Skip to content
Textbook Guides

Solution Manual for Single Variable Calculus: Concepts and Contexts, 4th Edition

Solution Manual for Single Variable Calculus: Concepts and Contexts, 4th Edition provides you with expert textbook solutions that ensure you understand every concept thoroughly.

Preview 16 of 1377 pages

See the first 16 pages exactly as they appear in the document.

  1. Page 1
    Read page 1 as text

    COMPLETE SOLUTIONS MANUAL for Stewart’s SINGLE VARIABLE CALCULUS CONCEPTS AND CONTEXTS FOURTH EDITION JEFFERY A. COLE Anoka Ramsey Community College TIMOTHY J. FLAHERTY Carnegie Mellon University

  2. Page 2
  3. Page 3
    Read page 3 as text

    ABBREVIATIONS AND SYMBOLS CD concave downward CU concave upward D the domain of  FDT First Derivative Test HA horizontal asymptote(s) I interval of convergence IP inection point(s) R radius of convergence VA vertical asymptote(s) CAS = indicates the use of a computer algebra system. H = indicates the use of l’Hospital’s Rule.  = indicates the use of Formula  in the Table of Integrals in the back endpapers. s = indicates the use of the substitution {  = sin   = cos   } . c = indicates the use of the substitution {  = cos   =  sin   } . I/D Increasing/Decreasing Test

  4. Page 4
    Read page 4 as text

    vii CONTENTS ■ DIAGNOSTIC TESTS 1 1 ■ FUNCTIONS AND MODELS 9 1.1 Four Ways to Represent a Function 9 1.2 Mathematical Models: A Catalog of Essential Functions 20 1.3 New Functions from Old Functions 26 1.4 Graphing Calculators and Computers 38 1.5 Exponential Functions 45 1.6 Inverse Functions and Logarithms 51 1.7 Parametric Curves 59 Laboratory Project ■ Running Circles Around Circles 71 Review 73 Principles of Problem Solving 83 2 ■ LIMITS AND DERIVATIVES 87 2.1 The Tangent and Velocity Problems 87 2.2 The Limit of a Function 90 2.3 Calculating Limits Using the Limit Laws 96 2.4 Continuity 104 2.5 Limits Involving Infinity 113 2.6 Derivatives and Rates of Change 123 2.7 The Derivative as a Function 134 2.8 What Does Say about ? 146 Review 152 Focus on Problem Solving 163 3 ■ DIFFERENTIATION RULES 167 3.1 Derivatives of Polynomials and Exponential Functions 167 Applied Project ■ Building a Better Roller Coaster 177 3.2 The Product and Quotient Rules 179 f f  Q Q

  5. Page 5
    Read page 5 as text

    viii ■ CONTENTS 3.3 Derivatives of Trigonometric Functions 188 3.4 The Chain Rule 194 Laboratory Project ■ Bézier Curves 207 Applied Project ■ Where Should a Pilot Start Descent? 208 3.5 Implicit Differentiation 209 3.6 Inverse Trigonometric Functions and Their Derivatives 219 3.7 Derivatives of Logarithmic Functions 225 Discovery Project ■ Hyperbolic Functions 230 3.8 Rates of Change in the Natural and Social Sciences 232 3.9 Linear Approximations and Differentials 241 Laboratory Project ■ Taylor Polynomials 247 Review 249 Focus on Problem Solving 261 4 ■ APPLICATIONS OF DIFFERENTIATION 271 4.1 Related Rates 271 4.2 Maximum and Minimum Values 278 Applied Project ■ The Calculus of Rainbows 288 4.3 Derivatives and the Shapes of Curves 289 4.4 Graphing with Calculus and Calculators 310 4.5 Indeterminate Forms and l’Hospital’s Rule 328 4.6 Optimization Problems 340 Applied Project ■ The Shape of a Can 358 4.7 Newton’s Method 359 4.8 Antiderivatives 368 Review 375 Focus on Problem Solving 393 5 ■ INTEGRALS 403 5.1 Areas and Distances 403 5.2 The Definite Integral 412 5.3 Evaluating Definite Integrals 421 Discovery Project ■ Area Functions 428 5.4 The Fundamental Theorem of Calculus 430 5.5 The Substitution Rule 436 5.6 Integration by Parts 444 5 .6 5.6 I ntegration by Parts 444 Integration by Parts 444

