(Solution Manual) Calculus Early Transcendentals

Briggs, Cochran, Gillett, Schulz, 3

Math & Statistics

Solution Manual

3rd Edition

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Solution Manual for Calculus Early Transcendentals, The downloadable file consists of 875 pages, and the total size of the file is 8.71 MB. Purchase now only for highly discounted price.

Author(s): Briggs, Cochran, Gillett, Schulz

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SOLUTIONS MANUAL MARK WOODARD CALCULUS EARLY TRANSCENDENTALS THIRD EDITION William Brigg s Lyle Cochran Bernhard Gillett Eric Schulz Get complete solutions manual all chapters download link below https://www.mediafire.com/file/tgs6s4n o2xh3ivs/SM+Calculus+Early+Transcend entals,+3e+Will iam+Briggs,+Lyle+Cochra n,+Bernard+Gillett,+Eric+Schulz.zip/file Contents 1 Functions 5 1 . 1 R e v i e wo fF u n c t i o n s......................................... 5 1.2 Representing Functions ....................................... 1 4 1.3 Inverse, Exponential and Logarithmic Functions ......................... 3 2 1 . 4 T r i g o n o m e t r i cF u n c t i o n sa n dT h e i rI n v e r s e s............................ 4 1C h a p t e rO n eR e v i e w............................................ 5 4 2 Limits 67 2 . 1 T h eI d e ao fL i m i t s.......................................... 6 72 . 2 D e fi n i t i o no faL i m i t......................................... 7 22 . 3 T e c h n i q u e sf o rC o m p u t i n gL i m i t s ................................. 8 72.4 Infinite Limits ............................................ 9 7 2.5 Limits at Infinity ...........................................1 0 6 2 . 6 C o n t i n u i t y ..............................................1 2 0 2 . 7 P r e c i s eD e fi n i t i o n so fL i m i t s ....................................1 3 3C h a p t e rT w oR e v i e w............................................1 4 2 3 Derivatives 155 3 . 1 I n t r o d u c i n gt h eD e r i v a t i v e .....................................1 5 53 . 2 T h eD e r i v a t i v ea saF u n c t i o n....................................1 6 43 . 3 R u l e so fD i ff e r e n t i a t i o n.......................................1 8 23 . 4 T h eP r o d u c ta n dQ u o t i e n tR u l e s..................................1 9 03 . 5 D e r i v a t i v e so fT r i g o n o m e t r i cF u n c t i o n s ..............................2 0 23 . 6 D e r i v a t i v e sa sR a t e so fC h a n g e...................................2 1 13 . 7 T h eC h a i nR u l e ...........................................2 2 63 . 8 I m p l i c i tD i ff e r e n t i a t i o n .......................................2 4 03.9 Derivatives of Logarithmic and Exponential Functions ......................2 5 8 3 . 1 0 D e r i v a t i v e so fI n v e r s eT r i g o n o m e t r i cF u n c t i o n s..........................2 7 03 . 1 1 R e l a t e dR a t e s ............................................2 7 9C h a p t e rT h r e eR e v i e w...........................................2 9 1 4 Applications of the Derivative 307 4 . 1 M a x i m aa n dM i n i m a ........................................3 0 74 . 2 M e a nV a l u eT h e o r e m........................................3 2 24 . 3 W h a tD e r i v a t i v e sT e l lU s......................................3 3 04 . 4 G r a p h i n gF u n c t i o n s.........................................3 4 84 . 5 O p t i m i z a t i o nP r o b l e m s.......................................3 8 24 . 6 L i n e a rA p p r o x i m a t i o na n dD i ff e r e n t i a l s..............................4 0 54.7 L’Hˆ o p i t a l ’ sR u l e...........................................4 1 5 4 . 8 N e w t o n ’ sM e t h o d ..........................................4 2 94 . 9 A n t i d e r i v a t i v e s............................................4 4 5C h a p t e rF o u rR e v i e w ...........................................4 5 6 12 Contents 5I n t e g r a t i o n 479 5.1 Approximating Areas under Curves . . . .............................4 7 9 5 . 2 D e fi n i t eI n t e g r a l s...........................................4 9 9 5.3 Fundamental Theorem of Calculus .................................5 1 9 5 . 4 W o r k i n gw i t hI n t e g r a l s .......................................5 3 65.5 Substitution Rule ..........................................5 4 6 C h a p t e rF i v eR e v i e w............................................5 5 7 6 Applications of Integration 573 6 . 1 V e l o c i t ya n dN e tC h a n g e......................................5 7 36 . 2 R e g i o n sB e t w e e nC u r v e s ......................................5 8 76 . 3 V o l u m eb yS l i c i n g..........................................6 0 26 . 4 V o l u m eb yS h e l l s...........................................6 1 06 . 5 L e n g t ho fC u r v e s...........................................6 2 06 . 6 S u r f a c eA r e a.............................................6 2 66 . 7 P h y s i c a lA p p l i c a t i o n s........................................6 3 4C h a p t e rS i xR e v i e w ............................................6 4 4 7 Logarithmic, Exponential, and Hyperbolic Functions 661 7.1 Logarithmic and Exponential Functions Revisited ........................6 6 1 7 . 2 E x p o n e n t i a lM o d e l s.........................................6 6 97 . 3 H y p e r b o l i cF u n c t i o n s ........................................6 7 5C h a p t e rS e v e nR e v i e w...........................................6 8 7 8 Integration Techniques 693 8 . 1 B a s i cA p p r o a c h e s ..........................................6 9 3 8 . 2 I n t e g r a t i o nb yP a r t s.........................................7 0 3 8 . 3 T r i g o n o m e t r i cI n t e g r a l s.......................................7 2 28.4 Trigonometric Substitutions ....................................7 3 1 8 . 5 P a r t i a lF r a c t i o n s...........................................7 4 88 . 6 I n t e g r a t i o nS t r a t e g i e s........................................7 6 38 . 7 O t h e rM e t h o d so fI n t e g r a t i o n....................................7 9 88 . 8 N u m e r i c a lI n t e g r a t i o n........................................8 0 78 . 9 I m p r o p e rI n t e g r a l s..........................................8 1 8C h a p t e rE i g h tR e v i e w...........................................8 3 3 9 Differential Equations 855 9 . 1 B a s i cI d e a s..............................................8 5 59 . 2 D i r e c t i o nF i e l d sa n dE u l e r ’ sM e t h o d................................8 6 19 . 3 S e p a r a b l eD i ff e r e n t i a lE q u a t i o n s..................................8 7 39 . 4 S p e c i a lF i r s t - O r d e rL i n e a rD i ff e r e n t i a lE q u a t i o n s.........................8 8 69 . 5 M o d e l i n gw i t hD i ff e r e n t i a lE q u a t i o n s ...............................8 9 4C h a p t e rN i n eR e v i e w ...........................................9 0 3 10 Sequences and Infinite Series 911 1 0 . 1 A nO v e r v i e w.............................................9 1 11 0 . 