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Unformatted text preview: 120 I: CHAPTERZ LIMITS AND DERIVATIVES 2.8 Derivatives The line from P(2. f(2)) to Q(2 + h. f(2 + h)) f(2 + h) — my is the line that has slope h 2. As It decreases. the line PQ becomes steeper. so its slope increases. So 0 < “4313(2) < “3; : gm < £312 “3": : 52). Thus. 0 < § [f(4) — f(2)] < f(3) — f(2) < f’(2). 3. g’(0) is the only negative value. The slope at a: : 4 is smaller than the slope at x : 2 and both are smaller than the
slope atav : —2. Thus. g’(0) < 0 < g'(4) < g’(2) < g'(—2). 4. Since (4. 3) is on y : f(4) = 3. The slope of the tangent line between (0, 2) and (4. 3) is i. so f’ (4) : i. 5. We begin by drawing a curve through the
origin at a slope of 3 to satisfy f(0) : 0 and
f’(0) : 3. Since f’(1) : 0. we will round
off our ﬁgure so that there is a horizontal tangent directly over z : 1. Lastly. we make sure that the curve has a slope of 71 as we
pass over :8 : 2. Two of the many possibilities are shown. 7. Using Deﬁnition 2 with f(:c) : 3:52 — 5:3 and the point (2. 2), we have
f(2 + h) — f<2> [3(2 + h)? e 5<2 + M] e 2 fmﬂa—ﬁf—Zﬂ h
12 12h 3h2~10—5h —2 . 2
— lim ( + + ) * lim 3h + 7h 2 lim (3h+ 7) : 7,
h—>0 h hao h hao So an equation of the tangent line at (2, 2) is y a 2 : 7(m — 2) or y : 7x 7 12. 8. Using Deﬁnition 2 with g(;c) : 1 7 3:3 and the point (07 1). we have 3 r 3
I _ ~ 9(0+h)'g(0)# ‘ [1i(0+h)]#1_~ (1h)_1:1‘ _h2 :0
9 <0> * Ila, ff—h  I12}. h ' I135 f}. m > ' So an equation of the tangent line is y a 1 : 0(36 — 0) or y : 1. SECTION 2.8 DERIVATIVES 9. (a) Using Deﬁnition 2 with : :53 i 53: + 1 and the point (17 —3). we have F(1+h) _ Fm [(1+ 103 V 5(1+h)+ 1] 7 (—3) , i. _.
“U113; h 9.11% h
. (1+3h+3h2+h375—5h+1)+3 . h3+3h2—2h
—11m —11m
hao h h—vO h
2 i
=1imW:1im(h2+3he2)=_2
h—rO h h—rO So an equation of the tangent line at (1. —3) is y ~ (—3) = 72(m ~ 1) <:> y = —2m — 1. (b) 8 a—l—h i a
— 1 2 h 1 2
10. (a) 0’01): 113%) L” *2 CW : £13) \+ (‘H h) + a
, a+202+h+2ah—a—2az~2ah . 1 ,2
— 1 1 _ 1 2
133%) h(1+2a+2h)(1+2a) 1.13%) (1+2a+2h)(1+20) ( J“ ‘1) So the slope of the tangent at the point (~ﬁ, — is (b) m 2 [1+ 2(—%)] _2 : 4. and thus an equation is ﬂow/2430+; _ 1+h _ 1
11. (a) f’(1) : gigw : (b) 3.4 So let F(h) 2 We calculate:
l h W h Fm 
01 3.484 ~01 3.121 
0.9 ‘ 1.1
0.01 3.314 —0.01 3.278 26
0.001 3.298 —0.001 3.294 From the graph. we estimate that the
0.0001 3.296 —0.0001 3.296 slope of the tangent is about 3.2 — 2.8 0.4 1.06 — 0.94 I 0.12 z 3'3‘ We estimate that f’(1) m 3.296. j 121 2! CHAPTER 2 LIMITS AND DERIVATIVES I W I E + h _ E
, 7r 7r 1.