  6. Page 6
    Read page 6 as text

    CONTENTS ■ ix 5.7 Additional Techniques of Integration 452 5.8 Integration Using Tables and Computer Algebra Systems 459 Discovery Project ■ Patterns in Integrals 465 5.9 Approximate Integration 467 5.10 Improper Integrals 479 Review 491 Focus on Problem Solving 505 6 ■ APPLICATIONS OF INTEGRATION 513 6.1 More about Areas 513 6.2 Volumes 523 Discovery Project ■ Rotating on a Slant 537 6.3 Volumes by Cylindrical Shells 538 6.4 Arc Length 546 Discovery Project ■ Arc Length Contest 554 6.5 Average Value of a Function 554 Applied Project ■ Where To Sit at the Movies 557 6.6 Applications to Physics and Engineering 557 Discovery Project ■ Complementary Coffee Cups 568 6.7 Applications to Economics and Biology 569 6.8 Probability 572 Review 575 Focus on Problem Solving 585 7 ■ DIFFERENTIAL EQUATIONS 593 7.1 Modeling with Differential Equations 593 7.2 Direction Fields and Euler’s Method 596 7.3 Separable Equations 604 Applied Project ■ How Fast Does a Tank Drain? 616 Applied Project ■ Which Is Faster, Going Up or Coming Down? 618 7.4 Exponential Growth and Decay 619 Applied Project ■ Calculus and Baseball 624 7.5 The Logistic Equation 625 7.6 Predator-Prey Systems 635 Review 640 Focus on Problem Solving 647 Focus on Problem Solving 647 Focus on Problem Solving 6

  7. Page 7
    Read page 7 as text

    x ■ CONTENTS 8 ■ INFINITE SEQUENCES AND SERIES 653 8.1 Sequences 653 Laboratory Project ■ Logistic Sequences 661 8.2 Series 665 8.3 The Integral and Comparison Tests; Estimating Sums 678 8.4 Other Convergence Tests 684 8.5 Power Series 691 8.6 Representations of Functions as Power Series 698 8.7 Taylor and Maclaurin Series 707 Laboratory Project ■ An Elusive Limit 721 8.8 Applications of Taylor Polynomials 722 Applied Project ■ Radiation from the Stars 733 Review 735 Focus on Problem Solving 747 ■ APPENDIXES 753 A Intervals, Inequalities, and Absolute Values 753 B Coordinate Geometry 756 C Trigonometry 762 D Precise Definitions of Limits 766 F Sigma Notation 771 G Integration of Rational Functions by Partial Fractions 774 H Polar Coordinates 785 Discovery Project ■ Conic Sections in Polar Coordinates 808 I Complex Numbers 809