2 S e q u e n c e s...............................................9 1 910.3 Infinite Series .............................................9 3 3 1 0 . 4 T h eD i v e r g e n c ea n dI n t e g r a lT e s t s.................................9 4 41 0 . 5 C o m p a r i s o nT e s t s ..........................................9 5 51 0 . 6 A l t e r n a t i n gS e r i e s..........................................9 6 31 0 . 7 T h eR a t i oa n dR o o tT e s t s .....................................9 7 01 0 . 8 C h o o s i n gaC o n v e r g e n c eT e s t....................................9 7 6C h a p t e rT e nR e v i e w............................................9 9 3 Copyright c/circlecopyrt2019 Pearson Education, Inc.Contents 3 11 Power Series 1005 1 1 . 1 A p p r o x i m a t i n gF u n c t i o n sW i t hP o l y n o m i a l s ........................... 1005 1 1 . 2 P r o p e r t i e so fP o w e rS e r i e s ..................................... 1023 1 1 . 3 T a y l o rS e r i e s............................................. 1032 1 1 . 4 W o r k i n gw i t hT a y l o rS e r i e s..................................... 1047 C h a p t e rE l e v e nR e v i e w .......................................... 1061 12 Parametric and Polar Curves 1071 1 2 . 1 P a r a m e t r i cE q u a t i o n s........................................ 1071 1 2 . 2 P o l a rC o o r d i n a t e s.......................................... 1088 1 2 . 3 C a l c u l u si nP o l a rC o o r d i n a t e s ................................... 1108 1 2 . 4 C o n i cS e c t i o n s............................................ 1123 C h a p t e rT w e l v eR e v i e w.......................................... 1143 13 Vectors and the Geometry of Space 1161 1 3 . 1 V e c t o r si nt h eP l a n e......................................... 1161 1 3 . 2 V e c t o r si nT h r e eD i m e n s i o n s.................................... 1169 1 3 . 3 D o tP r o d u c t s............................................. 1179 1 3 . 4 C r o s sP r o d u c t s............................................ 1187 1 3 . 5 L i n e sa n dP l a n e si nS p a c e...................................... 1196 1 3 . 6 C y l i n d e r sa n dQ u a d r i cS u r f a c e s .................................. 1204 C h a p t e rT h i r t e e nR e v i e w ......................................... 1219 14 Vector-Valued Functions 1233 1 4 . 1 V e c t o r - V a l u e dF u n c t i o n s ...................................... 1233 1 4 . 2 C a l c u l u so fV e c t o r - V a l u e dF u n c t i o n s................................ 1241 1 4 . 3 M o t i o ni nS p a c e........................................... 1247 1 4 . 4 L e n g t ho fC u r v e s........................................... 1266 1 4 . 5 C u r v a t u r ea n dN o r m a lV e c t o r s................................... 1272 C h a p t e rF o u r t e e nR e v i e w......................................... 1283 15 Functions of Several Variables 1299 1 5 . 1 G r a p h sa n dL e v e lC u r v e s...................................... 1299 1 5 . 2 L i m i t sa n dC o n t i n u i t y........................................ 1311 1 5 . 3 P a r t i a lD e r i v a t i v e s.......................................... 1317 1 5 . 4 T h eC h a i nR u l e ........................................... 1329 1 5 . 5 D i r e c t i o n a lD e r i v a t i v e sa n dt h eG r a d i e n t ............................. 1340 1 5 . 6 T a n g e n tP l a n e sa n dL i n e a rA p p r o x i m a t i o n............................ 1354 1 5 . 7 M a x i m u m / M i n i m u mP r o b l e m s................................... 1362 15.8 Lagrange Multipliers ......................................... 1373 C h a p t e rF i f t e e nR e v i e w.......................................... 1382 16 Multiple Integration 1393 1 6 . 1 D o u b l eI n t e g r a l so v e rR e c t a n g u l a rR e g i o n s ............................ 1393 1 6 . 2 D o u b l eI n t e g r a l so v e rG e n e r a lR e g i o n s............................... 1400 1 6 . 3 D o u b l eI n t e g r a l si nP o l a rC o o r d i n a t e s............................... 1416 1 6 . 4 T r i p l eI n t e g r a l s............................................ 1430 16.5 Triple Integrals in Cylindrical and Spherical Coordinates .................... 1442 1 6 . 6 I n t e g r a l sf o rM a s sC a l c u l a t i o n s................................... 1454 1 6 . 7 C h a n g eo fV a r i a b l e si nM u l t i p l eI n t e g r a l s............................. 1465 C h a p t e rS i x t e e nR e v i e w.......................................... 1477 Copyright c/circlecopyrt2019 Pearson Education, Inc.4 Contents 17 Vector Calculus 1491 1 7 . 1 V e c t o rF i e l d s............................................. 1491 1 7 . 2 L i n eI n t e g r a l s ............................................ 1504 1 7 . 3 C o n s e r v a t i v eV e c t o rF i e l d s ..................................... 1514 1 7 . 4 G r e e n ’ sT h e o r e m........................................... 1520 1 7 . 5 D i v e r g e n c ea n dC u r l......................................... 1534 1 7 . 6 S u r f a c eI n t e g r a l s........................................... 1545 1 7 . 7 S t o k e s ’T h e o r e m........................................... 1556 1 7 . 8 D i v e r g e n c eT h e o r e m......................................... 1564 C h a p t e rS e v e n t e e nR e v i e w......................................... 1573 D2 Second-Order Differential Equations 1585 D 2 . 1B a s i cI d e a s .............................................. 1585 D 2 . 2L i n e a rH o m o g e n e o u sE q u a t i o n s................................... 1604 D 2 . 3L i n e a rN o n h o m o g e n e o u sE q u a t i o n s................................. 1614 D 2 . 4A p p l i c a t i o n s ............................................. 1628 D 2 . 5C o m p l e xF o r c i n gF u n c t i o n s..................................... 1666 C h a p t e rD 2R e v i e w ............................................ 1685 Copyright c/circlecopyrt2019 Pearson Education, Inc.Chapter 1 Functions 1.1 Review of Functions 1.1.1A function is a rule that assigns each to each value of the independent variable in the domain a unique value of the dependent variable in the range. 1.1.2The independent variable belongs to the domain, while the dependent variable belongs to the range. 1.1.3GraphAdoes not represent a function, while graph Bdoes. Note that graph Afails the vertical line test, while graph Bpasses it. 1.1.4The domain of fis [1,4), while the range of fis (1,5]. Note that the domain is the “shadow” of the graph on the x-axis, while the range is the “shadow” of the graph on the y-axis. 1.1.5Item i. is true while item ii. isn’t necessarily true. In the definition of function, item i. is stipulated. However, item ii. need not be true – for example, the function f(x)=x2has two different domain values associated with the one range value 4, because f(2) =f(−2) = 4. 1.1.6g(x)=x2+1 x−1=(x+1)(x−1) x−1=x+1 ,x/negationslash=1 . T h e domain is {x:x/negationslash=1}and the range is {x:x/negationslash=2}. -



