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Unformatted text preview: 336 C CHAPTER4 APPLICATIONS OF DIFFERENTIATION 3 thatofy:x. E. f’(w):3$2~1/m2>0 <:> w4>§ <=> ac > %. so f is increasing on (*00, —%§) and (cf/Foo) and decreasing on (if/5,0) and (O, . F. Local maximum value __1_ __ , 5/4  1 _ 7
f( 3/3) 7 4 3 .localm1mmumvaluef(q\/—§) —43 5/4 G f”($) : 6$+2/$3 > 0 <=> a: > 0.sofisCUon (0.00) and
CDon (—00.0). No IP 70. lim — cos cc] : lim 1/902 : 0. so the graph off is za:,oo z—vioo asymptotic to that of cos x. The intercepts can only be found approximately f(:c) : f(iac). so f is even. lim (cossc + : 00. so arc—+0 3:2 m = 0 is a vertical asymptote. We don‘t need to calculate the derivatives. ,2 y = cosx since we know the asymptotic behavior of the curve. 4.6 Graphing with Calculus and Calculators 1. fps) : 4954 i 3223 + 899:2 — 95m + 29 => )“(zr):16903 — 96952 +1782: — 95 =>
f”(m) : 48302 71922 + 178. ﬂat) 2 0 <=> :c a: 0.5. 1.60; f’(;t) : 0 <=> as x 0.92. 2.5. 2.58 and
f"(m) : 0 <:> m z 1.46, 2.54. 10 10 H) 4.04 0 4
2.4 2.7 6 72 70.2 3.96 From the graphs of f’. we estimate that f' < 0 and that f is decreasing on (—00. 0.92) and (2.5. 2.58). and that
f' > O and f is increasing on (0.92. 2.5) and (2.58. 00) with local minimum values f(0.92) m —5.12 and
f(2.58) % 3.998 and local maximum value f(2.5) = 4. The graphs of f/ make it clear that f has a maximum and a minimum near :B : 2.5, shown more clearly in the fourth graph. From the graph of f", we estimate that f” > 0 and that f is CU 50 on (700, 1.46) and (2.54. and that f" < 0 and f is CD on
(1.46, 2.54). There are inflection points at about (1.467 —1.40) and (2.54., 3.999).
20 SECTION 4.6 GRAPHING WITH CALCULUS AND CALCULATORS C 337 2. f(a:) = m6 — 15105 + 75x4 — 125333 — .z =>
f’(m) : 6x5 — 75x4 + 300:53 — 3755c2 — 1 => f”(;r:) = 30x4 — 300273 + 9003c2 — 7509c.
f(:c) : 0 <=> m = O orm % 5.33; f’(x) : O <=> 2; m 2.50. 4.95, or 5.05;
f"(a:) = 0 <=> at = 0. Sorx % 1.38, 3.62. um v250 4.8 5.2 From the graphs of f’, we estimate that f is decreasing on (700, 2.50). increasing on (2.50, 4.95). decreasing on
(4.95, 5.05), and increasing on (5.0500), with local minimum values f(2.50) % —246.6 and f(5.05) % 75.03.
and local maximum value f (4.95) m —4.965 (notice the second graph of f ). From the graph of f ". we estimate
that f is CU on (—00, 0). CD on (O. 1.38), CU on (1.38. 3.62). CD on (3.62. 5), and CU on (5, 00). There are
inﬂection points at (0,0) and (5, —5). and at about (1.38, —126.38) and (3.62, *12862). i 2x—3 : f,,(m)__g w2—3x+24
3(122 ~3av—5)2/3 9(;c2 —3;tc75)5/3 m2—3x—5 1/3 Note: With some CAS’s. including Maple. it is necessary to deﬁne f(a:) : m [:02 — 3x 7 5] . since the CAS does not compute real cube roots of negative numbers. We estimate from the graph of f’ that f is
increasing on (1.5, 00). and decreasing on (700. 1.5). f has no maximum. Minimum value: f(1.5) z 71.9. From the graph of f". we estimate that f is CU on (—1.2. 4.2) and CD on (—00, —1.2) and (4.2, IP at
(—1.20) and (42,0). 338 :l CHAPTEB4 APPLICATIONS OF DIFFERENTIATION 4 3 2
$+x —2;1c +2 , 5 4i 3, 2 C
4 m2+x 2 f(x)‘2m +2:c(23:c 4;)2+2x 1
_ :0 +567 _> 6 57 4k 3 2 _
f,,($):2a: +3zc 3:1: 11$ +1230 +1853 2 (:02 +56 7 2)3 We estimate from the graph of f’ that f is increasing on (72.4., 72). (72, 41.5) and (1.5. 00) and decreasing
on (—00. —2.4). (—1.5.1) and (1.1.5). Local maximum value: f(~1.5) m 0.7.
Local minimum values: f(72.4) % 7.2. f(1.5) m 3.4. From the graph of f”. we estimate that f is CU on
(—00, —2), (—1.1.0.1) and (1. 00) and CD on (72. 71.1) and (0.1, 1).
f has I? at (—1.1,0.2) and (0.1. —1.1). 3: —2x3 + x2 + 1 5.f(w):m 3 (“PM : 2(3ac5 — 3z4 + 5:133 7 6302 + 336 + 4)
(:63 i 362 —4x +1)3 f"(1:) — 3 We estimate from the graph of f that y : 0 is a horizontal asymptote, and that there are vertical asymptotes at
I : —1.7. m : 0.24. and 3c : 2.46. From the graph of f’. we estimate that f is increasing on (—00, 71.7). (71.7., 0.24). and (0.24, 1). and that f is decreasing on (1.2.46) and (2.46, 00). There is a local maximum value
at f(1) : 7%. From the graph of f”. we estimate that f is CU on (—007 ~17), (—0.506, 0.24). and (2.46. 00). and that f is CD on (71.7, —0.506) and (0.24, 2.46). There is an inﬂection point at (—0.506, —0.192). SECTION 4.6 GRAPHlNG WITH CALCULUS AND CALCULATORS E 6. f(m) : tansc + Boosx => f'(:c) = sec2 2; — 58mm => f"(a:) = 2sec2m tanm — 5 cosm. Since f is periodic with period 277. and deﬁned for all :1: except odd multiples of we graph f and its derivatives on l2 12 12
17' J 317 1r i 317 71' I 371
‘2 r 2 2 V 2 2 2
—l2 —l2 —12 We estimate from the graph of f’ that f is increasing on (7%.021). (1.07. 2.07). and (2.93. and decreasing on (021,107) and (2.07. 2.93). Local minimum values: f(1.07) % 4.23. f (2.93) z —5.10. Local maximum values: f(0.21) m 5.10. f(2.07) m —4.23.