  8. Page 8
    Read page 8 as text

    DIAGNOSTIC TESTS Test A Algebra 1. (a) (  3) 4 = (  3)(  3)(  3)(  3) = 81 (b)  3 4 =  (3)(3)(3)(3) =  81 (c) 3  4 = 1 3 4 = 1 81 (d) 5 23 5 21 = 5 23  21 = 5 2 = 25 (e)  2 3   2 =  3 2  2 = 9 4 (f ) 16  3  4 = 1 16 3  4 = 1  4  16  3 = 1 2 3 = 1 8 2. (a) Note that  200 =  100 · 2 = 10  2 and  32 =  16 · 2 = 4  2 . Thus  200   32 = 10  2  4  2 = 6  2 . (b) (3  3 3 )(4  2 ) 2 = 3  3 3 16  2 4 = 48  5 7 (c)  3  3  2 3  2  1  2   2 =   2  1  2 3  3  2 3  2 = (  2  1  2 ) 2 (3  3  2 3 ) 2 =  4  1 9  3 6 =  4 9  3 6 =  9 7 3. (a) 3(  + 6) + 4(2   5) = 3  + 18 + 8   20 = 11   2 (b) (  + 3)(4   5) = 4  2  5  + 12   15 = 4  2 + 7   15 (c)    +         =     2     +        2 =   Or: Use the formula for the difference of two squares to see that    +        =     2     2 =   . (d) (2  + 3) 2 = (2  + 3)(2  + 3) = 4  2 + 6  + 6  + 9 = 4  2 + 12  + 9 . Note: A quicker way to expand this binomial is to use the formula (  + ) 2 =  2 + 2  + 2 with  = 2  and = 3 : (2  + 3) 2 = (2  ) 2 + 2(2  )(3) + 3 2 = 4  2 + 12  + 9 (e) See Reference Page 1 for the binomial formula (  + ) 3 =  3 + 3  2 + 3  2 + 3 . Using it, we get (  + 2) 3 =  3 + 3  2 (2) + 3  (2 2 ) + 2 3 =  3 + 6  2 + 12  + 8 . 4. (a) Using the difference of two squares formula,  2  2 = (  + )(   ) , we have 4  2  25 = (2  ) 2  5 2 = (2  + 5)(2   5) . (b) Factoring by trial and error, we get 2  2 + 5   12 = (2   3)(  + 4) . (c) Using factoring by grouping and the difference of two squares formula, we have  3  3  2  4  + 12 =  2 (   3)  4(   3) = (  2  4)(   3) = (   2)(  + 2)(   3) . (d)  4 + 27  =  (  3 + 27) =  (  + 3)(  2  3  + 9) This last expression was obtained using the sum of two cubes formula,  3 + 3 = (  + )(  2   + 2 ) with  =  and = 3 . [See Reference Page 1 in the textbook.] (e) The smallest exponent on  is  1 2 , so we will factor out   1  2 . 3  3  2  9  1  2 + 6   1  2 = 3   1  2 (  2  3  + 2) = 3   1  2 (   1)(   2) (f )  3  4  =  (  2  4) =  (   2)(  + 2) 1

  9. Page 9
    Read page 9 as text

    2 ¤ DIAGNOGSTIC TESTS 5. (a)  2 + 3  + 2  2    2 = (  + 1)(  + 2) (  + 1)(   2) =  + 2   2 (b) 2  2    1  2  9 ·  + 3 2  + 1 = (2  + 1)(   1) (   3)(  + 3) ·  + 3 2  + 1 =   1   3 (c)  2  2  4   + 1  + 2 =  2 (   2)(  + 2)   + 1  + 2 =  2 (   2)(  + 2)   + 1  + 2 ·   2   2 =  2  (  + 1)(   2) (   2)(  + 2) =  2  (  2    2) (  + 2)(   2) =  + 2 (  + 2)(   2) = 1   2 (d)    1  1  =    1  1  ·   = 2   2   = (   )( +  )  (   ) = +   1 =  (  + ) 6. (a)  10  5  2 =  10  5  2 ·  5 + 2  5 + 2 =  50 + 2  10   5  2  2 2 = 5  2 + 2  10 5  4 = 5  2 + 2  10 (b)  4 +  2 =  4 +  2 ·  4 + + 2  4 + + 2 = 4 +  4   4 + + 2  =   4 + + 2  = 1  4 + + 2 7. (a)  2 +  + 1 =   2 +  + 1 4  + 1  1 4 =   + 1 2  2 + 3 4 (b) 2  2  12  + 11 = 2(  2  6  ) + 11 = 2(  2  6  + 9  9) + 11 = 2(  2  6  + 9)  18 + 11 = 2(   3) 2  7 8. (a)  + 5 = 14  1 2    + 1 2  = 14  5  3 2  = 9   = 2 3 · 9   = 6 (b) 2   + 1 = 2   1   2  2 = (2   1)(  + 1)  2  2 = 2  2 +   1   = 1 (c)  2    12 = 0  (  + 3)(   4) = 0   + 3 = 0 or   4 = 0   =  3 or  = 4 (d) By the quadratic formula, 2  2 + 4  + 1 = 0   =  4 ±  4 2  4(2)(1) 2(2) =  4 ±  8 4 =  4 ± 2  2 4 = 2   2 ±  2  4 =  2 ±  2 2 =  1 ± 1 2  2 . (e)  4  3  2 + 2 = 0  (  2  1)(  2  2) = 0   2  1 = 0 or  2  2 = 0   2 = 1 or  2 = 2   = ± 1 or  = ±  2 (f ) 3 |   4 | = 10  |   4 | = 10 3    4 =  10 3 or   4 = 10 3   = 2 3 or  = 22 3 (g) Multiplying through 2  (4   )  1  2  3  4   = 0 by (4   ) 1  2 gives 2   3(4   ) = 0  2   12 + 3  = 0  5   12 = 0  5  = 12   = 12 5 . 9. (a)  4 5  3   17   9  3   12  3    4 or  4   3 . In interval notation, the answer is [  4  3) . (b)  2 2  + 8   2  2   8 0  (  + 2)(   4) 0 . Now, (  + 2)(   4) will change sign at the critical values  =  2 and  = 4 . Thus the possible intervals of solution are (    2) , (  2  4) , and (4   ) . By choosing a single test value from each interval, we see that (  2  4) is the only interval that satises the inequality.

