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Unformatted text preview: 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) 711312) ft : glint) h =i1:n)hsm(1/h).$mce —1 3 sin b S 1. we have
. 1 , 1
— h g hsm h S Ihl => * h S hsm h g Ih Because flLirnO(i h) = 0 and iIimo h : 0. we know that . . 1
213%) (h 8111 h) = 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) z —2 (b) f'(2) % 0.8
(c) f’(3) w —1 (d) f’(4) e —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) 431
(c) f’(—1) w 0 (d) f’(0) m #4
(e) f’(1)z 0 (f) 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. SECTION 2.9 THE DERIVATIVE AS A FUNCTION 127 Hints for Exercises 5—13: First plot :cintercepts on the graph of f’ for any horizontal tangents on the graph of f , Look for any corners on
the graph of f ~there will be a discontinuity on the graph of f’. On any interval where f has a tangent with positive (or negative) slope,
the graph of f’ will be positive (or negative). If the graph of the function is linear, the graph of f’ will be a horizontal line. 5. 11. 128 CHAPTER 2 LIMITS AND DERIVATIVES 14. The slopes of the tangent lines on the graph of y = P(t) are always
positive. so the yvalues ofy = P'(t) are always positive. These
values start out relatively small and keep increasing. reaching a
maximum at about t : 6. Then the y—values of y = P’(t) decrease and get close to zero. The graph of P’ tells us that the yeast culture grows most rapidly after 6 hours and then the growth rate declines. 15. It appears that there are horizontal tangents on the graph of M for
t = 1963 and t : 1971. Thus. there are zeros for those values oft
on the graph of [W . The derivative is negative for the years 1963
to l97l. 1950 1960 I970 1980 1990 16. See Figure l in Section 3.4. 17. _ 18. As at increases toward 1. f '(z) decreases from
The slope at 0 appears to be 1 and the slope at 1 very large numbers to 1. As cc becomes large.
appears to be 27. As x decreases. the slope gets f’(:13) gets closer to 0' As a guess. f’(:c) 2 1/352 closer to 0. Since the graphs are so similar. we or f’(;t) : 1/:c make sense. might guess that f’(;c) : ex, 19. (a) By zooming in. we estimate that f/(O) : 0. f'(%) : 1. f’(1) : 2‘ 25
and f’(2) : 4.
(b) By symmetry. f’(—;r:) : —f'(x). So f’(%) : —1.
f'(71): 2. and f’(~2) : #4.
(c) It appears that f’(m) is twice the value of :5. so we guess that f'(:c) : 230. 2.5 SECTION 2.9 THE DERIVATIVE AS A FUNCTION I: 129 . x+h —f(3: . (av—i—h)2—m2
@H222“ Z )=mt h
2 2 7 2 2
2lim(m ”WWII 3” _limm—+h=limw:hm(2m+h)=2x
11—»0 h h—>0 h h—»0 fix~90 20. (a) By zooming in. we estimate that f’(0) : 0. f’(§) m 0.75.
f’(1)m 3. f’(2) 2 12. and f’(3) m 27.
(b) By symmetry. f’(—m) 2 f’(ac). So f’(—%) z 0.75. f'(~1) % 3.
f’(—2) z 12. and f’(~3) z 27.
(d) Since f’(0) 2 0. it appears that f’ may have the form f/(m) 2 (1272.
Using f’(1) 2 3. we have a 2 3. so f’(m) 2 3:32. (:1: + m3 _ x3 (2:3 + 32% + 3th + h3) 2 $3 2 lim — lim (e) f’(:v) : mm W 1120 h h—»0 h h—>0 h
2 2 3 h 3 2 3 h h2 .
_lim3$h+3mh +h —lim (5” + m + )2lim(3x2+3mh+h2):3$2
h—>0 h h—>0 h h—>O
, _. f(:c+h)—f(m)_. 37—37 . 9 .
“fW132—7——Aﬂ h—Mmﬂﬂko Domain off 2 domain off’ 2 Re [12 + 7(5c + m} 2 (12 + 7x) , . x+h~ a: .
waﬂﬂi—%ﬁiwﬂ h
2lim12+7x+7h_12h7m2limﬂzlim727
h2o h h—toh 1120 Domain off 2 domain of fI 2 R. We +11) — f(x) 2 hm [1— 3(m + hf] — (1 — 3w?) nﬂﬂﬂﬂ h at h
_ 2 2 _ ¥ 2 _ 2 _ _ 2 y 2
¥ lim [1 3(m +2xh+h )] (1 3x ) lim 1 3w 62h 3h 1+3z
1220 h h—»O h
‘ l _ 2 _ i
2 lim M 2 lim m 2 lim(26x — 3h) 2 —62?