From the graph of f”. we estimate that f is CU on (0.76. g) and (2.38. égi). and CD on (4—; , 0.76) and (g, 2.38). f has 1P at (0.76. 4.57) and (2.38. —4.57). 7. f(zc):m274w+7cosx. —4£:c§4. f'(a:)=2:c—4—7sinrc => f”(w)=2—7cosw. f(x) : 0 <:> m z 1.10; f’(a:) : 0 <=> m % v1.49. —1.07. or 2.89; f”(x) = 0 <=> : :cos’1(§)m i128. From the graphs of f/. we estimate that f is decreasing (f' < 0) on (—41.7149). increasing on (—1.49, 71.07).
decreasing on (—1.07, 2.89). and increasing on (2.89. 4). with local minimum values f(—1.49) % 8.75 and f(2.89) m —9.99 and local maximum value f(—1.07) m 8.79 (notice the second graph of f). From the graph 339 340 C CHAPTER4 APPLICATIONS OF DIFFERENTIATION of f". we estimate that f is CU (f" > 0) on (41, —1.28). CD on (—1.28, 1.28). and CU on (1.28. 4). There are
inﬂection points at about (—1.28, 8.77) and (1.28. —1.48). was): ex :> mazw : f”(w):&W There are vertical asymptotes at :c = :I:3. It is difﬁcult to show all the pertinent features in one viewing rectangle, so we‘ll show f. f’. and f" for m < 3 and also form > 3. 0.5 0.5 0.5
FOHE < 3: 4% —4 ﬂ: —4  3
*05 —0.5 ~05
15 15 15
For m > 3:
75 —5 —5 We estimate from the graphs of f’ and f that f is increasing on (—00. —3). (—3, —2.16). and (4.16., 00) and
decreasing on (~2.16, 3) and (3‘ 4.16). There is a local maximum value of f(—2.16) % —0.03 and a local
minimum value of f(4.16) % 7.71. From the graphs of f”. we see that f is CU on (—00., —3) and (3., co) and
CD on (73, 3). There is no inﬂection point. 9. my) 2 8:63 — 3:32 — 10 :> f’(m) 2 24:22 — 6x => f”(;c) 2 48m — 6 40 20 ’97 _1,5 2
‘ *1 _ l —()'3 740 —1 —10.5 .0
Ln From the graphs, it appears that f(a:) : 8x3 — 31.2 — 10 increases on (700. 0) and (0.25, 00) and decreases on
(0. 0.25); that f has a local maximum value of f(0) : 710.0 and a local minimum value of f(0.25) % 710.1; that f is CU on (0.1, 00) and CD on (700. 0.1); and that f has an IP at (0.1, —10). To ﬁnd the exact values. note that
f’(m) : 24:1:2 — 6x = 69c(4ac i 1). which is positive (f is increasing) for (—00, 0) and (i, 00). and negative (f is decreasing) on (0, By the FDT. f has a local maximum at (13 = 0: f(0) = —10; and f has a local 10. 11. SECTION 4.6 GRAPHING WITH CALCULUS AND CALCULATORS 3 341 minimumatizfel) : {13— % ‘10: ( i 161 16 1 "(33) : 48m — 6 : 6(815 — 1). which is positive (f is CU) on 0 =< r%) OOIH
Ooh—A 100) and negative (f is CD)on (—ooi f has an [P at (§.f( OOIH 20
—6 From the graphs, it appears that f increases on (07 3.6) and decreases on (—00, 0) and (3.6, 00); that f has a local
maximum of f(3.6) z 2.5 and no local minima; that f is CU on (5.5, 00) and CD on (—00, 0) and (0. 5.5); and
11 20 m2 z2+11x720:1+ that f has an IP at (5.5, 2,3). f(m) 332 x : f’(;c) : 7112f2 + 4050—3 —:c’3(11x i 40). which is positive (f is increasing) on (0 ﬂ). and negative ’11 40. 11‘ 40 (f is decreasing) on (—00. 0) and on (ﬁt 00). By the FDT. f has a local maximum at a: = 2 1
%; and f has no local minimum. (ﬁf +11%) — 20 1600+ 11 ~ 11 i 40 — 20 . 121
(g)2 _ 1600 —11x_2 + 4093—3 => f”(a:) = 2231—3 — 12051244 : 2x’4(11x i 60). which is positive (f is CU) on =<6—1€—1>. 40
f(—1) :
f’(w) =
($00). and negative (f is CD) on (—oo,0) and (0. g). f has an [P at (%,f(%))
From the graph, it appears that f increases on (—2.1, 2.1) and decreases on (—3, —2.1) and (2.1, 3); that f has a local maximum of f(2.1) z 45
and a local minimum of f(~2.1) z 74.5; that f is CU on (—3.0. 0) and CD on (03.0), and that f has an IP at (0,0). f(.z) = :c 9 — 1‘2 => 2 i i 2
_ a: +F—9 aE2:92:I: Hm) . which is positive \/9 — :02 v9 — x2
(f is increasing) on (—355, and negative (f is decreasing) on (—3. ’32‘5) and 3) By the FDT.