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) 56 Chapter 1. Functions 1.1.7The domain of this function is the set of a real numbers. The range is [ −10,∞).-
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---- 1.1.8The independent variable tis elapsed time and the dependent variable dis distance above the ground. The domain in context is [0 ,8] 1.1.9The independent variable his the height of the water in the tank and the dependent variable Vis the volume of water in the tank. The domain in context is [0 ,50] 1.1.10f(2) =1 23+1=1 9.f(y2)=1 (y2)3+1=1 y6+1. 1.1.11f(g(1/2)) =f(−2) =−3;g(f(4)) =g(9) =1 8;g(f(x)) =g(2x+1)=1 (2x+1)−1=1 2x. 1.1.12O n ep o s s i b l ea n s w e ri s g(x)=x2+1 and f(x)=x5, because then f(g(x)) =f(x2+1)=(x2+1)5. Another possible answer is g(x)=x2andf(x)=(x+1)5, because then f(g(x)) =f(x2)=(x2+1)5. 1.1.13The domain of f◦gconsists of all xin the domain of gsuch that g(x) is in the domain of f. 1.1.14(f◦g)(3) =f(g(3)) =f(25) =√ 25 = 5. (f◦f)(64) =f(√ 64) =f(8) =√ 8=2√ 2. (g◦f)(x)=g(f(x)) =g(√x)=x3/2−2. (f◦g)(x)=f(g(x)) =f(x3−2) =√ x3−2 1.1.15 a. (f◦g)(2) =f(g(2)) =f(2) = 4. b.g(f(2)) =g(4) = 1. c.f(g(4)) =f(1) = 3. d. g(f(5)) =g(6) = 3. e.f(f(8)) =f(8) = 8. f. g(f(g(5))) =g(f(2)) =g(4) = 1. 1.1.16 a.h(g(0)) =h(0) =−1. b.g(f(4)) =g(−1) =−1. c.h(h(0)) =h(−1) = 0. d. g(h(f(4))) =g(h(−1)) =g(0) = 0. e.f(f(f(1))) =f(f(0)) =f(1) = 0. f. h(h(h(0))) =h(h(−1)) =h(0) =−1. g.f(h(g(2))) =f(h(3)) =f(0) = 1. h. g(f(h(4))) =g(f(4)) =g(−1) =−1. i.g(g(g(1))) =g(g(2)) =g(3) = 4. j. f(f(h(3))) =f(f(0)) =f(1) = 0. 1.1.17f(5)−f(0) 5−0=83−6 5=1 5.4; the radiosonde rises at an average rate of 15.4 ft/s during the first 5 seconds after it is released. Copyright c/circlecopyrt2019 Pearson Education, Inc.1.1. Review of Functions 7 1.1.18f(0) = 0. f(34) = 127852 .4−109731 = 18121 .4.f(64) = 127852 .4−75330.4 = 52522. f(64)−f(34) 64−34=52522−18121.4 30≈1146.69ft/s. 1.1.19f(−2) =f(2) = 2;g(−2) =−g(2) =−(−2) = 2;f(g(2)) =f(−2) =f(2) = 2;g(f(−2)) =g(f(2)) = g(2) =−2. 1.1.20The graph would be the result of leaving the portion of the graph in the first quadrant, and then also obtaining a portion in the third quadrant which would be the result of reflecting the portion in the first quadrant around the y-axis and then the x-axis. 1.1.21Function Ais symmetric about the y-axis, so is even. Function Bis symmetric about the origin so is odd. Function Cis symmetric about the y-axis, so is even. 1.1.22Function Ais symmetric about the y-axis, so is even. Function Bis symmetric about the origin, so is odd. Function Cis also symmetric about the origin, so is odd. 1.1.23f(x)=x2−5x+6 x−2=(x−2)(x−3) x−2=x−3,x/negationslash= 2. The domain of fis{x:x/negationslash=2}. The range is {y:y/negationslash=−1}. 1.1.24f(x)=x−2 2−x=x−2 −(x−2)=−1,x/negationslash= 2. The domain is {x:x/negationslash=2}. The range is {−1}. 1.1.25The domain of the function is the set of numbers xwhich satisfy 7 −x2≥0. This is the interval [−√ 7,√ 7]. Note that f(√ 7) = 0 and f(0) =√ 7. The range is [0 ,√ 7]. 1.1.26The domain of the function is the set of numbers xwhich satisfy 25 −x2≥0. This is the interval [−5,5]. Note that f(0) =−5a n df(5) = 0. The range is [ −5,0]. 1.1.27Because the cube root function is defined for all real numbrs, the domain is R, the set of all real numbers. 1.1.28The domain consists of the set of numbers wfor which 2 −w≥0, so the interval ( −∞,2]. 1.1.29The domain consists of the set of numbers xfor which 9 −x2≥0, so the interval [ −3,3]. 1.1.30Because 1+ t2is never zero for any real numbered value of t, the domain of this function is R,t h e set of all real numbers. 1.1.31a. The formula for the height of the rocket is valid from t= 0 until the rocket hits the ground, which is the positive solution to −16t2+96t+80 = 0, which the quadratic formula reveals is t=3+√ 14. Thus, the domain is [0 ,3+√ 14].b.



