x4 – 4x3
on the interval[-1, 2].
f’(x) = 12x3 – 12x2
0 = 12x3 – 12x2
0 = 12x2(x – 1)
x = 0, 1 are the critical numbers
Evaluate f at the
f(-1) =
endpoints of [-1, 2]
f(0) =
and the critical #’s.
f(1) =
f(2) =
7
0
-1
16
minimum
maximum
Ex. Find the absolute extrema of f(x) = 2x – 3x2/3
on [-1, 3]
2
-1 3
f ' ( x) = 2 - 2 x
= 2- 13
x
(
)
2 x1 3 - 2 2 x1 3 - 1
=
=
=0
13
13
x
x
C. N. ‘s are x = 1 because f’(1) = 0
and x = 0 because f’ is undefined
Evaluate f at the
endpoints of [-1, 3]
and the critical #’s.
f(-1) = -5 minimum
f(0) = 0 maximum
f(1) = -1
f(3) = -.24
Ex.
Find the absolute extrema of
f(x) = 2sin x – cos 2x on [0,2p ].
f’(x) = 2cos x + 2sin 2x
sin 2x = 2 sin x cos x
0 = 2cos x + 2(2sin x cos x)
Factor
0 = 2cos x(1 + 2sin x)
p
3p
@ and
2
2
7p
11p
sin x = -1/2 @ and
6
6
Evaluate f at the
endpoints of [0,2p ]
and the critical #’s.
2cos x = 0
cos x = 0
f (0) =
Êp ˆ
fÁ ˜=
Ë2¯
Ê 7p ˆ
fÁ
˜=
Ë 6 ¯
Ê 3p ˆ
fÁ ˜=
Ë 2 ¯
Ê 11p ˆ
fÁ
˜=
Ë 6 ¯
f (2p ) =
-1
3 max
-3 min
2
-1
-3 min
2
-1
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