2
f has a local maximum value off(¥) : 3—2—‘5‘ /9 — : g; and f has a local minimum value of
2 —3 2 Z 79 ~ ~ ~ I _ ‘33 _ 2 f( 2 ) 2 (smce f IS an odd function). f (m) — W + v9 :16 =>
,, \/9—$2(*2$)+$2(%)(97I2)_1/2(—2TC) 2 ,1/2 —2$—$3(9~m2)‘1 —:c
f A 22(9 :17 ) :
9 i x2 x/9——:c2
_ 73m x3 z(2m2 — 27) W w (9—39)“? 2 (97952?” which is positive (f is CU) on (73., 0) and negative (f is CD) on (0, 3). f has an IP at (0, 0). 342 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 12. From the graph. it appears that f increases on (—5.2. —1.0) and (1.0, 5.2)
and decreases on (~27r, 75.2), (71.0, 1.0). and (5.2. 27r); that f has local
maximum values of f(—1.0) "a 0.7 and f(5.2) z 7.0 and local minimum values of f(75.2) z —7.0 and f(1.0) % ~07; that f is CU on (7271’. —3.1) and (0.3.1) and CD on (~31, 0) and (3.1, 2%). and that f
has IP at (0,0), (—3.1, 73.1) and (3.1,3.1). f(a:) = a: * 2sinw :> @570 f’(a:) = 1 * 2cos cc, which is positive (f is increasing) when cossc < that is. on (v—, ——) and (
and negative (f is decreasing) on (~27r, and By the FDT. f has local maximum values
off(—§) : g + x/gand = 5?” + andlocal minimum values off(—5?") : i5?” — x/gand : 7% i f’(ac) : 1 7 20053: => f”(ac) = 2sin93. which is positive (f is CU) on (—27r. in) and
(0., 7r) and negative (f is CD) on (771'. 0) and (7r, 27r). f has H" at (0, 0). (771’, 771') and (71', 7r). 13. (a) f(:c) : $2 In x. The domain off is (0. 0,25 ° 1,75
IE7 0.25 . 11156 H . 1/ac . m2 .
‘ 2 _ _ l _ l —— =0.Th isaholeat 0.0.
(b) $133+ at hm $112314? 1/an2 $331+ —2/ac3 $351+ ( 2 ere ( ) (c) It appears that there is an [P at about (0.2, —0.06) and a local minimum at (0.6, —O.18). f(;c) = $2 lnw i 1 f’(ac):w2(1/:c)+(lnw)(2x):m(21nm+1)>0 <2) lnac>—§ <=> :c>e’1/2 , so f is increasing on (1/\/e_37 oo). decreasing on (0,1/\/E). By the FDT. f(1/\/E) : —1/(2e) is a local minimum value. This point is approximately (0.6065, 70.1839). which agrees with our estimate. f"(x)=z(2/z)+(2lnx+1)=2lnm+3>0 <=> 1nx>—% <=> x>e’3/2.sofisCUon (e_3/2,oo) and CD on (meW). IP is (KS/2, —3/(2e3)) x (0.2231, 70.0747). 4" 1/x(71/x2)
' 1/‘”: ‘ e :1‘ e——= lim el/xzoo.s0:c:0isaVA. zlirg+ me 1213+ $11,ng —1/$2 x—)O+ 14. (a) f(:c) : wel/I. The domain off is (700,0) U (0. 1/m Also lim steel/1c :OSince l/mHoo => 61/9” —>0. waO— SECTION 4.6 GRAPHING WITH CALCULUS AND CALCULATORS 343 (c) It appears that there is a local minimum at (1, 2.7). There are no IP and f is CD on (—00. 0) and CU on (0, 00) 10(53):”‘31/9E => fiiml=$€1/x(*;§>+el/ =el/ (17;)>0 <=> E<1 <=> $<Oor a: > 1, so f is increasing on (—00.0) and (1,00). and decreasing on (0.1). By the FDT. f(1) : e is a local minimum value. which agrees with our estimate.
f”(a:) : el/$(1/m2)+(1— 1/x)e1/z(—1/$2) = (el/x/zc2)(1 7 1 + 1/17): eve/$3 > 0 <=> :c > 0. so f is CU 0n (0. 00) and CD on (—007 0). No 1?. (x + 4)(w — 3)? 15. has VA atm = Oand atcv : 1 since 1%) : x4(a: — 1)
lirrbﬂx) : foo. lim f(9:) : —00 and lim+ f(x) : oo.
2—» m—il’ seal :0 + 4 (m — 3)2 m ' x2 [dividing numerator and
ﬂat) = 4—— . 3 a: denominator by a: ] ‘—3 ' (:1! f 1) a: g 2
= W —> 0 as :0 —> too. so f is asymptotic
:c(x — 1) to the xaxis. Since f is undeﬁned at w = 0. it has no y—intercept. f(m) : 0 => (:6 + 4)(ac A 3)2 = 0 =>
m : —4 or a: : 3, so f has sic—intercepts —4 and 3. Note. however. that the graph of f is only tangent to the m—axis and does not cross it at .7: = 3. since f is positive as w —> 3— and as a: —> 3+. 500
0.02 0,03
“ 4 2
8 —3,5
2,5 
70.04 71500 0 8 From these graphs. it appears that f has three maximum values and one minimum value The maximum values are approximately f(—5.6) : 0.0182. f(0.82) : —281.5 and f(5.2) : 0.0145 and we know (since the graph is tangent to the sc—axis at :c : 3) that the minimum value is f(3) : 0. i 10$(x — 1)4
f”) ‘ (z i 2W + 1>2
zlirnlﬂm) : 00. In; f(m) —00 and 111’;le f(w) = 00. 10 1 — 1 4 n 10 as w 6 ice, so f is asymptotic to has VA atac : —1 and atzr : 2 since ll ﬂan) = the line y : 10. f(0) : 0. so f has a y—intercept at 0. f(w) = 0 => 10$(x — 1)4 : 0 => :3 : Dora: 2 1. So f has m—intercepts 0 and 1. 344 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Note. however. that f does not change sign at :c : 1. so the graph is tangent to the ac—axis and does not cross it. We know (since the graph is tangent to the sic—axis at :v : 1) that the maximum value is f(1) 2 0. From the graphs it appears that the minimum value is about f(0.2) = 70.1.