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) The maximum appears to occur at t=3 . The height at that time would be 224. 1.1.32 a.d(0) = (10 −(2.2)·0)2= 100. b .T h et a n ki sfi r s te m p t yw h e n d(t)=0 ,w h i c hi sw h e n1 0 −(2.2)t=0 ,o rt=5 0/11. c. An appropriate domain would [0 ,50/11]. Copyright c/circlecopyrt2019 Pearson Education, Inc.8 Chapter 1. Functions 1.1.33g(1/z)=( 1/z)3=1 z3 1.1.34F(y4)=1 y4−3 1.1.35F(g(y)) =F(y3)=1 y3−31.1.36f(g(w)) =f(w3)=(w3)2−4=w6−4 1.1.37g(f(u)) =g(u2−4) = (u2−4)3 1.1.38f(2+h)−f(2) h=(2+h)2−4−0 h=4+4h+h2−4 h=4h+h2 h=4+h 1.1.39F(F(x)) =F/parenleftbigg1 x−3/parenrightbigg =1 1 x−3−3=1 1 x−3−3(x−3) x−3=1 10−3x x−3=x−3 10−3x 1.1.40g(F(f(x))) =g(F(x2−4)) =g/parenleftbigg1 x2−4−3/parenrightbigg =/parenleftbigg1 x2−7/parenrightbigg3 1.1.41f(√x+4)=(√x+4)2−4=x+4−4=x. 1.1.42F((3x+1)/x)=1 3x+1 x−3=1 3x+1−3x x=x 3x+1−3x=x. 1.1.43g(x)=x3−5a n df(x)=x10. 1.1.44g(x)=x6+x2+1 and f(x)=2 x2. 1.1.45g(x)=x4+2 and f(x)=√x. 1.1.46g(x)=x3−1a n df(x)=1√x. 1.1.47(f◦g)(x)=f(g(x)) =f(x2−4) =|x2−4|. The domain of this function is the set of all real numbers. 1.1.48(g◦f)(x)=g(f(x)) =g(|x|)=|x|2−4=x2−4. The domain of this function is the set of all real numbers. 1.1.49(f◦G)(x)=f(G(x)) =f/parenleftBig 1 x−2/parenrightBig =/vextendsingle/vextendsingle/vextendsingle1 x−2/vextendsingle/vextendsingle/vextendsingle= 1 |x−2|. The domain of this function is the set of all real numbers except for the number 2. 1.1.50(f◦g◦G)(x)=f(g(G(x))) =f/parenleftBig g/parenleftBig 1 x−2/parenrightBig/parenrightBig =f/parenleftbigg/parenleftBig 1 x−2/parenrightBig2 −4/parenrightbigg =/vextendsingle/vextendsingle/vextendsingle/vextendsingle/parenleftBig 1 x−2/parenrightBig2 −4/vextendsingle/vextendsingle/vextendsingle/vextendsingle. The domain of this function is the set of all real numbers except for the number 2. 1.1.51(G◦g◦f)(x)=G(g(f(x))) =G(g(|x|)) =G(x 2−4) =1 x2−4−2=1 x2−6. The domain of this function is the set of all real numbers except for the numbers ±√ 6. 1.1.52(g◦F◦F)(x)=g(F(F(x))) =g(F(√x)) =g(/radicalbig√x)=√x−4. The domain is [0 ,∞). 1.1.53(g◦g)(x)=g(g(x)) =g(x2−4) = (x2−4)2−4=x4−8x2+16−4=x4−8x2+12. The domain is the set of all real numbers. 1.1.54(G◦G)(x)=G(G(x)) =G(1/(x−2)) =1 1 x−2−2=1 1−2(x−2) x−2=x−2 1−2x+4=x−2 5−2x.T h e nG◦Gis defined except where the denominator vanishes, so its domain is the set of all real numbers except for x=5 2. 1.1.55Because ( x2+3)−3=x2,w em a yc h o o s e f(x)=x−3. 1.1.56Because the reciprocal of x2+3is1 x2+3,w em a yc h o o s e f(x)=1 x. 1.1.57Because ( x2+3)2=x4+6x2+9,w ema yc hoose f(x)=x2. Copyright c/circlecopyrt2019 Pearson Education, Inc.1.1. Review of Functions 9 1.1.58Because ( x2+3 )2=x4+6x2+ 9, and the given expression is 11 more than this, we may choose f(x)=x2+11. 1.1.59Because ( x2)2+3=x4+ 3, this expression results from squaring x2and adding 3 to it. Thus we may choose f(x)=x2. 1.1.60Because x2/3+3=(3√x)2+3,wema yc hoose f(x)=3√x. 1.1.61 a. True. A real number zcorresponds to the domain element z/2+19, because f(z/2+19) = 2( z/2+ 19)−38 =z+38−38 =z. b. False. The definition of function does not require that each range element comes from a unique domain element, rather that each domain element is paired with a unique range element. c. True. f(1/x)=1 1/x=x,a n d1 f(x)=1 1/x=x. d. False. For example, suppose that fis the straight line through the origin with slope 1, so that f(x)=x. Thenf(f(x)) =f(x)=x, while (f(x))2=x2. e. False. For example, let f(x)=x+2 andg(x)=2x−1. Thenf(g(x)) =f(2x−1) = 2x−1+2=2 x+1, whileg(f(x)) =g(x+2)=2( x+2)−1=2x+3. f. True. This is the definition of f◦g. g. True. If fis even, then f(−z)=f(z) for all z,s ot h i si st r u ei np a r t i c u l a rf o r z=ax.S o i f g(x)=cf(ax), theng(−x)=cf(−ax)=cf(ax)=g(x), sogis even. h. False. For example, f(x)=xis an odd function, but h(x)=x+ 1 isn’t, because h(2) = 3, while h(−2) =−1 which isn’t −h(2). i. True. If f(−x)=−f(x)=f(x), then in particular −f(x)=f(x), so 0 = 2 f(x), sof(x) = 0 for all x. 1.1.62f(x+h)−f(x) h=10−10 h=0 h=0 . 1.1.63f(x+h)−f(x) h=3(x+h)−3x h=3x+3h−3x h=3h h=3. 1.1.64f(x+h)−f(x) h=4(x+h)−3−(4x−3) h=4x+4h−3−4x+3 h=4h h=4. 1.1.65f(x+h)−f(x) h=(x+h)2−x2 h=(x2+2hx+h2)−x2 h=h(2x+h) h=2x+h. 1.1.66f(x+h)−f(x) h=2(x+h)2−3(x+h)+1−(2x2−3x+1) h= 2x2+4xh+2h2−3x−3h+1−2x2+3x−1 h=4xh+2h2−3h h=h(4x+2h−3) h=4x+2h−3. 1.1.67f(x+h)−f(x) h=2 x+h−2 x h=2x−2(x+h) x(x+h) h=2x−2x−2h hx(x+h)=−2h hx(x+h)=−2 x(x+h). 1.1.68f(x+h)−f(x) h=x+h x+h+1−x x+1 h=(x+h)(x+1)−x(x+h+1) (x+1)(x+h+1) h=x2+x+hx+h−x2−xh−x h(x+1)(x+h+1)= h h(x+1)(x+h+1)=1 (x+1)(x+h+1) Copyright c/circlecopyrt2019 Pearson Education, Inc.10 Chapter 1. Functions 1.1.69f(x)−f(a) x−a=x2+x−(a2+a) x−a=(x2−a2)+(x−a) x−a=(x−a)(x+a)+(x−a) x−a= (x−a)(x+a+1) x−a=x+a+1. 1.1.70 f(x)−f(a) x−a=4−4x−x2−(4−4a−a2) x−a=−4(x−a)−(x2−a2) x−a=−4(x−a)−(x−a)(x+a) x−a =(x−a)(−4−(x+a)) x−a=−4−x−a. 1.1.71f(x)−f(a) x−a=x3−2x−(a3−2a) x−a=(x3−a3)−2(x−a) x−a=(x−a)(x2+ax+a2)−2(x−a) x−a= (x−a)(x2+ax+a2−2) x−a=x2+ax+a2−2. 1.1.72f(x)−f(a) x−a=x4−a4 x−a=(x2−a2)(x2+a2) x−a=(x−a)(x+a)(x2+a2) x−a=(x+a)(x2+a2). 1.1.73f(x)−f(a) x−a=−4 x2−−4 a2 x−a=−4a2+4x2 a2x2 x−a=4(x2−a2) (x−a)a2x2=4(x−a)(x+a) (x−a)a2x2=4(x+a) a2x2. 1.1.74f(x)−f(a) x−a=1 x−x2−(1 a−a2) x−a=1 x−1 a x−a−x2−a2 x−a=a−x ax x−a−(x−a)(x+a) x−a=−1 ax−(x+a). 1.1.75 a. The slope is12227−10499 3−1= 864 ft/h. The hiker’s elevation increases at an average rate of 874 feet per hour. b. The slope is12144−12631 5−4=−487 ft/h. The hiker’s elevation decreases at an average rate of 487 feet per hour. c. The hiker might have stopped to rest during this interval of time or the trail is level on this portion of the hike. 1.1.76 a. The slope is11302−9954 3−1= 674 ft/m. The elevation of the trail increases by an average of 674 feet per mile for 1 ≤d≤3. b. The slope is12237−12357 6−5=−120 ft/m. The elevation of the trail decreases by an average of 120 feet per mile for 5 ≤d≤6. c. The elevation of the trail doesn’t change much for 4 .5≤d≤5. 1.1.77 a.