  10. Page 10
    Read page 10 as text

    TEST B ANALYTIC GEOMETRY ¤ 3 (c) The inequality  (   1)(  + 2) 0 has critical values of  2  0  and 1 . The corresponding possible intervals of solution are (    2) , (  2  0) , (0  1) and (1   ) . By choosing a single test value from each interval, we see that both intervals (  2  0) and (1   ) satisfy the inequality. Thus, the solution is the union of these two intervals: (  2  0) (1   ) . (d) |   4 | 3   3   4 3  1  7 . In interval notation, the answer is (1  7) . (e) 2   3  + 1  1  2   3  + 1  1  0  2   3  + 1   + 1  + 1  0  2   3    1  + 1  0    4  + 1  0 . Now, the expression   4  + 1 may change signs at the critical values  =  1 and  = 4 , so the possible intervals of solution are (    1) , (  1  4] , and [4   ) . By choosing a single test value from each interval, we see that (  1  4] is the only interval that satises the inequality. 10. (a) False. In order for the statement to be true, it must hold for all real numbers, so, to show that the statement is false, pick  = 1 and  = 2 and observe that (1 + 2) 2 6 = 1 2 + 2 2 . In general, (  +  ) 2 =  2 + 2  +  2 . (b) True as long as  and are nonnegative real numbers. To see this, think in terms of the laws of exponents:   = (  ) 1  2 =  1  2 1  2 =    . (c) False. To see this, let  = 1 and  = 2 , then  1 2 + 2 2 6 = 1 + 2 . (d) False. To see this, let  = 1 and  = 2 , then 1 + 1(2) 2 6 = 1 + 1 . (e) False. To see this, let  = 2 and = 3 , then 1 2  3 6 = 1 2  1 3 . (f ) True since 1     ·   = 1   , as long as  6 = 0 and   6 = 0 . Test B Analytic Geometry 1. (a) Using the point (2   5) and  =  3 in the point-slope equation of a line,  1 =  (    1 ) , we get  (  5) =  3(   2)  + 5 =  3  + 6  =  3  + 1 . (b) A line parallel to the  -axis must be horizontal and thus have a slope of 0 . Since the line passes through the point (2   5) , the -coordinate of every point on the line is  5 , so the equation is =  5 . (c) A line parallel to the -axis is vertical with undened slope. So the  -coordinate of every point on the line is 2 and so the equation is  = 2 . (d) Note that 2   4 = 3   4 =  2  + 3  = 1 2   3 4 . Thus the slope of the given line is  = 1 2 . Hence, the slope of the line we’re looking for is also 1 2 (since the line we’re looking for is required to be parallel to the given line). So the equation of the line is  (  5) = 1 2 (   2)  + 5 = 1 2   1  = 1 2   6 . 2. First we’ll nd the distance between the two given points in order to obtain the radius,  , of the circle:  =  [3  (  1)] 2 + (  2  4) 2 =  4 2 + (  6) 2 =  52 . Next use the standard equation of a circle, (   ) 2 + (   ) 2 =  2 , where (   ) is the center, to get (  + 1) 2 + (  4) 2 = 52 . TEST B

