Unformatted text preview: f (x) = lim
âˆ†xâ†’0 âˆ†x[(x + âˆ†x)2/3 + (x + âˆ†x)1/3 x1/3 + x2/3 ] f (x) = lim = lim âˆ†xâ†’0 21. 1
1
= 2/3
(x + âˆ†x)2/3 + (x + âˆ†x)1/3 x1/3 + x2/3
3x f (t + h) âˆ’ f (t)
[4(t + h)2 + (t + h)] âˆ’ [4t2 + t]
= lim
hâ†’0
hâ†’0
h
h f (t) = lim 4t2 + 8th + 4h2 + t + h âˆ’ 4t2 âˆ’ t
hâ†’0
h = lim 8th + 4h2 + h
= lim (8t + 4h + 1) = 8t + 1
hâ†’0
hâ†’0
h = lim 22. 4
4
4
Ï€ (r + h)3 âˆ’ Ï€r3
Ï€ (r3 + 3r2 h + 3rh2 + h3 âˆ’ r3 )
dV
3
3
3
= lim
= lim
hâ†’0
hâ†’0
dr
h
h
4
= lim Ï€ (3r2 + 3rh + h2 ) = 4Ï€r2
hâ†’0 3 23. (a) D (b) F (c) B (d) C (e) A (f ) E 24. Any function of the form f (x) = x + k has slope 1, and thus the derivative must be equal to 1
everywhere.
y 3 1 2 x 2 25. (a) (b) y (c) y x y x 1 2 x 1 26. (a) (b) y x 27. (a) f (x) = x2 and a = 3 (c) y x (b) f (x) = âˆš y x x and a = 1 71 28.
29. 30. Chapter 3 (a) (b) f (x) = x7 and a = 1 f (x) = cos x and a = Ï€ [4(x + h)2 + 1] âˆ’ [4x2 + 1]
4x2 + 8xh + 4h2 + 1 âˆ’ 4x2 âˆ’ 1
dy
= lim
= lim
= lim (8x + 4h) = 8x
hâ†’0
hâ†’0
dx hâ†’0
h
h
dy
= 8(1) = 8
dx x=1 dy
= lim
dx hâ†’0 5
+1 âˆ’
x+h
h = limhâ†’0
dy
dx x=âˆ’2 5
+1
x 5x âˆ’ 5(x + h)
5
5
âˆ’
x(x + h)
= lim x + h x = lim
hâ†’0
hâ†’0
h
h 5x âˆ’ 5x âˆ’ 5h
âˆ’5
5
= lim
=âˆ’ 2
hâ†’0 x(x + h)
hx(x + h)
x =âˆ’ 5
5
=âˆ’
(âˆ’2)2
4 31. y = âˆ’2x + 1 32. 1.5 5 2 2
0 2.5
0 3 33. (b) h
(f (1 + h) âˆ’ f (1))/h 0.5
0.1
0.01
0.001 0.0001 0.00001
1.6569 1.4355 1.3911 1.3868 1.3863 1.3863 34. (b) h
(f (1 + h) âˆ’ f (1))/h 0.5
0.1
0.01
0.001
0.0001 0.00001
0.50489 0.67060 0.70356 0.70675 0.70707 0.70710 35. (a) dollars/ft (b)
(c) As you go deeper the price per foot may increase dramatically, so f (x) is roughly the price per
additional foot.
If each additional foot costs extra money (this is to be expected) then f (x) remains positive. (d) From the approximation 1000 = f (300) â‰ˆ
so the extra foot will cost around $1000. 36. f (301) âˆ’ f (300)
we see that f (301) â‰ˆ f (300) + 1000,
301 âˆ’ 300 (a) gallons/dollar (b) The increase in the amount of paint that would be sold for one extra dollar. (c) It should be negative since an increase in the price of paint would decrease the amount of paint
sold.
f (11) âˆ’ f (10)
we see that f (11) â‰ˆ f (10) âˆ’ 100, so an increase of one
From âˆ’100 = f (10) â‰ˆ
11 âˆ’ 10
dollar would decrease the amount of paint sold by around 100 gallons. (d) 37. (a) F â‰ˆ 200 lb, dF/dÎ¸ â‰ˆ 60 lb/rad (b) Âµ = (dF/dÎ¸)/F â‰ˆ 60/200 = 0.3 38. (a) dN/dt â‰ˆ 34 million/year; in 1950 the world population was increasing at the rate of about 34
million per year. (b) dN/dt
34
â‰ˆ
â‰ˆ 0.014 = 1.4 %/year
N
2490 Exercise Set 3.2 72 41. (a) T â‰ˆ 120â—¦ F, dT /dt â‰ˆ âˆ’4.5â—¦ F/min (b) 39. k = (dT /dt)/(T âˆ’ T0 ) â‰ˆ (âˆ’4.5)/(120 âˆ’ 75) = âˆ’0.1 lim f (x) = lim âˆš
3 x = 0 = f (0), so f is continuous at x = 0.