_ lim “m5 + h) 1 “1(3)
h—»0 h.
t 1 h —1
SoletG(h) = “(4: )
weca'w'a‘e: i—
073 0.83
 h C(h) h C(h) 0.85
0.1 2.2305 ~01 1.8237 From the graph. we estimate that the slope of the
0.01 2.0203 —0.01 1.9803 . 1.07 — 0.91 _ 0.16 i
tangent is about W _ m —
0.001 2.0020 90.001 1.9980 ‘ — ‘ '
0.0001 2.0002 70.0001 1.9998 We estimate that g’ = 2. 13. Use Deﬁnition 2 with ﬂat) : 3 — 2w + 4332. f(a + h) — f(a.) [3 — 2((1 + h) + 4(a. + M2] — (3 — 2a + 4112) I u .
f (a) * 11.11% h ' 112% h
. (3—2a72h+4a2+8ah+4h2)—(3—2a+4a2)
— hm
1H0 h
2 _
: lim W A lim M 2+8a+4h)=1im( 2+8a+4h): 2+8a
h—»0 h 11—.0 h h—vO
h— _ a+h4—5a+h #a4—5a
14' Ma) _ hm f(a+ ) f(a) i hm I( ) ( )I ( )
h_.o h we h
(a4 + 4a3h I 6a2h2 + 4ah3 + h4 — 5a — 5h) — (a4 — 5a)
— lim
h—>0 h
4.1% + 6a2h2 + 4ah3 +114 9 5h. _ h(4a3 + 6a2h + 4ah2 + h3 — 5)
i 11111 — 11m
11—»0 h h—»0 h = lim (4a3 + 6a2h + 4ah2 + h3 i 5) : 4a3 — 5 h—>O 15. Use Deﬁnition 2 with f(t) : (215+ 1)/(t + f’(a) = 11111
hat) Wﬁaﬂ
f(a+h)f(a) (a+h)+3 a—I—3
h h
(2a+2h+1)(a+3)#(2a+1)(a+h+3) : lim
hao lim
h>0 h(a+h+3)(a+3)
(2a.2+6a+2ah—I6h+a+3) —(2a2+2ah+6a+a+h+3) 11m
h—tO 11111
haO h(a+h+3)(a+3)
5h _ mm 5 2 5
h(a+h+3)(a+3) #12a0(a+h+3)(a+3) (a+3)2 SECTION 2.3 DERIVATIVES 3 123
(a+h)2+1 a2+1 15If/(a)zimwzggw+—M_ﬁ_2
. (a2+2ah+h2+1)(a—2)—(a2+1)(a+h—2)
* 333% h(a + h. w 2)(a 7 2)
. (a3—2a2+2a2h—4ah+ah2—2h2+a—2)—(a3+a2h—2a2+a+h—2)
731113) h(a+h—2)(a—2)
a2h—4ah+ah2—2h2_h . h(a2—4a+ah~2h—1)
253% h(a+h—2)(a—2) ‘23}; h(a+h~2)(a~2)
:hma2—4a—l—ah—2h—1:a2—4a71
h—»0 (a+h—2)(ai2) ((1—2)2 17. Use Deﬁnition 2 with f(;v) : 1/\/z + 2. 1 l
, *1. f(a+h)—f(a)_lim (a+h)+2 ¢a+2_
f(a)—hl—>Inﬂ h *h—vO h
Va+ *\/a+h+2
Him x/a+h+2\/a+2 ehm x/a+2—\/a+h+2 x/a+2+\/(_l+h+2
_h—»0 h — ’HO hx/(T+h+2\/a—+2 'x/Zz+2+\/a+h+2
ﬁlim (a+2)—(a+h+2)
’HOhx/a+h+2\/a+2(\/a+2+\/a+h+2) 42
=l'm
ill—+0hx/a+h+2\/a+2(\/(T+2+\/(T+h+2)
—1
*1.
11% \/a+h+2\/a_+2(\/a+2+\/E+h+2)
*1 1 (Mfg/m) 2(a+2)3/2
‘/3(a+h)+1~ V3a+1 13. f’(a):gﬂw:gg h
1‘ (x/ga+3h+1~\/3a+1)(\/3a+3h+1+\/3Ta+1)
* 1m
hen h(3a+3h+1+\/?T+1) _ hm (3a+3h+ 1) e (3a+ 1) hm 3h
ha0h(\/3a+3h+1+\/§a+1) h—>0h(\/3a+3h+1+\/3a+1) — lim 3 3
h~0\/3a+3h+1+\/3a+1 7 2x/3a+1 Note that the answers to Exercises 19 —24 are not unique. 10 _
19. By Deﬁnition 2.}1Lirrb W : f’(1).where = $10 and a : 1.
—> L
10 _
0r: By Deﬁnition 2. lim (1 + h) 1 hﬁo T : f'(0). where : (1 +3010 anda = 0‘ 124 CHAPTER 2 LIMITS AND DERIVATIVES 6/16 + h — 2 ——'h— : f/(IG). where : ﬁand a :16.