  11. Page 11
    Read page 11 as text

    4 ¤ DIAGNOGSTIC TESTS 3. We must rewrite the equation in standard form in order to identify the center and radius. Note that  2 + 2  6  + 10 + 9 = 0   2  6  + 9 + 2 + 10 = 0 . For the left-hand side of the latter equation, we factor the rst three terms and complete the square on the last two terms as follows:  2  6  + 9 + 2 + 10 = 0  (   3) 2 + 2 + 10 + 25 = 25  (   3) 2 + ( + 5) 2 = 25 . Thus, the center of the circle is (3   5) and the radius is 5 . 4. (a)  (  7  4) and  (5   12)    =  12  4 5  (  7) =  16 12 =  4 3 (b)  4 =  4 3 [   (  7)]   4 =  4 3   28 3  3  12 =  4   28  4  + 3 + 16 = 0 . Putting = 0 , we get 4  + 16 = 0 , so the  -intercept is  4 , and substituting 0 for  results in a -intercept of  16 3 . (c) The midpoint is obtained by averaging the corresponding coordinates of both points:   7+5 2  4+(  12) 2  = (  1   4) . (d)  =  [5  (  7)] 2 + (  12  4) 2 =  12 2 + (  16) 2 =  144 + 256 =  400 = 20 (e) The perpendicular bisector is the line that intersects the line segment  at a right angle through its midpoint. Thus the perpendicular bisector passes through (  1   4) and has slope 3 4 [the slope is obtained by taking the negative reciprocal of the answer from part (a)]. So the perpendicular bisector is given by + 4 = 3 4 [   (  1)] or 3   4 = 13 . (f ) The center of the required circle is the midpoint of  , and the radius is half the length of  , which is 10 . Thus, the equation is (  + 1) 2 + ( + 4) 2 = 100 . 5. (a) Graph the corresponding horizontal lines (given by the equations =  1 and = 3 ) as solid lines. The inequality   1 describes the points (  ) that lie on or above the line =  1 . The inequality  3 describes the points (  ) that lie on or below the line = 3 . So the pair of inequalities  1   3 describes the points that lie on or between the lines =  1 and = 3 . (b) Note that the given inequalities can be written as  4  4 and  2 2 , respectively. So the region lies between the vertical lines  =  4 and  = 4 and between the horizontal lines =  2 and = 2 . As shown in the graph, the region common to both graphs is a rectangle (minus its edges) centered at the origin. (c) We rst graph = 1  1 2  as a dotted line. Since 1  1 2  , the points in the region lie below this line.

  12. Page 12
    Read page 12 as text

    TEST C FUNCTIONS ¤ 5 (d) We rst graph the parabola =  2  1 using a solid curve. Since   2  1 , the points in the region lie on or above the parabola. (e) We graph the circle  2 + 2 = 4 using a dotted curve. Since   2 + 2 2 , the region consists of points whose distance from the origin is less than 2, that is, the points that lie inside the circle. (f ) The equation 9  2 + 16 2 = 144 is an ellipse centered at (0  0) . We put it in standard form by dividing by 144 and get  2 16 + 2 9 = 1 . The  -intercepts are located at a distance of  16 = 4 from the center while the -intercepts are a distance of  9 = 3 from the center (see the graph). Test C Functions 1. (a) Locate  1 on the  -axis and then go down to the point on the graph with an  -coordinate of  1 . The corresponding -coordinate is the value of the function at  =  1 , which is  2 . So,  (  1) =  2 . (b) Using the same technique as in part (a), we get  (2) 2  8 . (c) Locate 2 on the -axis and then go left and right to nd all points on the graph with a -coordinate of 2 . The corresponding  -coordinates are the  -values we are searching for. So  =  3 and  = 1 . (d) Using the same technique as in part (c), we get   2  5 and  0  3 . (e) The domain is all the  -values for which the graph exists, and the range is all the -values for which the graph exists. Thus, the domain is [  3  3] , and the range is [  2  3] . 2. Note that  (2 + ) = (2 + ) 3 and  (2) = 2 3 = 8 . So the difference quotient becomes  (2 + )   (2) = (2 + ) 3  8 = 8 + 12 + 6 2 + 3  8 = 12 + 6 2 + 3 = (12 + 6 + 2 ) = 12 + 6 + 2 . 3. (a) Set the denominator equal to 0 and solve to nd restrictions on the domain:  2 +   2 = 0  (   1)(  + 2) = 0   = 1 or  =  2 . Thus, the domain is all real numbers except 1 or  2 or, in interval notation, (    2) (  2  1) (1   ) . (b) Note that the denominator is always greater than or equal to 1 , and the numerator is dened for all real numbers. Thus, the domain is (    ) . (c) Note that the function is the sum of two root functions. So is dened on the intersection of the domains of these two root functions. The domain of a square root function is found by setting its radicand greater than or equal to 0 . Now,