1:20 h i120 h 1120 Domain off 2 domain off’ 2 IR. 24. f'(x) : lim f(x + h) — f(:):) [5m + h)2 + 3(m + h) — 2] — (5562 + 32: — 2) h—ao h — III—IR) h
~1th 522+1Owh+5h2+3x+3h—2—522 —3:c+2 11 10xh+5h2+3h
i120 h, _ hi3) h
. h 10 5h 3
211m $2 : lin1(10:c + 5h + 3) = 103: +3
h—+0 h h—oD Domain off 2 domain of f' 2 IR. 130 3 CHAPTERZ LIMITS AND DERIVATIVES f(x+h)—f($) [($+h)3!3($+h)+5]—(x3#3x+5) 25. f’(:c) : £13) h : ’13) h _ (m3+3x2h+3mh2+h3—3:ci3h+5)—(a:3#3a:+5)
_ 11m h—>0 h . 3x2h+3xh2 +h3 —3h , M3352 +3mh+h2 —3)
_ 11m — 11m h—rO h h_.0 h
ziltini)(3x2+3mh+h2 —3) =3m2 —3 Domain off : domain of fI : IR. 26. fun) i nmw_ lim (“MW”) #(wh/E)
h hHO h h—>0
—lim(ﬂ' x+h*\/E. $+h+ﬁ> lim 1+_______(zv+h)—m
h—+0 h I h T—I—hI—ﬁ 11—»0 h(,/$+h+\/§) :lim =1+——— h~o<l+m>1+Tif M Domain off : [0. 00). domain of f’ : (0,00). 27. I _# ' ————— 7 m
9(56) 11m h 11 h) .___ lim ——
ha°h[/—41+2(:c+h )+m] h—»0\/1+2:c+2h+\/1+2:c 2x/12I—23: \/1_+2x g(m+h)ﬁg(m) s/1+2(m+h) x/——1+2ac [I/1+2( ()Hh HITH21] Domain ofg : [~%.oo).domain ofg' : (—%oo). 3+(w+h,) _ 3+:c
173(m+h) 1—350 f(56+h)f($) = lim 28. f’(:c) : lim h—vO h h—~0 h
f lim (3 + a: + h)(1 # 3w) # (3 + x)(1 — 3:0 — 3h)
h—iO h(1 7 3m — 3h)(14 3m)
. (3#933+a:~3m2+h 3m) (3 9x 9h+x 3x2 3mg)
# :13?) h(1 — 3w , 3h)(1 — 3x) 1' 10h 11m 10 _ 10
133%) h(1~ 3x — 3h)(1 — 3:13) h—»0 (1 — 3x ; 3h)(1 , 3x) (1— 3:5)2 Domain off : domain off : (#00. %) U (i 00). 4(t + h) 4t 4(t + h)(t + 1) — 4t(t + h + 1)
. G(t+h)—G(t)7, (t+h)+1 t+1 . (t+h+1)(t+1)
29 Gm :m—r— 112:) h —m h
1, (4t2 + 4ht + 4t +4h) — (4252 + 4m + 4t)
' £3}; h(t+h+1)(t+1)
4h 4 4 , [—— lin ——’———
r1113)h(t—I—h+1)(t+1):h4»0(t+h+1)(t+1)z(IH—l)2 Domain ofG : domain of G' : (~oo,—1)U(71,oo). SECTION 2.9 THE DERIVATIVE AS A FUNCTION 1 1
g(x+h)—g(93)¥. ($+h)2—; l' W
30 9 (90> , 33%, h 1‘3?) h — i“ W + MW
1' :02 — (x2 + Zach + W) l 72$}; i h221m —2x — h —2;c
: _ l —
hlgh h Ill{91%) h(w + h)2:c h—vO (m + M232 x4
: —23:’3
Domain ofg : domain ofg' : {ac  cc 7S 0}.