{1/16 + h v 2 _
————h _ 20. By Deﬁnition 2. lim
—>0 0r: By Deﬁnition 2. illim
,—>0 21. By Equation 3. lim 2 _ 32
za5 x e 5 f’(0). where f(;c) : {1/16 + x and a : 0. = f/(S). where f(:c) : 2”” and a = 5. . . ta x i 1 ,
22. By Equation 3. 11m —n—— i f (7r/4). where f(.’L') : tanm and a = 7r/4. ac—nr/4 LE — 7T/4 i cos(7r + h) + 1 23. By Deﬁnition 2. lim : f’(7r). where f(3:) : COS$ and a : 7r. h—»0 h
. . . . cos(7r + h) + 1 ,
0r. By Deﬁnition 2.11mi) ——h—— : f where : cos(7r + :6) and a : 0.
t4 + t i 2 24. By Equation 3.1im = f’(1). where f(t) : t4 + t and a : 1. t—>1 f(2 + h) i f(2) [(2 + h)2 # 6(2 + h) — 5] — [22 7 6(2) — 5] li1r1 25. 0(2) — f’(2) # lim haO h h—»0 h
(4+4h+h2—12r6h#5)7(~13) h272h
_ 1' _  : ~ _ I _ h £3310 h 113%)“ 2) 2 W5
. 2 + h 7 2
“1 [2(2 + h)3 g (2 + h) + 1] k [2(2)3 i 2 + 1]
‘ h—g) h,
1, (2h3+12h2+24h+16*2’h+1)— 15
# hlirb h
2h3 12 2 2
: limb + I? + 3" = limb (2h2 + 12h + 23) = 23 m/s 27. (a) f '(m) is the rate of change of the production cost with respect to the number of ounces of gold produced. Its
units are dollars per ounce. (b) After 800 ounces of gold have been produced. the rate at which the production cost is increasing is $17/ ounce. So the cost of producing the 800th (or 801st) ounce is about $17. (c) In the short term. the values of f will decrease because more efﬁcient use is made of start—up costs as :c increases. But eventually f' might increase due to large—scale operations. 28. (a) f I(5) is the rate of growth of the bacteria population when t : 5 hours. Its units are bacteria per hour (b) With unlimited space and nutrients. f’ should increase as it increases; so f’(5) < f’(10). If the supply of
nutrients is limited. the growth rate slows down at some point in time. and the opposite may be true. 29. (a) f'(v) is the rate at which the fuel consumption is changing with respect to the speed. Its units are (gal/m/(mi/h).
(b) The fuel consumption is decreasing by 0.05 (gal/h)/(mi/h) as the car‘s speed reaches 20 mi/h. So if you
increase your speed to 21 mi / h. you could expect to decrease your fuel consumption by about 0.05 (gal/h)/(mi/h). SECTION 2.8 DERIVATIVES 125 30. (a) f’ (8) is the rate of change of the quantity of coffee sold with respect to the price per pound when the price is
$8 per pound. The units for f’(8) are pounds/(dollars/pound). (b) f '(8) is negative since the quantity of coffee sold will decrease as the price charged for it increases. People are generally less willing to buy a product when its price increases. 31. T’(10) is the rate at which the temperature is changing at 10:00 AM. To estimate the value ofT’(10), we will average the difference quotients obtained using the times t 2 8 and t 2 12. Let
T(8) — T(10) ‘ 72 — 81 T(12) — T(10) : 88 2 81 _ : = = .5. Th
— 8—10 _2 4.5andB 12710 2 3 en
. T(t) 2 T(lO) A + B 4.5 + 3.5 O
I z z = _ 4 F h.
T (10) 33110 t2 10 2 2 / 32. For 1910: We will average the difference quotients obtained using the years 1900 and I920.
