Thomas Calculus 13th [Solutions]

Section 4.3 Monotonic Functions and the First Derivative Test 239
f ( b) f ( a)
74. From the Mean Value Theorem we have
f ( c ) where c is between a and b. But f ( c) 2 pc q 0
b a
q
has only one solution c
2 p
. (Note: p 0 since f is a quadratic function.)
75. Proof that ln b ln b ln x : Note that d ln b 1 b 1 and d (ln b ln x ) 1;
by Corollary 2 of the
x
dx x b/
x 2
x x dx
x
Mean Value Theorem there is a constant C so that ln b ln b ln x C ; if x = b, then
x
ln 1 = ln b ln b + C C = 0 ln ln b ln x.
b
x
76. (a) d 1 1
(tan x cot x ) 1 1 0 by Corollary 2 of the Mean Value Theorem that
dx 2 2
1 x 1 x
1 1
tan x cot x C for some constant C; if x = 1, then
1 1
1 1
tan 1 cot 1 C tan x cot x .
4 4 2
2
(b)
d
dx
sec
1 1 1 1
(sec x csc x ) 0 by Corollary 2 of the Mean Value Theorem that
1 1
2 2
x x 1 x x 1
x csc x C for some constant C; if x 2, then
1 1
sec 2 csc 2 C
4 4 2
1 1
sec x csc x .
2
77.
x x ( x x) 0 x
1
1
x
e
e e e e e for all x;
x1
x1 1 x1 x2 x1 x
e e e e
2
x2 x2
e
e
e
x x x ln y x x x x x x x x
y ( e ) ln y x2 ln e x2 x1 x1 x2
e e y e ( e ) e . Likewise,
x2 x1 x2x1 x1 x
e e e 2
78. 1 2 1 1 2 1 2 1 2 1 2
( ) .
4.3 MONOTONIC FUNCTIONS AND THE FIRST DERIVATIVE TEST
1. (a) f ( x) x( x 1) critical points at 0 and 1
(b) f | | increasing on ( , 0) and (1, ), decreasing on (0, 1)
0 1
(c) Local maximum at x 0 and a local minimum at x 1
2. (a) f ( x) ( x 1)( x 2) critical points at 2 and 1
(b) f | | increasing on ( , 2) and (1, ), decreasing on ( 2, 1)
2 1
(c) Local maximum at x 2 and a local minimum at x 1
3. (a)
(b)
2
f ( x) ( x 1) ( x 2) critical points at 2 and 1
f | | increasing on ( 2, 1) and (1, ), decreasing on ( , 2)
2 1
(c) No local maximum and a local minimum at x 2
4. (a)
(b)
2 2
f ( x) ( x 1) ( x 2) critical points at 2 and 1
f | | increasing on ( , 2) ( 2, 1) (1, ), never decreasing
2 1
(c) No local extrema
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