4 4 4 3 2 2 3 4 4
— — +4h6h+4h+hisc
31. f (56) — lim ‘ﬂm—I—h) 10(1) : lim LIE,” m = lim (3: m + an x )
h—m h h—+O h h—>0 h
3 2 2 3 4
411m 4:“ h+6$ h +4“ +h _ lim (4x3 +6x2h+4xh2+h3) 24x3
hao h hao
Domain off 2 domain of f’ : R
32. (a) y (b) Note that the third graph in part (a) has small negative values for its slope. f’; but as x —> 6’. f’ —> —00. See the graph in part (d)‘
(C)fl(w)=limw_hm6 —(m+h) m m+m M h M W+ m
= 1m W _ hm
magma/m] h~0h(m+m)
1 —1
— lim h—>0\/6—$—h+\/f_3—m 2\/G—:c
Domain off 2 (—oo,6]~ domain of f’ = (—00.6). 131 132 :I CHAPTERZ LIMITS AND DERIVATIVES (b) Notice that when f has steep tangent lines. f'(:c) is very large. 10 When f is ﬂatter. f’(:c) is smaller. ll
44 6 6
. t h — 2 i 2 2 g — 2
34. (801%): 11m f( + ) f(t) I hm 1+(t+h) 1+1: 2 lim 6+6t 6 6(t+h)
h—>0 h h—>0 h h—»0 h [1 + (t + h)2] (1 + If?)
, 7121,}; — 6h2 . —12t — 6h —12t
7 11m , — 11m _
zHoh[1+(t+h)2](1+t2) h—v0 [1+(t+h)2](1+t2) (1+t2)2
(b) Notice that f has a horizontal tangent when t 2 0. This corresponds 6 to f’(0) : 0. f’ is positive when f is increasing and negative when f
is decreasing. A 35. (a) U '(t) is the rate at which the unemployment rate is changing with respect to time. Its units are percent per year. (b) To ﬁnd U’(t). we use lim U“ + h) i UIt) m U“ + h) _ U“) for small values of h. h~90 h h,
U(1992) — U(1991) 7.5 — 6.8
F 1991:U’ 19 1 :—’———:———:.
or I 9 I 1992 1 1991 1 0 70
For 1992: We estimate U'(1992) by using it : ~1 and h = 1. and then average the two results to obtain a ﬁnal estimate. h : ,1 : U’(1992) % U(1991) — U(1992) : 6.8 — 7.5 Z 070: 1991 — 1992
1 3 — U 19 2 .9—7. So we estimate that UI (1992) m %[0.70 + (—0.60)] : 0.05. ‘— t 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000
IU'(t) 0.70 0.05 #070 —0.65 —0.35 0.35 ~04?) —0.35 —0.25 —0.20 36. (a) P'(t) is the rate at which the percentage of Americans under the age of 18 is changing with respect to time. Its
units are percent per year (% / yr). — t P — P t .
(b) To ﬁnd P’(t). we use flint) PU: + hf: PI ) m (t + h; I I for small values of h. P 9 —P19 3.79311
For 1950: P’(1950) : W = 5—16— : 0.46 For 1960: We estimate P’(1960) by using h = —10 and h = 10. and then average the two results to obtain a
ﬁnal estimate. P(1950)  P(1960) : 31.1 — 35.7 : 0'46 _ _ ’ m
h 1 10 :> P (1960) 1950~ 1960 710 37. 38. 39. 40. SECTION 2.9 THE DERIVATIVE AS A FUNCTION 3 P(1970) , P(1960) _ 34.0 , 35.7
1970 — 1960 _ 10
So we estimate that P'(1960) % %[0.46 —I— (~0.17)] : 0.145. t 1950 1960 1970 1980 1990 2000
’(t) 0.460 0.145 ~0.385 —0.415 —O.115 0.000 h = 10 :> P’(1960) % _ —0.17 (C) PIU‘I
0.5"
A
PU) 0.4T
371L
0.3T
35
ll 0.2 29
—0.1
27 —0.2»
25
~0'3"
1 —1 1— 1 r 1 >
1950 1960 1970 1980 1990 2000 ’ —0.4i
—0,54r (d) We could get more accurate values for P'(t) by obtaining data for the mid—decade years l955. 1965. 1975,
1985. and 1995. f is not differentiable at z : —1 or at a: : 11 because the graph has vertical tangents at those points; at as : 4.
because there is a discontinuity there; and at a: = 8, because the graph has a corner there. (a) g is discontinuous at x : —2 (a removable discontinuity). at a: = 0 (g is not deﬁned there). and at It : 5
(ajump discontinuity). (b) g is not differentiable at the points mentioned in part (a) (by Theorem 4). nor is it differentiable at w : —1 (corner). x : 2 (vertical tangent). or :v = 4 (vertical tangent). As we zoom in toward (~1, 0). the curve appears more and more like 2 a straight line. so f(x) : x + Incl is differentiable at :1: = —1. But no matter how much we zoom in toward the origin. the curve doesn‘t straighten out—we can’t eliminate the sharp point (a cusp). So f is ‘2 r I not differentiable at m = 0. 
