3000 Solved Problems In Calculus - WordPress

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SCHAUM'SOUTLINE OF3000 SOLVEDPROBLEMS INCalculusElliot Mendelson, Ph.D.Professor of MathematicsQueens CollegeCity University of New YorkSchaum's Outline SeriesMCGrawHillNew York Chicago San Francisco LisbonLondon Madrid Mexico City Milan New DelhiSan Juan Seoul Singapore Sydney Toronto

CONTENTSChapter 1INEQUALITIES1Chapter 2ABSOLUTE VALUE5Chapter 3LINES9Chapter 4CIRCLES19Chapter 5FUNCTIONS AND THEIR GRAPHS23Chapter 6LIMITS35Chapter 7CONTINUITY43Chapter 8THE DERIVATIVE49Chapter 9THE CHAIN RULE56Chapter 10TRIGONOMETRIC FUNCTIONS AND THEIR DERIVATIVES62Chapter 11ROLLE'S THEOREM, THE MEAN VALUE THEOREM, AND THE SIGNOF THE DERIVATIVE69Chapter 12HIGHER-ORDER DERIVATIVES AND IMPLICIT DIFFERENTIATION75Chapter 13MAXIMA AND MINIMA81Chapter 14RELATED RATES88Chapter 15CURVE SKETCHING (GRAPHS)100Chapter 16APPLIED MAXIMUM AND MINIMUM PROBLEMS118Chapter 17RECTILINEAR MOTION133Chapter 18APPROXIMATION BY DIFFERENTIALS138Chapter 19ANTIDERIVATIVES (INDEFINITE INTEGRALS)142Chapter 20THE DEFINITE INTEGRAL AND THE FUNDAMENTAL THEOREM OFCALCULUS152Chapter 21AREA AND ARC LENGTH163Chapter 22VOLUME173Chapter 23THE NATURAL LOGARITHM185Chapter 24EXPONENTIAL FUNCTIONS195Chapter 25L'HOPITAL'S RULE208Chapter 26EXPONENTIAL GROWTH AND DECAY215iii

ivCONTENTSChapter 27INVERSE TRIGONOMETRIC FUNCTIONS220Chapter 28INTEGRATION BY PARTS232Chapter 29TRIGONOMETRIC INTEGRANDS AND SUBSTITUTIONS238Chapter 30INTEGRATION OF RATIONAL FUNCTIONS: THE METHODOF PARTIAL FRACTIONS245INTEGRALS FOR SURFACE AREA, WORK, CENTROIDS253Chapter 31Surface Area of a Solid of Revolution / Work / Centroid of a Planar Region /Chapter 32IMPROPER INTEGRALS260Chapter 33PLANAR VECTORS268Chapter 34PARAMETRIC EQUATIONS, VECTOR FUNCTIONS, CURVILINEARMOTION274Parametric Equations of Plane Curves / Vector-Valued Functions /Chapter 35POLAR COORDINATES289Chapter 36INFINITE SEQUENCES305Chapter 37INFINITE SERIES312Chapter 38POWER SERIES326Chapter 39TAYLOR AND MACLAURIN SERIES340Chapter 40VECTORS IN SPACE. LINES AND PLANES347FUNCTIONS OF SEVERAL VARIABLES361Chapter 41Multivariate Functions and Their Graphs / Cylindrical and Spherical Coordinates /Chapter 42PARTIAL DERIVATIVES376Chapter 43DIRECTIONAL DERIVATIVES AND THE GRADIENT.EXTREME VALUES392Chapter 44MULTIPLE INTEGRALS AND THEIR APPLICATIONS405Chapter 45VECTOR FUNCTIONS IN SPACE. DIVERGENCE AND CURL.LINE INTEGRALS425DIFFERENTIAL EQUATIONS431INDEX443Chapter 46