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)) b. The slope of the secant line is given by 400−64 5−2=336 3= 112 feet per second. The object falls at an average rate of 112 feet per second over the interval 2 ≤t≤5. Copyright c/circlecopyrt2019 Pearson Education, Inc.1.1. Review of Functions 11 1.1.78 a.



 (/  )) (  ))b. The slope of the secant line is given by 4−1 .5−2=3 −1.5=−2 cubic centimeters per at- mosphere. The volume decreases by an av- erage of 2 cubic centimeters per atmosphereover the interval 0 .5≤p≤2. 1.1.79This function is symmetric about the y-axis, because f(−x)=(−x) 4+5(−x)2−12 =x4+5x2−12 = f(x). 1.1.80This function is symmetric about the origin, because f(−x)=3 (−x)5+2 (−x)3−(−x)=−3x5− 2x3+x=−(3x5+2x3−x)=f(x). 1.1.81This function has none of the indicated symmetries. For example, note that f(−2) =−26, while f(2) = 22, so fis not symmetric about either the origin or about the y-axis, and is not symmetric about thex-axis because it is a function. 1.1.82This function is symmetric about the y-axis. Note that f(−x)=2|−x|=2|x|=f(x). 1.1.83This curve (which is not a function) is symmetric about the x-axis, the y-axis, and the origin. Note that replacing either xby−xoryby−y(or both) yields the same equation. This is due to the fact that (−x)2/3=( (−x)2)1/3=(x2)1/3=x2/3, and a similar fact holds for the term involving y. 1.1.84This function is symmetric about the origin. Writing the function as y=f(x)=x3/5, we see that f(−x)=(−x)3/5=−(x)3/5=−f(x). 1.1.85This function is symmetric about the origin. Note that f(−x)=(−x)|(−x)|=−x|x|=−f(x). 1.1.86This curve (which is not a function) is symmetric about the x-axis, the y-axis, and the origin. Note that replacing either xby−xoryby−y(or both) yields the same equation. This is due to the fact that |−x|=|x|and|−y|=|y|. 1.1.87 a.f(g(−2)) =f(−g(2)) =f(−2) = 4 b.g(f(−2)) =g(f(2)) =g(4) = 1 c.f(g(−4)) =f(−g(4)) =f(−1) = 3 d. g(f(5)−8) =g(−2) =−g(2) =−2 e.g(g(−7)) =g(−g(7)) =g(−4) =−1f . f(1−f(8)) =f(−7) = 7 1.1.88 a.f(g(−1)) =f(−g(1)) =f(3) = 3 b.g(f(−4)) =g(f(4)) =g(−4) =−g(4) = 2 c.f(g(−3)) =f(−g(3)) =f(4) =−4d . f(g(−2)) =f(−g(2)) =f(1) = 2 e.g(g(−1)) =g(−g(1)) =g(3) =−4f . f(g(0)−1) =f(−1) =f(1) = 2 g.f(g(g(−2))) =f(g(−g(2))) =f(g(1)) =f(−3) = 3 h. g(f(f(−4))) =g(f(−4)) =g(−4) = 2 i.g(g(g(−1))) =g(g(−g(1))) =g(g(3)) =g(−4) = 2 Copyright c/circlecopyrt2019 Pearson Education, Inc.12 Chapter 1. Functions 1.1.89We will make heavy use of the fact that |x|isxif x>0, and is −xifx<0. In the first quadrant wherexandyare both positive, this equation becomes x−y= 1 which is a straight line with slope 1 and y-intercept −1. In the second quad- rant where xis negative and yis positive, this equation becomes −x−y= 1, which is a straight linewithslope −1andy-intercept −1. Inthethird quadrant where both xandyare negative, we ob- tain the equation −x−(−y) = 1, or y=x+1 , and in the fourth quadrant, we obtain x+y=1 . Graphing these lines and restricting them to the appropriate quadrants yields the following curve:/Minus4 /Minus2 2 4x /Minus4/Minus224y 1.1.90We have y=1 0+√ −x2+10x−9, so by subtracting 10 from both sides and squaring we have (y−10)2=−x2+10x−9, which can be written as x2−10x+(y−10)2=−9. To complete the square in x, we add 25 to both sides, yielding x2−10x+25+(y−10)2=−9+25, or (x−5)2+(y−10)2=1 6. This is the equation of a circle of radius 4 centered at (5 ,10). Because y≥10, we see that the graph of fis the upper half of this circle. The domain of the function is [1 ,9] and the range is [10 ,14]. 1.1.91We have y=2−√ −x2+6x+16, so by subtracting 2 from both sides and squaring we have (y−2)2=−x2+6x+16, which can be written as x2−6x+(y−2)2=1 6. To complete the square in x, we add 9 to both sides, yielding x2−6x+9+(y−2)2=1 6+9 , or (x−3)2+(y−2)2=2 5. This is the equation of a circle of radius 5 centered at (3 ,2). Because y≤2, we see that the graph of fis the lower half of this circle. The domain of the function is [ −2,8] and the range is [ −3,2]. 1.1.92 a. No. For example f(x)=x2+3 is an even function, but f(0) is not 0. b. Yes. because f(−x)=−f(x), and because −0=0 ,w em u s th a v e f(−0) =f(0) =−f(0), so f(0) =−f(0), and the only number which is its own additive inverse is 0, so f(0) = 0. 1.1.93Becausethecompositionof fwithitselfhasfirstdegree, fhasfirstdegreeasw ell,solet f(x)=ax+b. Then (f◦f)(x)=f(ax+b)=a(ax+b)+b=a2x+(ab+b). Equating coefficients, we see that a2= 9 and ab+b=−8. Ifa=3 ,w eg e tt h a t b=−2, while if a=−3w eh a v e b= 4. So the two possible answers are f(x)=3x−2a n df(x)=−3x+4. Copyright c/circlecopyrt2019 Pearson Education, Inc.1.1. Review of Functions 13 1.1.94Sincethesquareofalinearfunctionisaquadratic,welet f(x)=ax+b.T h e nf(x)2=a2x2+2abx+b2. Equating coefficients yields that a=±3a n db=±2. However, a quick check shows that the middle term is correct only when one of these is positive and one is negative. So the two possible such functions fare f(x)=3x−2a n df(x)=−3x+2. 