—1 As we zoom in toward (0, 1). the curve appears more and more like a 3 straight line, so f is differentiable at :c : 0. But no matter how much we zoom in toward (1, 0) or (~1, 0). the curve doesn‘t straighten k ‘
out—we can’t eliminate the sharp point (a cusp). So f is not _2 ‘ 2
differentiable at x = i1.  134 CHAPTER 2 LIMITS AND DERIVATIVES
41. (a) Note that we have factored a: — a as the difference of two cubes in the third step.
_ 1/3 _ 1/3 1/3 _ 1/3
f’(a) : lim M : lim :c__—a_ # lim a: a
:c>ct LC — a, x—ta (E — a, 140.051/3 — a1/3)(m2/3 + $1/3a1/3 + a2/3)
. 1 _ 1 1 72/3
— $1131 352/3 + x1/3a1/3 + (12/3 3a2/3 0r 3d
.0+h70_3h~0.1. ..
(b) f ((0) = I133) W = $131) fh = ’ILIIIb W This function increases without bound, so the
limit does not exist. and therefore f I(0) does not exist.
. . 1 . i .
(c) i131) f'(a:) : i136 3362/3 2 00 and f is continuous at a: = 0 (root function). so f has a vertical tangent
at x = 0.
. — 0) . 952/3 — 0 1
42. I : 9(3)) 9( = : ' ‘ '
(a) 9(0) $131) —————$ _ 0 31:13) ———$ $131) $1/3 . which does not exrst.
_ 2/3_ 2/3 1/3_ 1/3 1/3 1/3
(b) 9/(a) = lim M : lim L—a— — lim (cc (1 )(sc + a )
m—va. CC 7 CL 2410. LC — a, m—ra(x1/3 — (11/3)(.’E2/3 + m1/3a1/3 + a2/3)
1' 501/3 +a1/3 201/3 2 2 4/3
* 1m — — or —a
z_,a 352/3 + $1/3a1/3 + a2/3 3&2/3 3a1/3 3
(c) g(:c) = $2/3 is continuous at :c : 0 and (d) 04
2
. , _ . _ .
213131) lg (cc) # i131) 3$1/3 7 00. This shows that g has a vertical tangent line at a: : 0. 70.2 0.2
0
—a:6) if:c<6 6—30 ifsc<6
43f(m):lr—6l= . = .
#6 iftc26 :c—b £31226
li f(:c)—f(6) — 11m Ian—(”#0: 11m 366: lim 1:1.
5346+ {E 7 6 x—>6+ SC — 6 $——>6+ 117 ‘ w>6+
But 1m M: lim W211“, 6ﬁx
I—a6_ 56—6 35—.6 LB6 m—a6_$—6
: lim (71):—1
2—»6’ . f($) f(6) . / —1 if CC < 6
’ : l ————— d te st. However. as =
Sof (6) m1111S z 6 oes no x1 f ( I 1 if a: > 6 . ﬂ 6
Another way of writing the answer 18 f’(m) : 1: _ 6. 44. f (m) = [[311]] is not continuous at any integer n, so f is not differentiable at n by the contrapositive of Theorem 4. If a is not an integer. then f is constant on an open interval containing a, so f’(a) : 0. Thus. f'(a:) = 0. at not an integer. SECTION 2.9 THE DERIVATIVE AS A FUNCTION 3 135 (b) Since f(:L') = $2 fora: 2 0, we have f’(:1:) : 253 form > 0.