To the StudentThis collection of solved problems covers elementary and intermediate calculus, and much of advancedcalculus. We have aimed at presenting the broadest range of problems that you are likely to encounter—theold chestnuts, all the current standard types, and some not so standard.Each chapter begins with very elementary problems. Their difficulty usually increases as the chapter progresses, but there is no uniform pattern.It is assumed that you have available a calculus textbook, including tables for the trigonometric, logarithmic, and exponential functions. Our ordering of the chapters follows the customary order found in manytextbooks, but as no two textbooks have exactly the same sequence of topics, you must expect an occasionaldiscrepancy from the order followed in your course.The printed solution that immediately follows a problem statement gives you all the details of one way tosolve the problem. You might wish to delay consulting that solution until you have outlined an attack in yourown mind. You might even disdain to read it until, with pencil and paper, you have solved the problemyourself (or failed gloriously). Used thus, 3000 Solved Problems in Calculus can almost serve as a supplement to any course in calculus, or even as an independent refresher course.V

2CHAPTER 1(1) holds when and only when x 10. But x 3 implies x 10, and, therefore, the inequality (1) holdsfor all x 3.Answer * 10 or x 3. As shown in Fig. 1-2, the solution is the union of the intervals (10, oo) and( »,3).Fig. 1-21.9Solve1. x 5 0 [This is equivalent to x -5.]. We multiply the inequality (1) by x 5. x I Case 1. x 5 0 [This is equivalent to x -5.]. We multiply the inequality (1) by x 5. x x 5, 0 5 [Subtract x.] This is always true. So, (1) holds throughout this case, that is, wheneverx 5, 0 5 [Subtract x.] This is always true. So, (1) holds throughout this case, that is, wheneverx -5. Case 2. x 5 0 [This is equivalent to x -5.]. We multiply the inequality ( 1 ) by x 5. Theinequality is reversed, since we are multiplying by a negative number. x x 5, 0 5 [Subtract*.] Butinequality is reversed, since we are multiplying by a negative number. x x 5, 0 5 [Subtract*.] But0 5 is false. Hence, the inequality (1) does not hold at all in this case.Answer x -5. In interval notation, the solution is the set (-5, ).1.10SolveCase 1. x 3 0 [This is equivalent to jc -3.]. Multiply the inequality (1) by x 3. x-7 2x 6, -7 x 6 [Subtract x.], -13 x [Subtract 6.] But x -13 is always false when * -3.Hence, this case yields no solutions. Case 2. x 3 0 [This is equivalent to x — 3.]. Multiply theinequality (1) by x 3. Since x 3 is negative, the inequality is reversed. x-7 2x 6, —7 x 6[Subtract x.] \3 x [Subtract 6.] Thus, when x —3, the inequality (1) holds when and only when* -13.Answer —13 x —3. In interval notation, the solution is the set (—13, —3).1.11Solve(2jt-3)/(3;t-5) 3.Case 1. 3A.-5 0 [This is equivalent to * §.]. 2x-3 9x-l5 [Multiply by 3jf-5.], -3 7x-15 [Subtract 2x.], I2 7x [Add 15.], T a * [Divide by 7.] So, when x f , the solutions mustsatisfy x " . Case 2. 3 x - 5 0 [This is equivalent to x .]. 2* - 3 9* - 15 [Multiply by 3*-5.Reverse the inequality.], -3 7jr-15 [Subtract 2*.], 12 7x [Add 15.], s x [Divide by 7.] Thus,when x f , the solutions must satisfy x ! f . This is impossible. Hence, this case yields no solutions.Answer f x s -y. In interval notation, the solution is the set (§, ].1.12Solve (2*-3)/(3*-5) 3.Remember that a product is positive when and only when both factors have the same sign. Casel. Jt-2 0and x 3 0. Then x 2 and jt —3. But these are equivalent to x 2 alone, since x 2 im-and x 3 0. Then x 2 and jt —3. But these are equivalent to x 2 alone, since x 2 implies x -3. Case 2. * - 2 0 and A: 3 0. Then x 2 and j c — 3 , which are equivalent tox —3, since x -3 implies x 2.Answer x 2 or x -3. In interval notation, this is the union of (2, ) and (— », —3).1.13Solve Problem 1.12 by considering the sign of the functionf(x) (x — 2)(x 3).Refer to Fig. 1-3. To the left of x — 3, both x-2 and x 3 are negative and /(*) is positive. Asone passes through x - — 3, the factor x - 3 changes sign and, therefore, f(x) becomes negative. f(x)remains negative until we pass through x 2, where the factor x — 2 changes sign and f(x) becomes andthen remains positive. Thus, f(x) is positive for x — 3 and for x 2. AnswerFig. 1-3