E(1900) 2 E(1910) _ 48.3 — 51.1 2 g : 0.28 d
W A 1900 2 1910 10 an
_ E(1920) 190910) 7 552 _ 51'1 , 0.41. Then
1920 21910 10 . _ . E(t)~E(1910)~A+B
“mm—1211113110 16—1910 "V 2 at about 01345 year/year in l9l0. 2 0.345. This means that life expectancy at birth was increasing For 1950: Using data for I940 and I960 in a similar fashion, we obtain E’(1950) % [0.31 + 0.10]/2 2 0.205. So
life expectancy at birth was increasing at about 0.205 year/year in 1950. 33. (a) S’ (T) is the rate at which the oxygen solubility changes with respect to the water temperature. Its units are
(mg/L)/°C (b) For T 2 16°C. it appears that the tangent line to the curve goes through the points (07 14) and (32, 6). So 6 — 14
S’(16) z 32 _ 0 2 —% 2 —0.25 (mg/L)/°C. This means that as the temperature increases past 16°C. the oxygen solubility is decreasing at a rate of 0.25 (mg/L)/QC. 34. (a) S'(T) is the rate of change of the maximum sustainable speed of Coho salmon with respect to the temperature.
Its units are (cm/s)/°C. (b) For T 2 15 0C. it appears the tangent line to the curve goes through the points (10, 25) and (20, 32). So , 32 — 25 O . .
S (15) m 20 _ 10 2 0.7 (cm/s)/ C. This tells us that at T 2 15 0C. the max1mum sustainable speed ofCoho salmon is changing at a rate of 0.7 (cm/s)/°C. In a similar fashion for T 2 25 OC. we can use the points . 25 — 35
(20, 35) and (25, 25) to obtain S/(25) % 25 _ 20 2 —2 (cm/s)/°C. As it gets warmer than 20 0C, the maximum sustainable speed decreases rapidly. 35. Since f(ac) 2 msin(1/;v) when :U 2 0 and f(0) 2 0. we have , . 0+h—0 ,h‘lh—O .. f (0) 2 513% W 2 21115 sum+ 2 IILII'I'IO s1n(1/h). This limit does not exist since sin(1/h) takes the values —1 and 1 on any interval containing 0. (Compare with Example 4 in Section 2.2.) 126 CHAPTER 2 LIMITS AND DERIVATIVES 36. Since f(w) = 3:2 sin(1/:c) when so 7é 0 and f(0) = 0. we have , , f(0 + h) — f(0) . h2 sin(1/h) — 0 , . . 1
f (0) 7,131.1?) h : grin) h zillirn)h51n(1/h).Smce —1 3 sin E S 1. we have
. 1 , 1
— h g hsm E S Ihl => * h S hsm E g Because flLirnO(i = 0 and llirno h : 0. we know that . . 1
2113) (h 8111 = 0 by the Squeeze Theorem. Thus. f'(0) : 0. 2.9 The Derivative as a Function 1. Note: Your answers may vary depending on your estimates. By estimating the slopes of tangent lines on the graph of f. it appears that
(a) f'(1) % —2 (b) f'(2) % 0.8
(c) f’(3) w —1 (d) f’(4) w —0.5 2. Note: Your answers may vary depending on your estimates. By estimating the slopes of tangent lines on the graph of f . it appears that (a) f'(0) m *3 (b) f’(1) m 0
(c) f’(2) % 15 (d) f’(3) % 2
(e) f’(4) % 0 (f) f’(5) z —1.2 3. It appears that f is an odd function, so f’ will be an even
function—that is. f’(—a) : f’(a). (a) f’(—3) e 1.5 (b) f’(—2) ~1
(c) f’(—1) w 0 (d) f’(0) m #4
(e) f’(1)z 0 (ﬂ f’(2) z 1 (g) f'(3) % 1‘5 4. (a)’ : [1. since from left to right. the slopes of the tangents to graph (a) start out negative. become 0. then positive,
then 0. then negative again. The actual function values in graph 11 follow the same pattern. (b)’ : IV. since from left to right. the slopes of the tangents to graph (b) start out at a ﬁxed positive quantity. then
suddenly become negative. then positive again. The discontinuities in graph IV indicate sudden changes in the
slopes of the tangents. (c)’ : 1. since the slopes of the tangents to graph (c) are negative for ac < 0 and positive for x > 0. as are the function values of graph 1. (d)’ : 111. since from left to ri ht. the slo es of the tan ents to gra h (d) are positive. then 0. then negative. then 0.
g P g P then positive. then 0. then negative again. and the function values in graph III follow the same pattern. ...
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This note was uploaded on 12/08/2009 for the course MATH 101 taught by Professor Dr.tahir during the Fall '08 term at King Fahd University of Petroleum & Minerals.
 Fall '08
 Dr.Tahir
 Math

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