20 1
‘05 ‘ 1.5
’9 _I_ '0
*5 —l
2 3 2 3 2
x(x+1) x(:c+1) (:5 +189: —44:c—16)
17. : — : ’ : ___—_—_ ‘
f”) (x ~ 2m: — 4)4 f (5”) (a: — 2)3(:c — 4)5 (“0m CA5)
00011 ()00015 5000
_]5 A“'l'.‘l 0
22 r 32
’3" y‘ "° 1 '
70.0002 70.0001 —2000 From the graphs of f’. it seems that the critical points which indicate extrema occur at a: z —20. —0.3, and 205. as estimated in Example 3. (There is another critical point at x : 71. but the sign of f’ does not change there.) 2 (a: + 1) (x6 + 36955 + 6904 — 628:1:3 + 684:52 + 672:1: + 64) We differentiate again, obtaining f"(:c) : (a: i 2)4($ _ 4>6 000001
5000
—40 7 40
6
—0.00001 0 From the graphs of f”. it appears that f is CU on (#35.3,—5.0). (—1.705). (—0.12). (2,4) and (4,00) and CD on ( 00, 35.3). ( 5.0, 1) and ( 0.5, 0.1). We check back on the graphs of f to ﬁnd the y—coordinates of the
inﬂection points. and ﬁnd that these points are approximately (—35.3, A0015). (75.0, —0.005). (—1.0).
(—0.50, 0.00001), and (#01, 0.0000066). SECTION 4.6 GRAPHING WITH CALCULUS AND CALCULATORS 3 345 1051: w 41)4 , _ (a: *1)3(5$ — 1) .mm 200 2 10 750 From the graphs of f’, we estimate that f is increasing on (—00. 71) and (0.2, 1) and decreasing on (—1, 0.2). 0(zr ~1)2(5:03 4 8m? + 17:1: — 6) (1,2) and (2,00). Differentiating f’(ac). we get f”(m) : 6 i 2)5(w +1)4 100 ’5 —k—— 5 From the graphs of f”. it seems that f is CU on (—00. —1.0). (—1.0. 0.4) and (2.0, 00). and CD on (0.4, 2). There is an inﬂection point at about (0.4., —0.06). 31112 :5 since [2(172 + 1) cossc — xsin m] 19. : x = — with 0 < a: < 371'. From a CAS. I = —————— and
y f( ) x2 +1 _ _ 2/ (322 “)3/2
I, (4.7:4 + 6:52 + 5)cos2 w — 450(ch +1)sinmcosx — 2x4 — 2:32 — 3
y * (x2 +1)5/2 ' 0. 75 From the graph of f’ and the formula for y’. we determine that y' 2 0 when w : 7r. 27r. 371’. or x z 1.3. 4.6. or 7.8.
So f is increasing on (0.1.3). (7T.4.6). and (27r. 7.8). f is decreasing on (1.3, 7r). (4.6, 27r). and (7.8,37r). Local
maximum values: f(1.3) % 0.6. f(4.6) % 0.21, and f(7.8) m 0.13. Local minimum values: f(7r) = f(27r) : 0.
From the graph of f”. we see that y” : 0 <=> m m 0.6. 2.1. 3.8. 5.4. 7.0. or 8.6. So f is CU on (0, 0.6). (2.1, 3.8). (5.4, 7.0). and (8.6, 371'). f is CD on (0.6. 2.1). (3.8. 5.4). and (7.0. 8.6). There are [P at (0.6, 0.25).
(2.1.0.31). (38,010). (54,011). (7.0.0.061). and (8.6. 0.065). 346 I: CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 2I—1
20. x =‘ :5
ﬂ ) \4/sc4+;1:+1
4x3+6zc+9
f’(m =—% :5
> 4(m4+x+1)5/4
,, 32x6+96x4+152$3—48x2+6x+21
f (90‘ 21. 16(204 + x + 1)9/4 From the graph of f’. f appears to be decreasing on (—00, 70.94) and increasing on (—0.94. There is a local
minimum value of f(70.94) z 7301. From the graph of f". f appears to be CU on (—1.25. —0.44) and CD on
(A00, —1.25) and (—0.44, 00). There are inﬂection points at (~1.25. —2.87) and (—0.44, —2.14). 1 _ 61/1: 2 1/2?
W. From a CAS. 3/ = e ” _ —2€1/x(1 ~ 61/1 + 2x + 23:61”)
m2<1+el/x)2 — ‘ : $4<1+el/m)3 f is an odd function deﬁned on (—00, 0) U (0., 00). Its graph has no :c— or y—interceptsi Since lim f(:c) = 0. the :c—>ioo
w—axis is a HA. f’(:c) > 0 for ac # 0. so f is increasing on (—00. 0) and (0, It has no local extreme values.
f”(z) : 0 for :c % i0.417, so f is CU on (—00, —O.417). CD on (—0.417,0). CU on (07 0.417). and CD on
(0.417, 00). f has IPs at (A0417, 0.834) and (0.417. 70.834). 1 / etanx
.y:f(.’13): W.FromaCAS.y 17W and
,, eta”[eta”(2sinwc0sw— 1)+2sin;ccosm+1] :  . is a eriodic function with eriod 7r that has
y cos4a:(1+eta“z)3 f p p
positive values throughout its domain. which consists of all real numbers except odd multiples of g (that is. :I: . 1
2
:I:§275. $57". and so on). f has y—intercept but no m—intercepts. We graph f. f’. and f” on one period. (—g, v I Since f’(w) < 0 for all ac in the domain of f, f is decreasing on the intervals between odd multiples of f”(w) : 0 for I : 0 + rm and for :I: % :I:1.124 + n7r. so f is CD on (—3, —1.124). CU on (—1.124, 0), CD on
(0, 1.124). and CU on (1.124, Since f is periodic. this behavior repeats on every interval of length 7r. f has IPs
at (—1.124 + n7r. 0.890). (m, §). and (1.124 + m. 0.110). SECTION 4.6 GHAPHING WITH CALCULUS AND CALCULATORS Cl 347 l/z : (b) Recall that a17 : eb 1"“. lim :3 lim Junk”. As at —> 0+. 23. (a) : gel/m 160+ IA“
2 12—9: —+ —00. so 3:1” : (EU/ml” —> O. This indicates that there is
 a hole at (0.0). As an a 00. we have the indeterminate form 000.