INEQUALITIES1.143Solve (x-l)(x 4) 0.The key points of the function g(x) (x - l)(x 4) are x — 4 and x l (see Fig. 1-4). To theleft of x -4, both x — 1 and x 4 are negative and, therefore, g(x) is positive. As we pass throughx — 4, jr 4 changes sign and g(x) becomes negative. When we pass through * 1, A: - 1 changes signand g(x) becomes and then remains positive. Thus, (x - \)(x 4) is negative for -4 x 1. AnswerFig. 1-4Fig. 1-51.15Solve x2 - 6x 5 0.Factor: x2 -6x 5 (x - l)(x - 5). Let h(x) (x - \)(x - 5). To the left of x 1 (see Fig. 1-5),both .* - 1 and jc - 5 are negative and, therefore, h(x) is positive. When we pass through x \, x-\changes sign and h(x) becomes negative. When we move further to the right and pass through x 5, x — 5changes sign and h(x) becomes positive again. Thus, h(x) is positive for x 1 and for x 5.Answer x 5 or x 1. This is the union of the intervals (5, ) and (— , 1).1.16Solve x2 Ix - 8 0.Factor: x2 Ix - 8 (x &)(x - 1), and refer to Fig. 1-6. For jc -8, both x 8 and x-lare negative and, therefore, F(x) (x 8)(x - 1) is positive. When we pass through x -8, x 8changes sign and, therefore, so does F(x). But when we later pass through x l, x-l changes sign andF(x) changes back to being positive. Thus, F(x) is negative for -8 x 1. AnswerFig. 1-6Fig. 1-71.172Solve 5x - 2x 0.Factor: 5x - 2x2 x(5 - 2x), and refer to Fig. 1-7. The key points for the function G(x) x(5 - 2x)are x 0 and * . For x Q, 5-2x is positive and, therefore, G(x) is negative. As we passthrough x 0, x changes sign and. therefore, G(x) becomes positive. When we pass through x ,5 — 2x changes sign and, therefore, G(x) changes back to being negative. Thus, G(x) is positive when and onlywhen 0 x . Answer1.18Solve (Jt-l) 2 (* 4) 0.(x — I)2 is always positive except when x 1 (when it is 0). So, the only solutions occur when* 4 0 and j c l .Answer x — 4 [In interval notation, (— , — 4).]1.19Solve x(x-l)(x l) 0.The key points for H(x) x(x - l)(x 1) are x 0, x l, and jc -l (see Fig. 1-8). For x tothe left of — 1, x, x — 1, and x 1 all are negative and, therefore, H(x) is negative. As we passthrough x — 1, x 1 changes sign and, therefore, so does H(x). When we later pass through x 0, xchanges sign and, therefore, H(x) becomes negative again. Finally, when we pass through x l, x-\changes sign and H(x) becomes and remains positive. Therefore, H(x) is positive when and only when— 1 A: 0 or x \. Answer

