Page 132 - Calculus Workbook For Dummies

116 Part III: Differentiation

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5
^
7
k Find the absolute extrema of p x = ^h x + 1h - . x on the interval - , 2 31A. Absolute max at
(–2, 2); absolute mins at (–1, .5) and (31, .5).
I think you know the steps by now.
/ 4 5
5
p x = ^h x + 1h - . x
^
4 - / 1 5
p x = ^ x + 1h - .5
l ^ h
5
4
= / 1 5 - .5
5^ x + 1h
4
0 = / 1 5 - .5
5^ x + 1h
4
.5 = / 1 5
5^ x + 1h
/ 1 5
. 2 5^ x + 1h = 4
/ 1 5 8
^ x + 1h =
5
5
8
1 = c m
^ x + h
5
x = . 9 48576
That’s one critical number, but x = –1 is also one because it produces an undefined derivative.
/ 4 5
1 = - +
^
^
p - h ^ 1 1h - .5 - 1h
= .5
/ 4 5
+
p ^ . 9 48576 = ^h . 9 48576 1h - .5 9 ^ .48576h
. . 1 81072
/ 4 5
2 = - +
2 =
^
Left endpoint: p - h ^ 2 1h - .5 - h 2
^
/ 4 5
^
h
Right endpoint: p 31 = ^h 31 + 1h - .5 31 = 16 - 15 .5 = .5
^
Your absolute max is at the left endpoint: (–2, 2). There’s a tie for the absolute min: At the cusp:
(–1, .5) and at the right endpoint: (31, .5).
π 5 π
2
l Find the absolute extrema of q x = 2 cos x + 4 sinx on the interval - ; 2 , 4 E. Absolute min at
^ h
c - π , - 6m; absolute maxes at c π , 3m and c 5 π , 3m.
2 6 6
2
q x = 2 cos x + 4 sinx
^ h
2 $
q x = - 2 sin x 2 + 4 cosx
l ^ h
2
0 = - 4 sin x + 4 cosx
0 = sin x - cosx _ dividing by - 4i
2
x
0 = 2 sin cosx - cosx _ trig identityi
0 = cosx 2 ^ sinx - 1h
2 sinx - = 0
1
cosx = 0
1
- π π or sinx =
x = , 2
2 2
,
x = π 5 π
6 6