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Unformatted text preview: ) can change sign only if the factor L − 2P changes sign,
L
1
L
=
ln A.
, 1 = Ae−kLt , t =
which it does when P = L/2. From (5) we have
2
1 + Ae−kLt
Lk SUPPLEMENTARY EXERCISES FOR CHAPTER 5
4. (a) x3 x2
−
on [−2, 2]; x = 0 is a relative maximum and x = 1 is a relative
3
2
1
minimum, but y = 0 is not the largest value of y on the interval, nor is y = − the smallest.
6
False; an example is y = Supplementary Exercises 168 (b)
(c)
6. true
False; for example y = x3 on (−1, 1) which has a critical point but no relative extrema (a) (b) y 4 (c) y 4 4 x
2 7. (a) y x x 2 2 7(x − 7)(x − 1)
; critical points at x = 0, 1, 7;
3x2/3
neither at x = 0, relative max at x = 1, relative min at x = 7 (ﬁrst derivative test)
f (x) = f (x) = 2 cos x(1 + 2 sin x); critical points at x = π/2, 3π/2, 7π/6, 11π/6;
relative max at x = π/2, 3π/2, relative min at x = 7π/6, 11π/6
√
3 x−1
; critical points at x = 5; relative max at x = 5
(c) f (x) = 3 −
2 (b) x−9
27 − x
, f (x) =
; critical point at x = 9; f (9) > 0, relative min at x = 9
18x3/2
36x5/2 (a) f (x) = (b) 8. x3 − 4
x3 + 8
, f (x) = 2
;
x2
x3
1/3
1/3
critical point at x = 4 , f (4 ) > 0, relative min at x = 41/3 (c) 9. f (x) = 2 f (x) = sin x(2 cos x + 1), f (x) = 2 cos2 x − 2 sin2 x + cos x; critical points at x = 2π/3, π, 4π/3;
f (2π/3) < 0, relative max at x = 2π/3; f (π ) > 0, relative min at x = π ; f (4π/3) < 0, relative
max at x = 4π/3 lim f (x) = +∞, lim f (x) = +∞ x→−∞ y x→+∞ f (x) = x(4x2 − 9x + 6), f (x) = 6(2x − 1)(x − 1)
relative min at x = 0,
points of inﬂection when x = 1/2, 1,
no asymptotes 4
3
2
1 (1,2)
(0,1) ( 1 , 23 )
2 16
1 10. lim f (x) = −∞, lim f (x) = +∞ x→−∞ x→+∞ f (x) = x3 (x − 2)2 , f (x) = x2 (5x − 6)(x − 2),
f (x) = 4x(5x2 − 12x + 6) √
8 ± 2 31
critical points at x = 0,
√5
8 − 2 31
= −0.63
relative max at x =
5
√
8 + 2 31
= 3.83
relative min at x =
5
√
6 ± 66
= 0, −0.42, 2.82
points of inﬂection at x = 0,
5
no asymptotes (0.42,0.16) 2 x y
(0,0) (0.63,0.27) x
1 1 2 3 4 100 200
(2.82,165.00) (3.83,261.31) 169 11. Chapter 5 y lim f (x) doesn’t exist x→±∞ f (x) = 2x sec2 (x2 + 1),
2 4 2 2 2 f (x) = 2 sec (x + 1) 1 + 4x tan(x + 1) 2 critical point at x = 0; relative min at x = 0
2 x
2 1 1 2 point of inﬂection when 1 + 4x tan(x + 1) = 0 12. 2 π (n + 1 ) − 1, n = 0, 1, 2, . . .
2 vertical asymptotes at x = ± 4 lim f (x) = −∞, lim f (x) = +∞ x→−∞ 2 y x→+∞ f (x) = 1 + sin x, f (x) = cos x 4 critical points at x = 2nπ + π/2, n = 0, ±1, ±2, . . ., 2
−π no extrema because f ≥ 0 and by Exercise 51 of Section 5.1, x π
2 f is increasing on (−∞, +∞) 4 inﬂections points at x = nπ + π/2, n = 0, ±1, ±2, . . . 6 no asymptotes 13. x(x + 5)
2x3 + 15x2 − 25
, f (x) = −2 2
(x2 + 2x + 5)2
(x + 2x + 5)3
critical points at x = −5, 0; y f (x) = 2 1 relative max at x = −5, 0.8 relative min at x = 0 0.6 points of inﬂection at x = −7.26, −1.44, 1.20 0.4
0.2 horizontal asymptote y = 1 as x → ±∞ 14. x
20 10 3x2 − 25
3x2 − 50
, f (x) = −6
x4
x5
√
critical points at x = ±5 3/3;
√
relative max at x = −5 3/3,
√
relative min at x = +5 3/3 10 f (x) = 3 inﬂection points at x = ±5 20 y 5
x
4 2/3 6
5 horizontal asymptote of y = 0 as x → ±∞,
vertical asymptote x = 0
15. lim f (x) = +∞, lim f (x) = −∞ x→−∞ y x→+∞ x
x≤0
if
−2x
x>0
critical point at x = 0, no extrema 2 f (x) = inﬂection point at x = 0 (f changes concavity)
no asymptotes 1
2 1
2 x Supplementary Exercises 16. 170 5 − 3x
,
3(1 + x)1/3 (3 − x)2/3
−32
f (x) =
9(1 + x)4/3 (3 − x)5/3
critical point at x = 5/3; y f (x) = 4
2 relative max at x = 5/3 4 2 x 2
1 cusp at x = −1;
3 point of inﬂection at x = 3
oblique asymptote y = −x as x → ±∞
17. lim f (x) = +∞ y x→+∞ f (x) = 1 + ln x, f (x) = 1/x
lim f (x) = 0, lim+ f (x) = −∞ x→0+ 1 x→0 critical point at x = 1/e;
relative min at x = 1/e
1 2 x no points of inﬂection, no asymptotes 18. lim f (x) = +∞ y x→+∞ f (x) = x(2 ln x + 1), f (x) = 2 ln x + 3 0.2 0.4 x 0.6 lim f (x) = 0, lim+ f (x) = 0 x→0+ x→0 critical point at x = e−1/2 ; 0.1 −1/2 relative min at x = e point of inﬂection at x = e−3/2 19. 1 − 2 ln x
6 ln x − 5
, f (x) =
x3
x4
critical point at x = e1/2 , y f (x) = relative max at x = e1/2 x
1 2 1 5/6 point of inﬂection at x = e 2 horizontal asymptote y = 0 as x → +∞ 20. lim f (x) = +∞ y x→±∞ 2x
1 − x2
, f (x) = 2 2
+1
(x + 1)2
critical point at x = 0; f (x) = 2 x2 1 relative min at x = 0
points of inﬂection at x = ±1
no asymptotes x
2 2 171 21. Chapter 5 lim f (x) = +∞ y x→+∞ x−1
x2 − 2x + 2
, f (x) = ex
2
x
x3
critical point at x = 1; f (x) = ex 6 relative min at x = 1 2 no points of inﬂection 2 1 2 x 3 vertical asymptote x = 0,
horizontal asymptote y = 0 for x → −∞
22. f (x) = (1 − x)e−x , f (x) = (x − 2)e−x y critical point at x = 1; relative max at x = 1 0.2 point o...
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This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.
 Spring '14
 The Land

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