0 8 lim acl/x : lim 6(1/z)l“2.but lim 1—12 2 lim 1/—$ = 0. so
z—wo sv—voo mace m mace 1 lim m1” : e0 = 1. This indicates that y : 1 is a HA. x—roo (c) Estimated maximum: (2.72. 1.45). No estimated minimum. We use logarithmic differentiation to ﬁnd any 1 I 1 1 1
critical numbers. y — gal/I ~> lny — Ina: ~> y— — —  — + (ln$)(i—2> =>
cc y m a: :10 '—m1/I 1—111“? 0 > lnx 1 > x e.F0rO<a:<e.y'>0andforsc>e.y'<0.so
3/ x2 f(e) : 61/8 is a local maximum value. This point is approximately (2.7183, 14447). which agrees with our estimate. ((11) 0.1 From the graph. we see that f"($) : 0 at x z 0.58 and x m 4.37. Since f ” changes sign at these values. they are m—coordinates of inﬂection points. 24. (a) f(m) : (sin sari” is continuous where sings > 0. that is. 12
on intervals of the form (27m, (2n + 1)7r). so we have " graphed f on (0. 7r). . w' . . 77
(b) y : (sm LEV” :> lny : smx 1n sm m. so 0
. . . . . ln sina: H cotzc
hm lny : 11m smxlnsmx : 11m : 11m — : lim (~ Sinai) : 0
1H0“? z—.o+ m—r0+ cscm z—>O+ * cscm cotm z—»O+ => lim y = 60 : 1.
xﬁO‘i' (c) It appears that we have a local maximum at (1.57. 1) and local minima at (0.38. 0.69) and (2.76.069). y : (sin$)“i"x => lny = sina: lnsina: => /
. c052: . . . 
y; : (sma:)(sinw) +(1nsmm)cosx = cosm(1+lnsm:c) => 3/ : (s1nsc)smm(cosx)(1+lnsinw).
y' : 0 => cosz : Oorlnsinw = —1 => 1:2 : g orsina: : e71. On (O.7r).sinx : e7l => 2:1 2 sin‘1 (6’1) and $3 : 7r 7 sin‘1 (e71). Approximating these points gives us 348 :1 CHAPTERll APPLICATIONS OF DIFFERENTIATION (m1, f(ml)) x (0.3767. 0.6922). (22. f(m2)) :2 (1.5708. 1). and ($3, fag» x (2.7649, 0.6922). The approximations conﬁrm our estimates. From the graph. we see that f"(;c) : 0 at a: z 0.94 and an m 2.20. Since f” changes sign at these values, they are m—coordinates of inﬂection points. From the graph of f(:c) = sin(:c + sin 39:) in the viewing rectangle [07 7r] by [—12, 1.2]. it looks like f has two
maxima and two minima. If we calculate and graph f’(m) = [cos(:c + sin 3x)] (1 + 3 cos 31) on [0, 2n].
we see that the graph of f ’ appears to be almost tangent to the ccaxis at about cc : 0.7. The graph of f" : — [sin(:c + sin 32)] (1 + 3 cos 32:)2 + cos(m + sin 3x)(—9 sin 3:0) is even more interesting near this :c—value: it seems to just touch the m—axis. 0.1 0.002 1
m
ass ‘1. . ‘
0 7 0.58 ‘ 0.7
' ‘ *Z'n'
0.55 0.73
0.3 *0004 0.9997 If we zoom in on this place on the graph of f”, we see that f” actually does cross the axis twice near a: : 0.65.
indicating a change in concavity for a very short interval. If we look at the graph of f’ on the same interval. we see
that it changes sign three times near 1: : 0.65. indicating that what we had thought was a broad extremum at about :c = 0.7 actually consists of three extrema (two maxima and a minimum). These maximum values are roughly
f(0.59) : 1 and f(0.68) : 1. and the minimum value is roughly f(0.64) : 0.99996. There are also a maximum
value of about f(1.96) : 1 and minimum values of about f(1.46) : 0.49 and f(2.73) : 70.51. The points of
inﬂection on (0. 7r) are about (0.61, 0.99998). (0.66, 0.99998). (1.17, 0.72). (1.75, 0.77), and (2.28, 0.34). On
(7r, 2n). they are about (4.01, 70.34). (4.54, —0.77). (5.11. —0.72). (5.62, #099998). and (5.67, —0.99998).
There are also IP at (0, 0) and (7r, 0). Note that the function is odd and periodic with period Zn. and it is also rotationally symmetric about all points of the form ((2n + 1)7r, 0), 72 an integer. 26. f(:c) : $3 +csc : m(:c2 +c) 27. 349 SECTION 4.6 GRAPHING WlTH CALCULUS AND CALCULATORS => f'(m) 23x2+c => f”(w) =61' 20 20 20 5 5 5 5 5 5
720 —20 720
c : _6 c 2 0 c : 6 z—intercepts: When 0 Z O, 0 is the only ale—intercept. When 0 < O. the m—intercepts are 0 and i\/——c.
y—intercept = f(0) : 0. f is odd. so the graph is symmetric with respect to the origin. f"(:c) < 0 for a: < 0 and
f”(a:) > 0 for as > 0. so f is CD on (—00. 0) and CU on (0. oo) . The origin is the only inflection point. Ifc > 0. then f’(ac) > 0 for all at. so f is increasing and has no local maximum or minimum. If c = 0. then f’(w) Z 0 with equality at x : 0. so again f is increasing and has no local maximum or minimum. Ifc < 0, then f'(:lc) : 3[x2 — (—c/3)] = 3 :0 + —c/3)<$ — —c/3).
so f’ (x) > Don (—ee. —\/T/3) and (\f—‘E/iec); f’ (m)
(j/T/zsﬂ/T/s). It follows thatf(— —c/3) : egg/TBS
gcmisalccal minimum value. As c decreases (toward more negative values). the local a local maximum value and f( —c/3> maximum and minimum move further apart. There is no absolute maximum or minimum value. The only C: 2% ‘20
. . . . =4
transmonal value of 0 corresponding to a change ln character of the graph 5 : 6
c' : 8
is c = 0. f(:Z‘) = x4 + c3:2 = 2152(2)? + 0). Note that f is an even function. For c 2 0. the only sicintercept is the point (0.0).