  13. Page 13
    Read page 13 as text

    6 ¤ DIAGNOGSTIC TESTS 4    0    4 and  2  1  0  (   1)(  + 1)  0     1 or   1 . Thus, the domain of is (    1] [1  4] . 4. (a) Reect the graph of  about the  -axis. (b) Stretch the graph of  vertically by a factor of 2 , then shift 1 unit downward. (c) Shift the graph of  right 3 units, then up 2 units. 5. (a) Make a table and then connect the points with a smooth curve:   2  1 0 1 2  8  1 0 1 8 (b) Shift the graph from part (a) left 1 unit. (c) Shift the graph from part (a) right 2 units and up 3 units. (d) First plot =  2 . Next, to get the graph of  (  ) = 4   2 , reect  about the x -axis and then shift it upward 4 units. (e) Make a table and then connect the points with a smooth curve:  0 1 4 9 0 1 2 3 (f ) Stretch the graph from part (e) vertically by a factor of two.

  14. Page 14
    Read page 14 as text

    TEST D TRIGONOMETRY ¤ 7 (g) First plot = 2  . Next, get the graph of =  2  by reecting the graph of = 2  about the x -axis. (h) Note that = 1 +   1 = 1 + 1  . So rst plot = 1  and then shift it upward 1 unit. 6. (a)  (  2) = 1  (  2) 2 =  3 and  (1) = 2(1) + 1 = 3 (b) For   0 plot  (  ) = 1   2 and, on the same plane, for  0 plot the graph of  (  ) = 2  + 1 . 7. (a) (   )(  ) =  (  (  )) =  (2   3) = (2   3) 2 + 2(2   3)  1 = 4  2  12  + 9 + 4   6  1 = 4  2  8  + 2 (b) (   )(  ) =  (  (  )) =  (  2 + 2   1) = 2(  2 + 2   1)  3 = 2  2 + 4   2  3 = 2  2 + 4   5 (c) (    )(  ) =  (  (  (  ))) =  (  (2   3)) =  (2(2   3)  3) =  (4   9) = 2(4   9)  3 = 8   18  3 = 8   21 Test D Trigonometry 1. (a) 300  = 300    180   = 300  180 = 5  3 (b)  18  =  18    180   =  18  180 =   10 2. (a) 5  6 = 5  6  180    = 150  (b) 2 = 2  180    = 360   114  6  3. We will use the arc length formula,  =  , where  is arc length,  is the radius of the circle, and  is the measure of the central angle in radians. First, note that 30  = 30    180   =  6 . So  = (12)   6  = 2  cm. 4. (a) tan(  3) =  3 You can read the value from a right triangle with sides 1, 2, and  3 . (b) Note that 7  6 can be thought of as an angle in the third quadrant with reference angle  6 . Thus, sin(7  6) =  1 2 , since the sine function is negative in the third quadrant. (c) Note that 5  3 can be thought of as an angle in the fourth quadrant with reference angle  3 . Thus, sec(5  3) = 1 cos(5  3) = 1 1  2 = 2 , since the cosine function is positive in the fourth quadrant. 5. sin  =  24   = 24 sin  and cos  =  24  = 24 cos  5 . 5. sin sin   = =   24 24     = 24 sin = 24 sin   and and cos cos   = =   24 24   = 24 cos = 24 co  

