Skip to content
Exam Guides

Test Bank for Calculus For Biology and Medicine , 4th Edition

Test Bank for Calculus For Biology and Medicine , 4th Edition provides an extensive collection of questions to test your knowledge.

Preview 31 of 1025 pages

See the first 31 pages exactly as they appear in the document.

  1. Page 1
    Read page 1 as text

     I NSTRUCTOR ’ S S OLUTIONS M ANUAL R OGER L IPSETT TO ACCOMPANY C ALCULUS FOR B IOLOGY AND M EDICINE F OURTH E DITION Claudia Neuhauser University of Minnesota Marcus L. Roper University of California—Los Angeles

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

    Contents 1 Preview and Review 5 1.1 Precalculus Skills Diagnostic Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.2 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 1.3 Elementary Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 1.4 Graphing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 Chapter 1 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78 2 Discrete-Time Models, Sequences, and Difference Equations 87 2.1 Exponential Growth and Decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 2.2 Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 2.3 Modeling with Recursion Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 Chapter 2 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126 3 Limits and Continuity 133 3.1 Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 3.2 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 3.3 Limits at Infinity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156 3.4 Trigonometric Limits and the Sandwich Theorem . . . . . . . . . . . . . . . . . . . . . . . 159 3.5 Properties of Continuous Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164 3.6 A Formal Definition of Limits (Optional) . . . . . . . . . . . . . . . . . . . . . . . . . . . 169 Chapter 3 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173 4 Differentiation 185 4.1 Formal Definition of the Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185 4.2 Properties of the Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191 4.3 The Power Rule, the Basic Rules of Differentiation, and the Derivatives of Polynomials . . 201 4.4 The Product and Quotient Rules, and the Derivatives of Rational and Power Functions . 214 4.5 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224 4.6 Implicit Functions and Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . . . . 232 4.7 Higher Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239 4.8 Derivatives of Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242 4.9 Derivatives of Exponential Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247 4.10 Derivatives of Inverse Functions, Logarithmic Functions, and the Inverse Tangent Function 254 4.11 Linear Approximation and Error Propagation . . . . . . . . . . . . . . . . . . . . . . . . . 262 Chapter 4 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266

  4. Page 4
    Read page 4 as text

    2 CONTENTS 5 Applications of Differentiation 275 5.1 Extrema and the Mean-Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275 5.2 Monotonicity and Concavity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 288 5.3 Extrema and Inflection Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302 5.4 Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 308 5.5 L’Hˆ opital’s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 320 5.6 Graphing and Asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331 5.7 Recurrence Equations: Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351 5.8 Numerical Methods: The Newton-Raphson Method . . . . . . . . . . . . . . . . . . . . . . 362 5.9 Modeling Biological Systems Using Differential Equations . . . . . . . . . . . . . . . . . . 369 5.10 Antiderivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 375 Chapter 5 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 386 6 Integration 399 6.1 The Definite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 399 6.2 The Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 418 6.3 Applications of Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 430 Chapter 6 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 457 7 Integration Techniques and Computational Methods 465 7.1 The Substitution Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 465 7.2 Integration by Parts and Practicing Integration . . . . . . . . . . . . . . . . . . . . . . . . 471 7.3 Rational Functions and Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 486 7.4 Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 509 7.5 Numerical Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 522 7.6 The Taylor Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 539 7.7 Tables of Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 546 Chapter 7 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 552 8 Differential Equations 565 8.1 Solving Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 565 8.2 Equilibria and Their Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 587 8.3 Differential Equation Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 615 8.4 Integrating Factors and Two-Compartment Models . . . . . . . . . . . . . . . . . . . . . . 634 Chapter 8 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 648 9 Linear Algebra and Analytic Geometry 655 9.1 Linear Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 655 9.2 Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 667 9.3 Linear Maps, Eigenvectors, and Eigenvalues . . . . . . . . . . . . . . . . . . . . . . . . . . 683 9.4 Demographic Modeling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 707 9.5 Analytic Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 714 Chapter 9 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 724 10 Multivariable Calculus 733 10.1 Functions of Two or More Independent Variables . . . . . . . . . . . . . . . . . . . . . . . 733 10.2 Limits and Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 743 10.3 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 748 10.4 Tangent Planes, Differentiability, and Linearization . . . . . . . . . . . . . . . . . . . . . . 757 10.5 The Chain Rule and Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . 766

  5. Page 5
    Read page 5 as text

    CONTENTS 3 10.6 Directional Derivatives and Gradient Vectors . . . . . . . . . . . . . . . . . . . . . . . . . 770 10.7 Maximization and Minimization of Functions . . . . . . . . . . . . . . . . . . . . . . . . . 777 10.8 Diffusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 805 10.9 Systems of Difference Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 808 Chapter 10 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 824 11 Systems of Differential Equations 831 11.1 Linear Systems: Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 831 11.2 Linear Systems: Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 863 11.3 Nonlinear Autonomous Systems: Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . 873 11.4 Nonlinear Systems: Lotka-Volterra Model for Interspecific Interactions . . . . . . . . . . . 895 11.5 More Mathematical Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 908 Chapter 11 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 924 12 Probability and Statistics 935 12.1 Counting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 935 12.2 What is Probability? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 940 12.3 Conditional Probability and Independence . . . . . . . . . . . . . . . . . . . . . . . . . . . 948 12.4 Discrete Random Variables and Discrete Distributions . . . . . . . . . . . . . . . . . . . . 956 12.5 Continuous Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 979 12.6 Limit Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1000 12.7 Statistical Tools . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1011 Chapter 12 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1020