We calculate f’(:1:) : 4a3 + 2cm : 4:6(m2 + §c) => f"(a:) : 12302 + 2c. Ifc 3 0. x : 0 is the only critical point and there is no inﬂection point. As we can see from the examples. there is no change in the basic shape of the graph for c 2 0; it merely becomes steeper as 0 increases. For c : 0. the graph is the simple curve y = 56*. For c < 0. there are m—intercepts at 0 and at ::\/—c. Also. there is a maximum at (0, 0). and there are minima at (it / —%c, — if). As c —> —00. the w—coordinates of these minima get larger in absolute value. and the minimum points move 1 5 downward. There are inﬂection points at (i —3c. —§662). which also move away from the origin as c a —00. 350 28. 29. I: CHAPTER 4 APPLICATIONS OF DIFFERENTIATION We need only consider the function ﬂay) : (IIZVCZ — m2 for c 2 0. because ifc is replaced by —c. the function is unchanged. For c = 0. the graph consists of the single point (0. O). The domain of f is [—c, c]. and the graph of f is symmetric about the y—axis. x3 2$(c2 — 1'2) — :53 2\/c_2 ./Cz_$2 So we see that all members of the family of curves have horizontal tangents at a: = 0. since f’ (0) 2 0 for all c > 0. 322(332 — 3‘62) 02_m2 f/(m):2cc\/c2~x2+z2 *2a:\/c2 x2 —223
_ x2 c2—x2 Also. the tangents to all the curves become very steep as a: —> :I:c. since lim f’(a:) : 00 and $H7(‘+ lim f’($) : —00. We set f’(x) : 0 <=> a: : 0 0r 2:2 — 302 : 0, so the absolute maximum values 3—H," 3 are f(:: (—30) = ﬁe? : (—9.102 —I— 2c2) V02 A m2 i (—3253 + 2c2w) (—9: x/c2 — m2) _ 6x4 — 902352 + 2C4
(62 _ $2)3/2 ' 2 _ 902 tea/33
_ 12 {9—mggt/E—ayc3) 144 f”(m) c2_x2 Using the quadratic formula. we ﬁnd that f"(;v) = O <:> :c . Since —c < at < c. we take 2:2 : #3. so the inﬂection points are (iv Ll‘gc. 3.5
From these calculations we can see that the maxima and the points of inﬂection get both horizontally and vertically further from the origin as 0 increases Since all of the functions have two maxima and two inﬂection points. we see that the basic shape of the curve does not change as 0 changes. c : O is a transitional value—we get the graph ofy : 1. For c > 0. we see that there is a HA at y = 1. and that the
graph spreads out as 0 increases. At ﬁrst glance there appears to be a minimum at (0. 0). but f (0) is undeﬁned. so there is no minimum or maximum For c < 0. we still have the HA at y : 1. but the range is (1, 00) rather
than (0.1). We also have a VA at m : 0. f(a;) : 6—6”2 => f’($) = 6—4“E2 (—2c/sc3) :> _ 2c(20 v 3332) 6 / 2 . f’(ac) aé 0 and f’(a:) exists for all :5 ¢ 0 (and 0 is not in the domain of f). so there are no
$ e!V I f”(w) maxima or minima. f”(a:) = 0 => :5 : :I: 20/3. so ifc > 0. the inﬂection points spread out as c increases. 30. 31. SECTION 4.6 GRAPHING WlTH CALCULUS AND CALCULATORS E 351 and if c < 0. there are no 1P. For c > 0. there are IP at (i: 20/3, (W2). Note that the y—coordinate of the IP is constant. We see that ifc S 0. f(x) : 1n(ac2 + c) is only deﬁned for x2 > 7c => > \/—c. and lim = lim f(z) : 700. since lny —? —00 as y —> 0. Thus. for c < 0. there are vertical
I—'\/——C+ x—y—‘/_—c asymptotes at as = :i:\/E, and as c decreases (that is. c increases), the asymptotes get further apart. For c = 0. lim f (x) : —00, so there is a vertical asymptote at as = 0. If c > 0. there are no asymptotes, To ﬁnd the extrema
m—AO 1 . . .
and inﬂection points. we differentiate: f(;1:) : 1n<x2 + c) :> f’(m) = $2 + C so by the First Derivative Test there is a local and absolute minimum at m : 0. Differentiating again. we get 1
w2+c f”(€c) : (2) + 2m[— (x2 + c)‘2 (295)] : Now if c S 0. f" is always negative. so f is concave down on both of
the intervals on which it is deﬁned. Ifc > 0. then f" Changes sign when
c : 9:2 <=> x : iﬁ. So for c > 0 there are inﬂection points at a: : iﬁ, and as c increases. the inﬂection points get further apart. Note that c : 0 is a transitional value at which the graph consists of the sic—axis. Also. we can see that if we substitute ~c for c, the function f(:c) = % will be reﬂected in the w—axis. so we investigate only positive
c a: values of c (except 0 : 71. as a demonstration of this reﬂective property). Also. f is an odd
function. In; f(:c) : 0, so y = 0 is a horizontal asymptote for all c. We calculate z—v 00 (1 + c2w2)c — c;c(2c2:c) C(C2I2 — 1)
far :————————. 'z :0 <=> 02x2—120 42> w=:l:1 c.Sothere ( ) (1+ c2ac2)2 (1+ czar2)2 f( ) / is an absolute maximum value of f(1/c) : % and an absolute minimum value of f(—1/c) : ~é. These extrema have the same value regardless of c. but the maximum points move closer to the y—axis as 0 increases. (—2c3cc) (1 + c2x2)2 — (ic3z2 + c)[2(1 + c2z2) (2c2$)] f (x) 7 (1+ 022:2)4
_ (—20313)(1 + 02232) + (03.272 — c) (482:) 2c3m(c2m2 — 3)
— (1 + c2x2)3 U (1 + c2332)3 f”(w) : 0 <:> m = O or ix/g/c. so there are inﬂection points at (0.0)
and at (ix/g/c, lag/4). Again. the ycoordinate of the inﬂection points does not depend on c. but as 0 increases. both inﬂection points approach the yaxis. 352 D CHAPTEB4 APPLICATIONS OF DIFFERENTIATION 1 . . .