1.1.95Letf(x)=ax2+bx+c.T h e n(f◦f)(x)=f(ax2+bx+c)=a(ax2+bx+c)2+b(ax2+bx+c)+c. Expanding this expression yields a3x4+2a2bx3+2a2cx2+ab2x2+2abcx+ac2+abx2+b2x+bc+c,w h i c h simplifies to a3x4+2a2bx3+(2a2c+ab2+ab)x2+(2abc+b2)x+(ac2+bc+c). Equating coefficients yields a3=1 ,s o a=1 . T h e n2 a2b=0 ,s o b= 0. It then follows that c=−6, so the original function was f(x)=x2−6. 1.1.96Because the square of a quadratic is a quartic, we let f(x)=ax2+bx+c. Then the square of f isc2+2bcx+b2x2+2acx2+2abx3+a2x4. By equating coefficients, we see that a2= 1 and so a=±1. Because the coefficient on x3must be 0, we have that b= 0. And the constant term reveals that c=±6. A quick check shows that the only possible solutions are thus f(x)=x2−6a n df(x)=−x2+6. 1.1.97f(x+h)−f(x) h=√ x+h−√x h=√ x+h−√x h·√ x+h+√x√ x+h+√x=(x+h)−x h(√ x+h+√x)= 1√ x+h+√x. f(x)−f(a) x−a=√x−√a x−a=√x−√a x−a·√x+√a√x+√a=x−a (x−a)(√x+√a)=1√x+√a. 1.1.98f(x+h)−f(x) h=/radicalbig 1−2(x+h)−√1−2x h= /radicalbig 1−2(x+h)−√1−2x h·/radicalbig 1−2(x+h)+√1−2x/radicalbig 1−2(x+h)+√1−2x= 1−2(x+h)−(1−2x) h(/radicalbig 1−2(x+h)+√1−2x)= −2/radicalbig 1−2(x+h)+√1−2x. f(x)−f(a) x−a=√1−2x−√1−2a x−a=√1−2x−√1−2a x−a·√1−2x+√1−2a√1−2x+√1−2a= (1−2x)−(1−2a) (x−a)(√1−2x+√1−2a)=(−2)(x−a) (x−a)(√1−2x+√1−2a)=−2 (√1−2x+√1−2a). 1.1.99f(x+h)−f(x) h=−3√x+h−−3√x h=−3(√x−√ x+h) h√x√ x+h=−3(√x−√ x+h) h√x√ x+h·√x+√ x+h√x+√ x+h= −3(x−(x+h)) h√x√ x+h(√x+√ x+h)=3√x√ x+h(√x+√ x+h). f(x)−f(a) x−a=−3√x−−3√a x−a=−3/parenleftBig√a−√x√a√x/parenrightBig x−a=(−3)(√a−√x) (x−a)√a√x·√a+√x√a+√x= (3)(x−a) (x−a)(√a√x)(√a+√x)=3√ax(√a+√x). 1.1.100f(x+h)−f(x) h=/radicalbig (x+h)2+1−√ x2+1 h= /radicalbig (x+h)2+1−√ x2+1 h·/radicalbig (x+h)2+1+√ x2+1/radicalbig (x+h)2+1+√ x2+1= (x+h)2+1−(x2+1) h(/radicalbig (x+h)2+1+√ x2+1)=x2+2hx+h2−x2 h(/radicalbig (x+h)2+1+√ x2+1)=2x+h/radicalbig (x+h)2+1+√ x2+1. Copyright c/circlecopyrt2019 Pearson Education, Inc.14 Chapter 1. Functions f(x)−f(a) x−a=√ x2+1−√ a2+1 x−a=√ x2+1−√ a2+1 x−a·√ x2+1+√ a2+1√ x2+1+√ a2+1= x2+1−(a2+1) (x−a)(√ x2+1+√ a2+1)=(x−a)(x+a) (x−a)(√ x2+1+√ a2+1)=x+a√ x2+1+√ a2+1. 1.1.101 This would not necessarily have either kind of symmetry. For example, f(x)=x2is an even function and g(x)=x3is odd, but the sum of these two is neither even nor odd. 1.1.102 This would be an odd function, so it would be symmetric about the origin. Suppose fis even and gis odd. Then ( f·g)(−x)=f(−x)g(−x)=f(x)·(−g(x)) =−(f·g)(x). 1.1.103 This would be an even function, so it would be symmetric about the y-axis. Suppose fis even and gis odd. Then g(f(−x)) =g(f(x)), because f(−x)=f(x). 1.1.104 This would be an even function, so it would be symmetric about the y-axis. Suppose fis even and gis odd. Then f(g(−x)) =f(−g(x)) =f(g(x)). 1.2 Representing Functions 1.2.1Functions can be defined and represented by a formula, through a graph, via a table, and by using words. 1.2.2The domain of every polynomial is the set of all real numbers. 1.2.3The slope of the line shown is m=−3−(−1) 3−0=−2/3. They-intercept is b=−1. Thus the function is given by f(x)=−2 3x−1. 1.2.4Because it is to be parallel to a line with slope 2, it must also have slope 2. Using the point-slope form of the equation of the line, we have y−0=2 (x−5), ory=2x−10. 1.2.5The domain of a rational functionp(x) q(x)is the set of all real numbers for which q(x)/negationslash=0 . 1.2.6A piecewise linear function is one which is linear over intervals in the domain. 1.2.7Forx<0, the graph is a line with slope 1 and y- intercept 3, while for x>0, it is a line with slope −1/2a n dy-intercept 3. Note that both of these lines contain the point (0 ,3). The function shown can thus be written f(x)=⎧ ⎨ ⎩x+3 if x<0; −1 2x+3 ifx≥0. 1.2.8The transformed graph would have equation y=√x−2+3. 1.2.9Compared to the graph of f(x), the graph of f(x+2) will be shifted 2 units to the left. 1.2.10Compared to the graph of f(x), the graph of −3f(x) will be scaled vertically by a factor of 3 and flipped about the xaxis. 1.2.11Compared to the graph of f(x), the graph of f(3x) will be compressed horizontally by a factor of1 3. 1.2.12To produce the graph of y=4 (x+3)2+6 from the graph of x2,o n em u s t 1. shift the graph horizontally by 3 units to left 2. scale the graph vertically by a factor of 43. shift the graph vertically up 6 units. Copyright c/circlecopyrt2019 Pearson Education, Inc.1.2. Representing Functions 15 1.2.13f(x)=|x−2|+3, because the graph of fis obtained from that of |x|by shifting 2 units to the right and 3 units up. g(x)=−|x+2|−1, because the graph of gis obtained from the graph of |x|by shifting 2 units to the left, then reflecting about the x-axis, and then shifting 1 unit down. 1.2.14 a.-
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 Copyright c/circlecopyrt2019 Pearson Education, Inc.16 Chapter 1. Functions 1.2.15 The slope is given by5−3 2−1= 2, so the equation of the line is y−3=2 (x−1), which can be written asf(x)=2x−2+3, or f(x)=2x+1. -