  15. Page 15
    Read page 15 as text

    8 ¤ DIAGNOGSTIC TESTS 6. sin  = 1 3 and sin 2  + cos 2  = 1  cos  = 1  1 9 = 2  2 3 . Also, cos = 4 5  sin = 1  16 25 = 3 5 . So, using the sum identity for the sine, we have sin(  + ) = sin  cos + cos  sin = 1 3 · 4 5 + 2  2 3 · 3 5 = 4 + 6  2 15 = 1 15  4 + 6  2  7. (a) tan  sin  + cos  = sin  cos  sin  + cos  = sin 2  cos  + cos 2  cos  = 1 cos  = sec  (b) 2 tan  1 + tan 2  = 2 sin  (cos  ) sec 2  = 2 sin  cos  cos 2  = 2 sin  cos  = sin 2  8. sin 2  = sin   2 sin  cos  = sin   2 sin  cos   sin  = 0  sin  (2 cos   1) = 0  sin  = 0 or cos  = 1 2   = 0 ,  3 ,  , 5  3 , 2  . 9. We rst graph = sin 2  (by compressing the graph of sin  by a factor of 2) and then shift it upward 1 unit.

  16. Page 16
    Read page 16 as text

    1 FUNCTIONS AND MODELS 1.1 Four Ways To Represent a Function In exercises requiring estimations or approximations, your answers may vary slightly from the answers given here. 1. (a) The point (1  3) is on the graph of  , so  (1) = 3 . (b) When  =  1 , is about  0  2 , so  (  1)  0  2 . (c)  (  ) = 1 is equivalent to = 1  When = 1 , we have  = 0 and  = 3 . (d) A reasonable estimate for  when = 0 is  =  0  8 . (e) The domain of  consists of all  -values on the graph of  . For this function, the domain is  2    4 , or [  2  4] . The range of  consists of all -values on the graph of  . For this function, the range is  1   3 , or [  1  3] . (f ) As  increases from  2 to 1 , increases from  1 to 3 . Thus,  is increasing on the interval [  2  1] . 2. (a) The point (  4   2) is on the graph of  , so  (  4) =  2 . The point (3  4) is on the graph of  , so  (3) = 4 . (b) We are looking for the values of  for which the -values are equal. The -values for  and  are equal at the points (  2  1) and (2  2) , so the desired values of  are  2 and 2 . (c)  (  ) =  1 is equivalent to =  1 . When =  1 , we have  =  3 and  = 4 . (d) As  increases from 0 to 4 , decreases from 3 to  1 . Thus,  is decreasing on the interval [0  4] . (e) The domain of  consists of all  -values on the graph of  . For this function, the domain is  4    4 , or [  4  4] . The range of  consists of all -values on the graph of  . For this function, the range is  2   3 , or [  2  3] . (f ) The domain of  is [  4  3] and the range is [0  5  4] . 3. From Figure 1 in the text, the lowest point occurs at about (   ) = (12   85) . The highest point occurs at about (17  115) . Thus, the range of the vertical ground acceleration is  85    115 . Written in interval notation, we get [  85  115] . 4. Example 1: A car is driven at 60 mi  h for 2 hours. The distance  traveled by the car is a function of the time  . The domain of the function is {  | 0    2 } , where  is measured in hours. The range of the function is {  | 0    120 } , where  is measured in miles. Example 2: At a certain university, the number of students  on campus at any time on a particular day is a function of the time  after midnight. The domain of the function is {  | 0    24 } , where  is measured in hours. The range of the function is {  | 0     } , where  is an integer and  is the largest number of students on campus at once.

1361 more pages in the full document

Unlock it once and it stays in your library, ready to chat with or turn into flashcards.

Unlock the full document

More Mathematics study materials

See all

Study this document with CramX

Once it is in your library, every tool can work from it.