IS an even function. and also that lim f(:1c) 2 0 for any value of c. so 3; : 0 32. Note that ac : ———
f( > (1 — x2)2 —I— cmz 1—»ioo 3 0:0 is a horizontal asymptote. We calculate the derivatives:
fl”) 74(1 — m2):c + 2210‘ 4$[$2 + (it: — 1)]
[(1 71:2)2 +0552] [(1~m2)2 +cm2]2 10:56 + (9c 718)x4 + (3c2 712C + 6);:22 + 2 ~ 0
[$4 + (c v 2):c2 +1]3 . We ﬁrst consider the case 0 > 0. Then the denominator of f' is . and f”($) : 2 positive. that is. (1 7 m2)2 + car2 > 0 for all m, so f has domain IR and also f > 0. If %c i 1 Z 0; that is. c 2 2.
then the only critical point is f (0) : 1. a maximum. Graphing a few examples for c 2 2 shows that there are two IP
which approach the y—axis as c ~> 00. c = 2 and c : 0 are transitional values ofc at which the shape of the curve changes. For 0 < c < 2. there are . . . . . 1
three critical p0ints: f(0) = 1. a minimum value. and f(:ti / 1 — éc) : both maximum values. As c
c — c decreases from 2 to 0. the maximum values get larger and larger. and the sc—values at which they occur go from O to :l:1. Graphs show that there are four inﬂection points for 0 < c < 2. and that they get farther away from the origin. both vertically and horizontally. as c —> 0+. For c = 0. the function is simply asymptotic to the maxis and to the lines I. : ::1. approaching +00 from both sides of each. The y—intercept is 1. and (0. 1) is a local minimum. There are no inﬂection points. Now if c < 0. we can write : 1 : 1 1 . So f has vertical (17m2)2+cw2 (1—f52)27(\/A—c;1;)2 (x2—\/——cw*1)(m2+\/——cx—1)
asymptoteswherexziﬂm—120 <=> 9:: (iﬂix/4—0V2orm: (\/——ci\/4TcV2Asc decreases. the two exterior asymptotes move away from the origin. while the two interior ones move toward it. We graph a few examples to see the behavior of the graph near the asymptotes. and the nature of the critical points ‘ 0:74 x:0and:c=:: 1v§cz We see that there is one local minimum value. f(0) : 1. and there are two local maximum values, f (i, / 1 — §c> : W as before. As c decreases. the :c—values at which these maxima occur get larger, and
C 1 the maximum values themselves approach 0. though they are always negative. SECTION 4.6 GHAPHING WITH CALCULUS AND CALCULATORS 2' 353
33. : cm + since :> : c + COSIE => f"(m) — * sinw
ﬂeas) = —f(:c). so f is an odd function and its graph is symmetric with respect to the origin
f(:c) : 0 <=> sinac = —cac. so 0 is always an sic—intercept.
f’(:c) : 0 <1: cosa: : *0. so there is no critical number when c > 1. If c g 1. then there are inﬁnitely
many critical numbers. If an is the unique solution of cosx = —c in the interval [0. 7r]. then the critical numbers are 34. 27m i $1. where n ranges over the integers. (Special cases: When c : 1. .731 : 0; when c = 0. a: = g; and when
c = —1.:c1 : 7r.)
f"(2:) < 0 <=> form ((2n 7 1)7r. 2n7r). The inﬂection points of f are the points (27177. 2mm). where n is an integer. sina: > 0. so f is CD on intervals of the form (27m, (211 + 1)7r). f is CU on intervals of the Ifc Z 1. then f’(x) Z 0 for all :3. so f is increasing and has no extremum. Ifc S —1. then f’($) S 0 for all z.
so f is decreasing and has no extremum. If cl < 1. then f’(:c) > 0 <=> cosm > —c <=> a: is in an interval of
the form (27m — $1. 27% + x1) for some integer n. These are the intervals on which f is increasing. Similarly. we
ﬁnd that f is decreasing on the intervals of the form (2mr + $1. 2(n + 1)7r — 2:1). Thus. f has local maxima at the
points 2n7r + 231. where f has the values C(2mr + $1) + sin 221 = C(Zmr + :01) + and f has local
minima at the points 2n7r — :61. where we have f(2mr — m1) : C(2mr — 3:1) — sin m1 : C(2mr 7 x1) 7 The transitional values of c are 71 and 1. The
inﬂection points move vertically. but not horizontally.