4CHAPTER 1Fig. 1-8Fig. 1-91.20Solve (2jt l)(jt-3)Cx 7) 0.See Fig. 1-9. The key points for the function K(x) (2x l)(x - 3)(x 7) are x -7, x -%, andx 3. For A: to the left of x--l, all three factors are negative and, therefore, AT(x) is negative. When wepass from left to right through x — 7, * 7 changes sign, and, therefore, K(x) becomes positive. Whenwe later pass through x - \, 2x 1 changes sign, and, therefore, K(x) becomes negative again. Finally,as we pass through x 3, x — 3 changes sign and K(x) becomes and remains positive. Hence, K(x) isnegative when and only when x -7 or 3 * 7. Answer1.21DoesimplyNo. Let a 1 and b -2.1.22Solve A- x2.x x2 is equivalent to x2-x 0, x(x-l) 0, 0 jc l.1.23Solve x2 x\jr .v3 is equivalent to x3 - x2 0, x'(x 1) 0, * 1, and x O.1.24Find all solutions ofThis is clearly true when x is negative and y positive, and false when x is positive and y negative. When .v andy are both positive, or x and y are both negative, multiplication by the positive quantity jcv yields the equivalentinequality y x.1.25Solve (x-l)(x-2)(x-3)(x-4) 0.Solve (x-l)(x-2)(x-3)(x-4) 0.When x 4, the product is positive. Figure 1-10 shows how the sign changes as one passes through thepoints 4, 3, 2,1. Hence, the inequality holds when l x 2 or 3 x 4 .Fig. 1-10

62.11CHAPTER 2Solve l 3/A- 2.This breaks up into two cases: Case 1. l 3/x 2. 3/x l [Hence, x 0.], 3 x. Case 2. 1 ilx -2. 3/x -3 [Hence, * 0.], 3 -3x [Reverse to .], -Kje [Reverse to .].So, either 0 A - 3 or -Kx 0. Answer2.12Solve * 2 -10 6.This is equivalent to -6 A- 2 -10 6, 4 jc 2 16, 2 j: s4.So, either 2 s j c 4 or — 4 s * — 2 . Answer2.13Solve 2*-3 * 2 .There are two cases: Case 1. 2*-3 .v 2. j c - 3 2, A- 5. Case 2. 2x - 3 -(jt 2). 2x - 3 -x-2, 3x-3 -2, 3.x 1, x \.So, either A- 5 or x j. Answer2.14Solve2x-l \x l\.Since an absolute value is never negative. 2 . v - l a O . There are two cases: Case 1. x 7 0. 2x — l A- 7, A - - 1 7, A- 8. Case 2. x 7 0. 2* - 1 - ( A - 7), 2*-l -jc-7, 3x - 1 -7, 3x -6,je -2. But then, 2jc-l -5 0.So, the only solution is x 8. Answer2.15Solve 2*-3 x 2 .This is equivalent to -\x 2\ 2x -3 x 2 . There are two cases: Case 1. A: 2 0. -(x 2) 2Ar-3 jc 2, - j c - 2 2 j c - 3 A : 2, K3A- and jc 5, x 5. Case 2: jc 2 0. -(jr 2) 2*-3 ;t 2, - x - 2 2 - 3 A : 2, l 3jc and A: 5, j j e and x 5 [impossible]. So, j A ' 5is the solution.2.16Solve\2x - 5 -4.There is no solution since an absolute value cannot be negative.2.17Solve 0 3* l iFirst solve 3* 1 5. This is equivalent to - 5 3 A 1 5 , - 3 A : - § [Subtract 1.], - ? x — [Divide by 3.] The inequality 0 \3x l excludes the case where 0 3* 1 , that is, where*--*.Answer All A: for which — 5 A- -1 except jc — 3.2.18The well-known triangle inequality asserts that « U S M U . Prove by mathematical induction that, forn 2, u, H2 un\ u, u z M,, .The case n 2 is the triangle inequality. Assume the result true for some n.and the inductive hypothesis,By the triangle inequality u, «2 un w n 1 s u, u2 «„! k 1 s ( M, u z M J) u,, 1 and, therefore, the result also holds for n 1.2.19Prove M — v\ \u\ — \v\ \.\u\ \u (u-v)\ \v\ \u-v\ [Triangleinequality.] Hence, \u - v\ a \u\ - \v\. Similarly, i -u s y - w . But, \v - u\ \u - v\. So, \u - v\ a (maximum of u - y and u - M ) u - \v\ \.2.20Solve *-l x-2 .Analytic solution. The given equation is equivalent to - A- -2 x - l \x -2\. Case 1. A--2 0.-(x-2) x-Kx-2. Then, -K-2, which is impossible. Case 2. A--2 0. -(x-2) x-l x-2, -x 2 x-l x-2, 3 2x, \ x. Thus, the solution consists of all A-such that A- .