() 1.2.16 The slope is given by0−(−3) 5−2=1 ,s ot h ee q u a t i o n of the line is y−0=1 (x−5), orf(x)=x−5.
 
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() 1.2.17We are looking for the line with slope 3 that goes through the point (3 ,2). Using the point-slope form of the equation of a line, we have y−2=3 (x−3 ) ,w h i c hc a nb ew r i t t e na s y=2+3x−9, ory=3x−7. 1.2.18We are looking for the line with slope −4 which goes through the point ( −1,4). Using the point-slope form of the equation of a line, we have y−4=−4(x−(−1)), which can be written as y=4−4x−4, or y=−4x. 1.2.19We have 571 = Cs(100), so Cs=5.71. Therefore N(150) = 5 .71(150) = 856 .5 million. 1.2.20We have 226 = Cs(100), so Cs=2.26. Therefore N(150) = 2 .26(150) = 339 million. 1.2.21Using price as the independent variable pand the average number of units sold per day as the dependent variable d, we have the ordered pairs (250 ,12) and (200 ,15). The slope of the line determined by these points is m=15−12 200−250=3 −50. Thus the demand function has the form d(p)=(−3/50)p+bfor some constant b. Using the point (200 ,15), we find that 15 = ( −3/50)·200 +b,s ob= 27. Thus the demand function is d=(−3p/50)+27. While the domain of this linear function is the set of all real numbers, the formula is only likely to be valid for some subset of the interval (0 ,450), because outside of that interval eitherp≤0o rd≤0. Copyright c/circlecopyrt2019 Pearson Education, Inc.1.2. Representing Functions 17











() 1.2.22The profit is given by p=f(n)=8n−175. The break-even point is when p= 0, which occurs when n= 175/8=2 1.875, so they need to sell at least 22 tickets to not have a negative profit.
   
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 1.2.23 a. Using the points (1986 ,1875) and (2000 ,6471) we see that the slope is about 328.3. At t=0 ,t h ev a l u e ofpis 1875. Therefore a line which reasonably approximates the data is p(t) = 328.3t+1875. b. Using this line, we have that p(9) = 4830 breeding pairs. 1.2.24The cost per mile is the slope of the desired line, and the intercept is the fixed cost of 3.5. Thus, the cost per mile is given by c(m)=2.5m+3.5. When m=9 ,w eh a v e c(9) = (2.5)(9)+3 .5=2 2.5+3.5=2 6 dollars.





 1.2.25Forx≤3, we have the constant function 3. For x≥3, we have a straight line with slope 2 that contains the point (3 ,3). So its equation is y−3=2 (x−3), ory=2x−3. So the function can be written asf(x)=⎧ ⎨ ⎩3i f x≤3; 2x−3i fx>3 Copyright c/circlecopyrt2019 Pearson Education, Inc.18 Chapter 1. Functions 1.2.26Forx<3 we have straight line with slope 1 and y-intercept 1, so the equation is y=x+1. For x≥3, we have a straight line with slope −1 3which contains the point (3 ,2), so its equation is y−2=−1 3(x−3), ory=−1 3x+3. Thus the function can be written as f(x)=⎧ ⎨ ⎩x+1 if x<3; −1 3x+3 ifx≥3 1.2.27 T h ec o s ti sg i v e nb y c(t)=⎧ ⎨ ⎩0.05t for 0≤t≤60 1.2+0.03tfor 60<t≤120.






 1.2.28 T h ec o s ti sg i v e nb y c(m)=⎧ ⎨ ⎩3.5+2.5mfor 0≤m≤5 8.5+1.5mform>5.
   

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- b. The function is a polynomial, so its domain is the set of all real numbers. c. It has one peak near its y-intercept of (0 ,6) and one valley between x= 1 and x=2 . I t s x-intercept is nearx=−4/3. 1.2.36 a. -
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-- b. The function’s domain is the set of all real numbers. c. Ithasavalleyatthe y-interceptof(0 ,−2), andisvery steep atx=−2a n dx=2w h i c ha r et h e x-intercepts. It is symmetric about the y-axis. Copyright c/circlecopyrt2019 Pearson Education, Inc.20 Chapter 1. Functions 1.2.37 a. -
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 b. The domain of the function is the set of all real num- bers except −3. c. There is a valley near x=−5.2a n dap e a kn e a r x=−0.8. Thex-intercepts are at −2 and 2, where the curve does not appear to be smooth. There is a vertical asymptote at x=−3. The function is never below the x-axis. The y-intercept is (0 ,4/3). 1.2.38 a. -
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 b. The domain of the function is ( −∞,−2]∪[2,∞) c.x-intercepts are at −2 and 2. Because 0 isn’t in the domain, there is no y-intercept. The function has a valley at x=−4. 1.2.39 a. -
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-- b. The domain of the function is ( −∞,∞) c. The function has a maximum of 3 at x=1/2, and a y-intercept of 2. Copyright c/circlecopyrt2019 Pearson Education, Inc.1.2. Representing Functions 21 1.2.40 a. -