[See Exercise 2.9.l9(d).] Similarly. since f(w) : —a:2 for z < 0. we have f'(:c) : —2w form < 0. At :6 : 0. we have
wlml f’(0) = lim M : lim — : lim m : 0. 1—»0 IE — 0 x—>0 ac z—rO So f is differentiable at 0‘ Thus. f is differentiable for all 2:. 2:1: if .77 2 0
(c) From part (b), we have f’(a:) : = 2 Incl —21: if x < O
f(4+h)~f(4) . 5—(4+h)~1 —h
I ‘ _ _ l — *1 ‘ d
4‘ (a) f <4) .1329. h .122}. h .32. a"
1 — 1
, _ f(4+h)—f(4)_ 5~(4+h) _ 1~(1—h) 1, 1 :1
f+(4) 7 333+ h Ill—1351+ h _ hl—l»r(1;1+ h(1 — h) hi1; — h
0 if a: < 0 (c)f(m): 5~w if0<sc<4
1/(5720) if 3624
These expressions show that f is continuous on the intervals
(~oo,0). (0. 4). (4, 5) and (5,00). Since
lim f(:c) 2 lim (5 — cc) = 5 71$ 0 : lim f(z),i1_rgf(a:) z—+O+ z—i0+ arr—’0— does not exist. so f is discontinuous (and therefore not
differentiable) at 0. At 4 we have lim f(:v) = lim (5 e :13) 2 1 and lim f(;z:) : lim ac—r4‘ x—>4’ z—>4+ 5. =1~somf<xi=1= f<4> and f is continuous at 4. Since f(5) is not deﬁned, f is discontinuous at 5‘
(d) From (a). f is not differentiable at 4 since f: (4) yé fl. (4) and from (c). f is not differentiable at 0 or 5. 47. (a) Iff is even. then
JCT—SE) : lim f(—:c + h) _ f(_$) : lim (fl—(m ‘ h)l _ f(—$) f($  h)  flSC) h~i0 h h—»o h 222% h
. f(w—h)—fx . +A A ,
=12. _,. ” "MPh]  .5152.” :2 f‘“—~f<> Therefore, f ’ is odd.
(1)) Iff is Odd. then f'(—;r) :23) W : £125 fl—(JE — hi31— f(‘$) : 213%) ‘f(x — Z) +f(m)
:32?) f(x~ 1— ﬁx) [mm _ _h] _ A130 ﬂaw?) — f(x) ﬂ ﬁx) Therefore, f’ is even. 136 I: CHAPTERZ LIMITS AND DERIVATIVES 48. (a) T (c) dT/dl 0 t (b) The initial temperature of the water is close to room temperature because of the water that was in the pipes.
When the water from the hot water tank starts coming out. dT/dt is large and positive as T increases to the
temperature of the water in the tank. In the next phase. dT/dt : 0 as the water comes out at a constant. high
temperature. After some time. dT/dt becomes small and negative as the contents of the hot water tank are
exhausted. Finally. when the hot water has nm out. dT/dt is once again 0 as the water maintains its (cold)
temperature. 49. In the right triangle in the diagram. let Ay be the side opposite angle (15 and Arc the side adjacent angle (15. Then the slope of the tangent line 6 is
m : Ay/Asc : tan (1). Note that 0 < qb < %. We know (see Exercise 19)
that the derivative of f(:c) = m2 is f’(ac) : 230. So the slope of the tangent to the curve at the point (1, 1) is 2. Thus. (15 is the angle between 0 and g whose tangent is 2; that is. d) : tan’1 2 3 63°. 2 Review
CONCEPT CHECK ____———————— 1. (a) lim f($) : L: See Deﬁnition 2.2.] and Figures 1 and 2 in Section 2.2. I—‘Hl (b) lim f(x) : L: See the paragraph after Deﬁnition 2.2.2 and Figure 9(b) in Section 2.2. CE—HZ+ (c) lim f(m) : L: See Deﬁnition 2.2.2 and Figure 9(a) in Section 2.2. 12—»(1’ (d) lim ﬂan) : 00: See Deﬁnition 2.2.4 and Figure 12 in Section 2.2. I—‘a (e) lim ﬂat) 2 L: See Deﬁnition 2.6.] and Figure 2 in Section 2.6. z—mo 2. In general. the limit of a function fails to exist when the function does not approach a ﬁxed number. For each of the following functions. the limit fails to exist at a: : 2. The left— and right—hand There is an There are an inﬁnite limits are not equal. inﬁnite discontinuity. number of oscillations. 3. (a)—(g) See the statements of Limit Laws 1—6 and l l in Section 2.3. ...
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 Fall '08
 Dr.Tahir
 Math

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