ABSOLUTE VALUE7Geometric solution. \u — v\ is the distance between u and v. So, the solution consists of all points A: that arecloser to 1 than to 2. Figure 2-1 shows that these are all points x such that x .Fig. 2-12.21Solve \x l/x\ 2.This is equivalent to2[Since jc2 l 0.], *2 l 2 * , x2 -2\x\ 1 0,222 * -2W 1 0 [Since x x .], ( ;t -l) 0, \x\*\.Answer All x except * ! and x —l.2.22Solve \x l/x\ 4.This is equivalent to[Completing the square],When x 0, 2-V3 x 2 V3, and, when x 0, -2-V3 x -2 V3. Answer2.23Solve x K jc .When x O, this reduces to x 1 x , which is impossible. When x 0, the inequality becomesx K-x, which is equivalent to 2x l 0, or 2x —l, or x — \. Answer2.24Prove afr a - fc .From the definition of absolute value, a a and \b\ b.Since a - ft is nonnegative, a - fe must be ab .2.25Hence, a \b\ ( a)- ( b) (ab).Solve 2(x-4) 10. 2 - *-4 2(*-4) 10, 2 *-4 10, x-4 5, -5 jt:-4 5, -Kx 9. Answer2.26Solve \x2 - 17 8.There are two cases. Case 1. x 2 -17 8. * 2 25, x 5. Case 2. x2-ll -8. x2 9, x 3.So, there are four solutions: 3, 5. Answer2.27Solve jt-l l.- K x - K l , 0 x 2.2.28Solve\3x 5\ 4.-4 3A; 5 4 ,2.29Solve-9 3*: -l,-3 x -j. 4 2.First solve the negation, \x 4 s 2: — 2 s x 4 2, — 6 s — 2. Hence, the solution of the originalinequality is x -6 or * — 2.2.30Solve 2x-5 3.First solve the negation \2x-5\ 3:the original inequality is x s 1 or x s 4.2.31Solve-3 2x-5 3, 2 2x 8, Kx 4.Hence, the solution of 7je-5 3* 4 .Case 1. 7x-5 3A: 4. Then 4 9, x \. Case 2. 7 ; -5 -(3* 4). Then 7* - 5 -3x - 4,WA; 1, x tb Thus, the solutions are 1 and

82.32CHAPTER 2Solve 3*-2 s x-l .This is equivalent to - x-1 3*-2 s *-1 . Case 1. J t - l 0 . Then -(x - I)s3x - 2 x - 1,-* l 3 * - 2 * - l ; the first inequality is equivalent to x and the second to x s j . But this isimpossible. Case 2. *-l 0. -x l 3 x - 2 s * - l ; the first inequality is equivalent to jc s f andthe second to jt . Hence, we have f & x s . Answer2.33Solve * - 2 x - 5 9.Case 1. x 5. Then j r - 2 jt-5 9, 2*-7 9, 2* 16, x 8. Case 2. 2 x 5 . Thenj t - 2 5 - x 9, 3 9, which is impossible. Case 3. x 2. Then 2-x 5-x 9, l-2x 9,2x —2, x —\. So, the solutions are 8 and-1.2.34Solve 4-*s: 5x l .Case 1. Sx laO, that is, J t a - j . Then 4-* 5j: l, 3 6 :, i :. Thus, we obtain thesolutions - *: !. Case 2. 5* l 0, that is, x -%. Then 4-x -5x-l, 4x -5, xs- .Thus, we obtain the solutions -! * - . Hence, the set of solutions is [- 5, I] U [- f , - j) [-1, ].2.35Prove a-& « & .By the triangle inequality, \a-b\ o (-fc) « \-b\ \a\ \b\.2.36Solve the inequality \x - 1 a jc -3 .We argue geometrically from Fig. 2-2. \x — 1 is the distance of x from 1, and \x — 3 is the distance of xfrom 3. The point x 2 is equidistant from 1 and 3. Hence, the solutions consist of all x a 2.Fig. 2-2