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 b. The domain of the function is ( −∞,∞) c. The function contains a jump at x=1 . T h em a x - imum value of the function is 1 and the minimum value is−1. 1.2.41 a. The zeros of fare the points where the graph crosses the x-axis, so these are points A,D,F,a n dI. b. The only high point, or peak, of foccurs at point E, because it appears that the graph has larger and largeryvalues as xincreases past point Iand decreases past point A. c. The only low points, or valleys, of fare at points BandH, again assuming that the graph of f continues its apparent behavior for larger values of x. d. Past point H, the graph is rising, and is rising faster and faster as xincreases. It is also rising between pointsBandE, but not as quickly as it is past point H. So the marked point at which it is rising most rapidly is I. e. Before point B, the graph is falling, and falls more and more rapidly as xb e c o m e sm o r ea n dm o r e negative. It is also falling between points EandH, but not as rapidly as it is before point B.S ot h e marked point at which it is falling most rapidly is A. 1.2.42 a. The zeros of gappear to be at x=0 ,x=1 ,x=1.6, andx≈3.15. b .T h et w op e a k so f gappear to be at x≈0.5a n dx≈2.6, with corresponding points ≈(0.5,0.4) and ≈(2.6,3.4). c. The only valley of gis at≈(1.3,−0.2). d. Moving right from x≈1.3, the graph is rising more and more rapidly until about x=2 ,a tw h i c h point it starts rising less rapidly (because, by x≈2.6, it is not rising at all). So the coordinates of the point at which it is rising most rapidly are approximately (2 .1,g(2))≈(2.1,2). Note that while the curve is also rising between x= 0 and x≈0.5, it is not rising as rapidly as it is near x=2 . e. To the right of x≈2.6, the curve is falling, and falling more and more rapidly as xincreases. So the point at which it is falling most rapidly in the interval [0 ,3] is atx= 3, which has the approximate coordinates (3 ,1.4). Note that while the curve is also falling between x≈0.5a n dx≈1.3, it is not falling as rapidly as it is near x=3 . Copyright c/circlecopyrt2019 Pearson Education, Inc.22 Chapter 1. Functions 1.2.43 a.-
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b. Thisappearstohaveamaximumwhen θ=0 .O u r vision is sharpest when we look straight ahead. c. For|θ|≤.19◦. We have an extremely narrow range where our eyesight is sharp. 1.2.44Because the line is horizontal, the slope is constantly 0. So S(x)=0 . 1.2.45The slope of this line is constantly 2, so the slope function is S(x)=2 . 1.2.46The function can be written as |x|=⎧ ⎨ ⎩−xifx≤0 xifx>0. The slope function is S(x)=⎧ ⎨ ⎩−1i fx<0 1i fx>0. 1.2.47The slope function is given by S(x)=⎧ ⎨ ⎩1i f x<0; −1/2i fx>0. 1.2.48The slope function is given by s(x)=⎧ ⎨ ⎩1i f x<3; −1/3i fx>3. 1.2.49 a. Because the area under consideration is that of a rectangle with base 2 and height 6, A(2) = 12. b. Because the area under consideration is that of a rectangle with base 6 and height 6, A(6) = 36. c. Because the area under consideration is that of a rectangle with base x and height 6, A(x)=6x. 1.2.50 a. Because the area under consideration is that of a triangle with base 2 and height 1, A(2) = 1. b. Because the area under consideration is that of a triangle with base 6 and height 3, the A(6) = 9. c. Because A(x) represents the area of a triangle with base xand height (1 /2)x, the formula for A(x)i s 1 2·x·x 2=x2 4. 1.2.51 a. Because the area under consideration is that of a trapezoid with base 2 and heights 8 and 4, we have A(2) = 2·8+4 2= 12. Copyright c/circlecopyrt2019 Pearson Education, Inc.1.2. Representing Functions 23 b. Notethat A(3)representstheareaofatrapezoidwithbase3andheights8and2, so A(3) = 3·8+2 2= 15. SoA(6) = 15+( A(6)−A(3)), and A(6)−A(3) represents the area of a triangle with base 3 and height 2. Thus A(6) = 15+6 = 21. c. Forxbetween 0 and 3, A(x) represents the area of a trapezoid with base x, and heights 8 and 8 −2x. Thus the area is x·8+8−2x 2=8x−x2.F o rx>3,A(x)=A(3)+A(x)−A(3) = 15+2( x−3) = 2x+9. Thus A(x)=⎧ ⎨ ⎩8x−x2if 0≤x≤3; 2x+9 if x>3. 1.2.52 a. Because the area under consideration is that of trapezoid with base 2 and heights 3 and 1, we have A(2) = 2·3+1 2=4 . b. Note that A(6) =A(2)+(A(6)−A(2), and that A(6)−A(2) represents a trapezoid with base 6 −2=4 and heights 1 and 5. The area is thus 4+/parenleftbig 4·1+5 2/parenrightbig = 4+12 = 16. c. Forxbetween 0 and 2, A(x) represents the area of a trapezoid with base x, and heights 3 and 3 −x. Thus the area is x·3+3−x 2=3x−x2 2.F o rx>2,A(x)=A(2)+A(x)−A(2) = 4+( A(x)−A(2)). Note thatA(x)−A(2) represents the area of a trapezoid with base x−2 and heights 1 and x−1. Thus A(x)=4+( x−2)·1+x−1 2=4+(x−2)/parenleftbigx 2/parenrightbig =x2 2−x+4.Th us A(x)=⎧ ⎨ ⎩3x−x2 2if 0≤x≤2; x2 2−x+4 ifx>2. 1.2.53 a. True. A polynomial p(x) can be written as the ratio of polynomialsp(x) 1, so it is a rational function. However, a rational function like1 xis not a polynomial. b. False. For example, if f(x)=2x,t h e n(f◦f)(x)=f(f(x)) =f(2x)=4xis linear, not quadratic. c. True. In fact, if fis degree mandgis degree n, then the degree of the composition of fandgism·n, regardless of the order they are composed. d. False. The graph would be shifted two units to the left. 1.2.54We complete the square for −x2+8x+ 9. Call this quantity z.T h e n z=−(x2−8x−9), so z=−(x2−8x+16+(−16−9)) =−((x−4)2−25 = 25−(x−4)2.T h u sf(x)isobtainedfromthegraph ofg(x)=√ 25−x2byshifting4unitstotheright. Thus the graph of fis the upper half of a circle of radius 5 centered at (4 ,0).

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