âˆš
3
f (0 + h) âˆ’ f (0)
hâˆ’0
1
= lim
= lim 2/3 = +âˆž, so
lim
hâ†’0
hâ†’0
hâ†’0 h
h
h
f (0) does not exist.
xâ†’0 y xâ†’0 2 2 42. lim f (x) = lim(x âˆ’ 2)2/3 = 0 = f (2) so f is continuous at x = 2. xâ†’2 xâ†’2 x 2 y f (2 + h) âˆ’ f (2)
h âˆ’0
1
= lim
= lim 1/3 which does not exist
hâ†’0
hâ†’0 h
h
h
so f (2) does not exist.
2/3 5 lim hâ†’0 x 2 43. lim f (x) = lim+ f (x) = f (1), so f is continuous at x = 1. xâ†’1âˆ’ xâ†’1 y f (1 + h) âˆ’ f (1)
[(1 + h) + 1] âˆ’ 2
= limâˆ’
= limâˆ’ (2 + h) = 2;
hâ†’0
hâ†’0
hâ†’0
h
h
f (1 + h) âˆ’ f (1)
2(1 + h) âˆ’ 2
= lim+
= lim+ 2 = 2, so f (1) = 2.
lim
hâ†’0+
hâ†’0
hâ†’0
h
h
2 5 limâˆ’ 3 44. lim f (x) = lim+ f (x) = f (1) so f is continuous at x = 1. xâ†’1âˆ’ x y xâ†’1 f (1 + h) âˆ’ f (1)
[(1 + h)2 + 2] âˆ’ 3
= limâˆ’
= limâˆ’ (2 + h) = 2;
hâ†’0
hâ†’0
hâ†’0
h
h
f (1 + h) âˆ’ f (1)
[(1 + h) + 2] âˆ’ 3
= lim+
= lim+ 1 = 1, so f (1)
lim
hâ†’0+
hâ†’0
hâ†’0
h
h
does not exist. 5 limâˆ’ 3 45. 3 3 x f is continuous at x = 1 because it is diï¬€erentiable there, thus lim f (1 + h) = f (1) and so f (1) = 0
hâ†’0 f (1 + h)
f (1 + h) âˆ’ f (1)
f (1 + h)
because lim
exists; f (1) = lim
= lim
= 5.
hâ†’0
hâ†’0
hâ†’0
h
h
h
f (x + h) âˆ’ f (x)
, but (with
h
f (h) + 5xh
y = h) f (x + h) = f (x) + f (h) + 5xh so f (x + h) âˆ’ f (x) = f (h) + 5xh and f (x) = lim
=
hâ†’0
h
f (h)
lim (
+ 5x) = 3 + 5x.
hâ†’0
h 46. Let x = y = 0 to get f (0) = f (0) + f (0) + 0 so f (0) = 0. f (x) = lim 47. f (x) = limhâ†’0 hâ†’0 f (x + h) âˆ’ f (x)
f (x)f (h) âˆ’ f (x)
f (x)[f (h) âˆ’ 1]
f (h) âˆ’ f (0)
= lim
= lim
= f (x) lim
hâ†’0
hâ†’0
hâ†’0
h
h
h
h
= f (x)f (0) = f (x) 73 Chapter 3 EXERCISE SET 3.3
1. 28x6 2. 5. 0 6. 9. 3ax2 + 2bx + c 13. f (x) = (3x2 + 6) 3. 24x7 + 2 4. 2x3 âˆš 1
7. âˆ’ (7x6 + 2)
3 8. 2
x
5 1
a 2 2x + âˆš
11. 24xâˆ’9 + 1/ x 1
b âˆ’3xâˆ’4 âˆ’ 7xâˆ’8 15. 10. âˆ’36x11 1
d
2x âˆ’
dx
4
= 18x2 âˆ’ 3 x + 12
2 14.
+ 2x âˆ’ 1
4 5
12. âˆ’42xâˆ’7 âˆ’ âˆš
2x 1
1
âˆšâˆ’2
2x x 1
d
(3x2 + 6) = (3x2 + 6)(2) + 2x âˆ’
dx
4 (6x) d
d
(7 + x5 ) + (7 + x5 ) (2 âˆ’ x âˆ’ 3x3 ) = (2 âˆ’ x âˆ’ 3x3 )(5x4 ) + (7 + x5 )(âˆ’1 âˆ’ 9x2 )
dx
dx
= âˆ’24x7 âˆ’ 6x5 + 10x4 âˆ’ 63x2 âˆ’ 7 16. f (x) = (2 âˆ’ x âˆ’ 3x3 ) d
d
(2xâˆ’3 + xâˆ’4 ) + (2xâˆ’3 + xâˆ’4 ) (x3 + 7x2 âˆ’ 8)
dx
dx
= (x3 + 7x2 âˆ’ 8)(âˆ’6xâˆ’4 âˆ’ 4xâˆ’5 ) + (2xâˆ’3 + xâˆ’4 )(3x2 + 14x) = âˆ’15xâˆ’2 âˆ’ 14xâˆ’3 + 48xâˆ’4 + 32xâˆ’5 17. f (x) = (x3 + 7x2 â...
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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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