CHAPTER 3Lines3.1Find the slope of the line through the points (—2, 5) and (7,1).Remember that the slope m of the line through two points (xlt y j and (x2, y2) is given by the equationHence, the slope of the given line is3.2Find a point-slope equation of the line through the points (1, 3) and (3, 6).The slope m of the given line is (6 - 3)/(3 - 1) . Recall that the point-slope equation of the line throughpoint (x1, y ) and with slope m is y — yt tn(x — *,). Hence, one point-slope equation of the given line, usingthe point (1, 3), is y — 3 \(x — 1). AnswerAnother point-slope equation, using the point (3,6), is y - 6 \(x — 3). Answer3.3Write a point-slope equation of the line through the points (1,2) and (1,3).The line through (1,2) and (1,3) is vertical and, therefore, does not have a slope.point-slope equation of the line.3.4Thus, there is noFind a point-slope equation of the line going through the point (1,3) with slope 5.y -3 5(* - 1). Answer3.5Find the slope of the line having the equationy - 7 2(x - 3) and find a point on the line.y — 7 2(x - 3) is a point-slope equation of the line. Hence, the slopethe line.3.6m 2, and (3, 7) is a point onFind the slope-intercept equation of the line through the points (2,4) and (4,8).Remember that the slope-intercept equation of a line is y mx b, where m is the slope and b is they-intercept (that is, the v-coordinate of the point where the line cuts the y-axis). In this case, the slopem (8-4)7(4-2) 2 .Method 1. A point-slope equation of the line is y - 8 2(* — 4). This is equivalent to y - 8 2* — 8, or,finally, to y 2x. AnswerMethod 2. The slope-intercept equation has the form y 2x b. Since (2,4) lies on the line, we maysubstitute 2 for x and 4 for y. So, 4 2 - 2 6 , and, therefore, b 0. Hence, the equation is y 2x.Answer3.7Find the slope-intercept equation of the line through the points (—1,6) and (2,15).The slope m (15 -6)/[2- (-1)] 1 3. Hence, the slope-intercept equation looks like y 3x b.Since (-1, 6) is on the line, 6 3 (— \) b, and therefore, b 9. Hence, the slope-intercept equation isy 3x 9.3.8Find the slope-intercept equation of the line through (2, —6) and the origin.The origin has coordinates (0,0). So, the slope m (-6 - 0) 1(2 - 0) -1 -3. Since the line cuts they-axis at (0, 0), the y-intercept b is 0. Hence, the slope-intercept equation is y -3x.3.9Find the slope-intercept equation of the line through (2,5) and (—1, 5).The line is horizontal. Since it passes through (2,5), an equation for it isslope-intercept equation, since the slope m 0 and the y-intercept b is 5.y 5 . But, this is the9