when G changes. When lc Z 1. there is no extremum. For
[cl < 1, the maxima are spaced 27r apart horizontally. as are the minima. The horizontal spacing between maxima and adjacent minima is regular (and equals 7r) when 0 : 0. but the horizontal space between a local maximum and the nearest local minimum shrinks as c approaches 1. For f(t) = C(e’“ — 6—“). C affects only vertical stretching. so we let C = 1. From the ﬁrst ﬁgure. we notice that the graphs all pass through the origin. approach the t—axis as t increases, and approach —oo as t —> —00. Next we let a : 2 and produce the second ﬁgure. Here. as b increases. the slope of the tangent at the origin increases and the local maximum value increases. ﬁt) : 6—2t ’ 671” => f/(t) : 56—“ ~ 2e_2‘. f’(0) : b — 2. which increases as 17 increases. 354 35. 36. 3 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
b 2 b 1112*“) 2)t lnb—ln2 bi2 f'(t) = 0 :> be‘“ : 26—2t : 6(1’72” :> —> t — t1 — . which
(b i 2)22/<H> decreases as (2 increases (the maximum is getting closer to the y—axis). f(t1) ~ b1+2/(b 2) . We can show that this value increases as 1) increases by considering it to be a function of b and graphing its derivative with respect to b, which is always positive. 1' H . 1
11m
3—» 700 C653” lim
maioo 6C1 Ifc < 0. then lim f(ac) xﬁioo = 0. and lim : oo. mace —00, and lim f(:z) 2 lim 1 :0. w—voo z—voo cecz Ifc > 0. then lim f(m) $—’—OO Ifc : 0, then f(:c) = 9:. so lim f(x) = :I:oo respectively. x—»ioo
So we see that c : 0 is a transitional value. We now exclude the case c = 0. since we know how the function
behaves in that case. To ﬁnd the maxima and minima of f. we differentiate: f(a:) : :ce’” => f'(x) : x(—ce’“) + e‘” = (1 — cage—CI. This is 0 when 1 M cm : 0 <=> x =1/c.Ifc < 0 then this represents a minimum value of f(1/c) : 1/(ce). since f'(:1:) changes from negative to positive at a; : 1/0. 3. For c = 0, there is no inﬂection point; the curve is CU everywhere. If c increases. the curve simply becomes steeper. and ifc > 0. it represents a maximum value. As c increases. the maximum or minimum point gets closer to the origin. To ﬁnd the inﬂection
points. we differentiate again: f’(:c) = e‘”(1 — cw) f"(x) : e”’m(—c) + (1 i cm)(~ce_°$) : (cm i 2)ce_”". This :> changes sign when car i 2 : 0 <:> m = 2/c. So as [cl increases. the points of inﬂection get closer to the origin. and there are still no inﬂection points. If c starts at 0 and decreases. a slight upward bulge appears near ac : 0. so
that there are two inﬂection points for any 0 < 0. This can be seen algebraically by calculating the second
derivative: = x4 + c222 + :0 ¢ : 4x3 —I— 2cm + 1 => f"(x) 2 12:1:2 + 20. Thus. f"(:c) > 0 _1 60. For c z 0. the graph has one critical when 0 > 0‘ For c < 0, there are inﬂection points when 3c : i number, at the absolute minimum somewhere around as = —0.6. As c increases, the number of critical points does
not change. If (3 instead decreases from 0, we see that the graph eventually sprouts another local minimum. to the
right of the origin, somewhere between a: : 1 and x : 2. Consequently, there is also a maximum near as : 0.
After a bit of experimentation, we ﬁnd that at c = —1.5. there appear to be two critical numbers: the absolute
minimum at about a: : —1, and a horizontal tangent with no extremum at about 3 = 0.5. For any 0 smaller than this there will be 3 critical points, as shown in the graphs with c = —3 and with c : 75. To prove this algebraically. we calculate f’(:):) = 4x3 + 2cm + 1. Now if we substitute our value ofc = —1.5, the formula for f’(:c) becomes 4303 ~ 3x + 1 : (a: —I— 1)(2m  1)2. This has a double root at an : indicating that the function has two critical Nh—I points: 9: = —1 and x : ,just as we had guessed from the graph. SECTION 4.6 GRAPHING WITH CALCULUS AND CALCULATORS :l 355 37. (a) : csc4 7 2x2 + 1. For c : 0. : —2:c2 + 1. a parabola whose vertex. (0. 1). is the absolute maximum. For c > 0. f(:c) : 0504 — 2x2 + 1 opens upward with two minimum points. As c —> 0. the minimum
points spread apart and move downward; they are below the w—axis for 0 < c < 1 and above for c > 1. For c < 0. the graph opens downward~ and has an absolute maximum at a: : 0 and no local minimum. (b) f'(x) = 4cx3 7 4x = 4c:c(x2 A 1/c) (c 9é 0). Ifc g 0. 0 is the F4 (:1
c=0,5
c=0.2 1 “1! vii“ —12 l Al :1: : :tl/\/Ei Heref(::1/\/E) : c(1/c2) i 2/c+ 1 : *1/c+1i C=_ only critical number. f"($) : 12cm2 7 4. so f”(0) : i4 and there is a local maximum at (0, f(0)) : (0, 1). which lies on y : 1 7 :32. If c > 0, the critical numbers are 0 and :l:1/\/Ei As before. there is a local maximum at (0. f(0)) : (0. 1). which lies on y : 1 — $2. f”(i1/\/E) :12 i 4 : 8 > 0. so there is a local minimum at But (il/ﬁ, —1/c+ 1) lies ony =1— 3:2 since
17 (:l:1/\/E)2 : 1 — 1/c. 38. (a) : 21E3 —l—c:c2 +2.16 :> : 6$2 + 2cm+ 2 = 2(3m2 +0.1“ + 1). : O <=> —c :l: \/C2 — 12
6 9: : . So f has critical points <:> c2 — 12 Z 0 ¢> c Z 2V3. For c : ::2\/§. f’(m) Z 0 on (—00, 00). so f' does not change signs at *c/6, and there is no extremumi If C2 — 12 > 0. then 7c e V02 — 12
6
—c+\/c2—12 *C*\/C2——12 6 . So f has a local maximum at x : ——6— and a local minimum at f ' changes from positive to negative at a: : and from negative to positive at m: —c+ x/c2 — 12
——6 . (b) Let .160 be a critical number for Then f’($o) : 0 :> _ i 2
f(m0) 22.733 +0353 +2$o : 2x3 +mg<ﬂ) +2150 3 3
:2m0—xoi3m0+2x0:xoi$g So the point is ($0,310) : ($0.210 — 3:8); that is. the point lies on the curve y : x 7 2:3. ...
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 Fall '08
 Dr.Tahir
 Math

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