103.10CHAPTER 3Find the slope and y-intercept of the line given by the equation 7x 4y 8.If we solve the equation Ix 4y 8 for y, we obtain the equation y — \x 2, which is theslope-intercept equation. Hence, the slope m — I and the y-intercept b 2.3.11Show that every line has an equation of the form Ax By C, where A and B are not both 0, and that,conversely, every such equation is the equation of a line.If a given line is vertical, it has an equation x C. In this case, we can let A 1 and B 0. If thegiven line is not vertical, it has a slope-intercept equation y mx b, or, equivalently, — mx y b. So,let A — — m, 5 1, and C b. Conversely, assume that we are given an equation Ax By C, withA and B not both 0. If B 0, the equation is equivalent to x CIA, which is the equation of a verticalline. If B 0, solve the equation for y:with slope3.12This is the slope-intercept equation of the lineand y-interceptFind an equation of the line L through (-1,4) and parallel to the line M with the equation3x 4y 2.Remember that two lines are parallel if and only if their slopes are equal. If we solve 3x 4y 2 for y,namely, y — f * i, we obtain the slope-intercept equation for M. Hence, the slope of M is — and,therefore, the slope of the parallel line L also is - . So, L has a slope-intercept equation of the formy -\x b. Since L goes through (-1,4), 4 -\ (-1) b, and, therefore, fc 4-i "- Thus, theequation of L is y - \x T 3.13Show that the lines parallel to a line Ax By C are those lines having equations of the formfor some E. (Assume that B 0.)Ax By ESo, the slope isIf we solve Ax By C for y, we obtain the slope-intercept equation-A/B. Given a parallel line, it must also have slope —A/B and, therefore, has a slope-intercept equationand, thence to Ax By bB. Conversely, a line withwhich is equivalent toequation Ax By E must have slope -A/B (obtained by putting the equation in slope-intercept form) andis, therefore, parallel to the line with equation Ax By C.3.14Find an equation of the line through (2, 3) and parallel to the line with the equation 4x — 2y 7.By Problem 3.13, the required line must have an equation of the form 4x - 2y E. Since (2, 3) lies on theline, 4(2) - 2(3) E. So, 8-6 2. Hence, the desired equation is 4x - 2y 2.3.15Find an equation of the line through (2,3) and parallel to the line with the equationy 5.Since y 5 is the equation of a horizontal line, the required parallel line is horizontal. Since it passesthrough (2, 3), an equation for it is y 3.3.16Show that any line that is neither vertical nor horizontal and does not pass through the origin has an equation ofthe formwhere b is the y-intercept and a is the -intercept (Fig. 3-1).Fig. 3-1In Problem 3.11, set CIA a and CIB b. Notice that, when y 0, the equation yields the valuex a, and, therefore, a is the x-intercept of the line. Similarly for the y-intercept.

LINES3.1711Find an equation of the line through the points (0,2) and (3,0).The y-intercept is b 2 and the -intercept is a 3. So, by Problem 3.16, an equation of the line is3.18If the point (2, k) lies on the line with slopem 3 passing through the point (1, 6), find k.A point-slope equation of the line is y — 6 3(x — 1). Since (2, k) lies on the line, k — 6 3(2-1).Hence, k 9.3.19Does the point (-1, -2) lie on the line L through the points (4,7) and (5,9)?The slope of L is (9 - 7)7(5 - 4) 2. Hence, a point-slope equation of L is y-7 2(x- 4). If wesubstitute —1 for x and -2 for y in this equation, we obtain —2 — 7 2(-l — 4), or —9 -10, which isfalse. Hence, (—1, —2) does not lie on L.3.20Find the slope-intercept equation of the line M through (1,4) that is perpendicular to the line L with equation2x - 6y 5.Solve 2x - 6y 5 for y, obtaining y \x — f . So,

CONTENTS Chapter 1 INEQUALITIES Chapter 2 ABSOLUTE VALUE Chapter 3 LINES Chapter 4 CIRCLES Chapter 5 FUNCTIONS AND THEIR GRAPHS Chapter 6 LIMITS Chapter 7 CONTINUITY Chapter 8 THE DERIVATIVE Chapter 9 THE CHAIN RULE Chapter 10 TRIGONOMETRIC FUNCTIONS AND THEIR DERIVATIVES Chapter 1

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