  6. Page 6
    Read page 6 as text

    Chapter 1 Preview and Review 1.1 Precalculus Skills Diagnostic Test 1. (a) y = x + 5 (b) y = 2 x − 1 (c) y = 3 x − 5 (Review 1.2.2) 2. (a) 5 9 (80 − 32) = 240 9 ≈ 26 . 7 ◦ C (b) Inverting the given formula gives x = 9 5 y + 32, so that 9 5 ( − 10) + 32 = − 18 + 32 = 14 ° F (c) Set y = x in the conversion formula and solve for x : x = 5 9 ( x − 32) = 5 9 x − 160 9 ⇒ 4 9 x = − 160 9 ⇒ 4 x = − 160 ⇒ x = − 40 . (Review 1.2.2) 3. A circle with center ( x, y ) = ( − 1 , 5) and with radius 3. (Review 1.2.3) 4. (a) 180 π · π 7 = 180 7 ≈ 25 . 7 ° . (b) x = − π 3 + 2 nπ and x = − 2 π 3 + 2 nπ for any n ∈ Z . (c) Divide both sides of the identity sin 2 θ + cos 2 θ = 1 by cos 2 θ : sin 2 θ + cos 2 θ cos 2 θ = 1 cos 2 θ so tan 2 θ + 1 = sec 2 θ. (d) π 12 , 7 π 12 , 3 π 4 (Review 1.2.4) 5. (a) (i) 2 7 / 3 (ii) 2 8 / 3 (b) Take base 2 logarithms of both sides: x log 2 4 = log 2 1 2 , so 2 x = − 1 and therefore x = − 1 2 .

  7. Page 7
    Read page 7 as text

    6 Preview and Review (c) Since 10 000 = 10 4 , we have log 10 10 000 = 4. (d) log 10 3 x + log 10 5 x = log 10 (3 x · 5 x ) = log 10 (15 x 2 ). (e) ln ( x 2 ) + ln x = 2 ln x + ln x = 3 ln x . Solving 3 ln x = 2 gives ln x = 2 3 , so that x = e 2 / 3 . (f ) If ln x = 3, then x = e 3 , so that log 10 x = log 10 ( e 3 ) = 3 log 10 e . (Review 1.2.5) 6. (a) Using the quadratic formula, the roots are − 1 ± √ 1 2 − 4 · 1 · 1 2 · 1 = − 1 2 ± i √ 3 2 . (b) (1 + i )(2 − i ) = 1 · 2 + 1 · ( − i ) + i · 2 − i 2 = 3 + i . (c) z + z = ( a + bi ) + ( a − bi ) = 2 a ∈ R (Review 1.2.6) 7. (a) (i) f ( x ) ∈ [0 , 1] (ii) f ( x ) ∈ [0 , ∞ ) (b) (i) f (3) = √ 3 (ii) ( f ◦ g )( x ) = f ( ( x + 1) 2 ) = √ ( x + 1) 2 = | x + 1 | . (iii) ( g ◦ f )(4) = g ( √ 4) = g (2) = (2 + 1) 2 = 9. (c) Since f ( x ) = | x | ≥ 0, then ( f ◦ f )( x ) = | f ( x ) | = f ( x ) (Review 1.3.1) 8. (a) For large values of x , (iii) grows much faster due to the presence of the x 3 term. For example, when x = 6, p 1 ( x ) = 6, p 2 ( x ) = 18, and p 3 ( x ) = 72. (b) (i) The highest power of r that appears is 2, so this is a degree 2 polynomial in r . (ii) Since the velocity should not be negative we must have 1 − r 2 a 2 ≥ 0, so that r ≤ a . Since r is a distance, it must be nonnegative. Putting these together gives 0 ≤ r ≤ a for the domain. (iii) Since 1 − r 2 a 2 ≤ 1 for the allowable values of r , the range is 0 ≤ u ( r ) ≤ u 0 . (iv) The maximum value occurs when r = 0; that value is u 0 . u ( r ) = 1 2 u 0 when 1 − r 2 a 2 = 1 2 , which happens when r = a √ 2 . (Review 1.3.2) 9. (a) For the three values of c , we have r (1) = 1 1 + 10 ≈ 0 . 091 , r (2) = 2 2 + 10 ≈ 0 . 167 , r (3) = 3 3 + 10 ≈ 0 . 231 . The rate of metabolization is largest when c = 3. (b) Since r (5) = 5 5 + 10 = 1 3 , r (10) = 10 10 + 10 = 1 2 , the rate of metabolization increases by less than double.

  8. Page 8
    Read page 8 as text

    1.1 Precalculus Skills Diagnostic Test 7 (c) From part (b), r (5) = 1 3 ; to get r ( c ) = 2 3 we need r ( c ) = c c + 10 = 2 3 , so that 3 c = 2( c + 10). Solving gives c = 20. (d) No. r (10) = 1 2 . Doubling the rate of metabolization would increase it to 1. But since the denominator of r ( c ) is always strictly greater than the numerator, its value can never be 1. (Review 1.3.3) 10. The relationship is N = kA 1 / 5 . (a) We want to find A so that kA 1 / 5 = 1 2 kA 1 / 5 0 . Cancelling the k ’s gives A 1 / 5 = 1 2 A 1 / 5 0 . Now raise both sides to the fifth power, giving A = A 0 32 . The new habitat area is A 0 32 . (b) We want to find A so that kA 1 / 5 = 1 3 kA 1 / 5 0 . Cancelling the k ’s gives A 1 / 5 = 1 3 A 1 / 5 0 . Now raise both sides to the fifth power, giving A = A 0 243 . The new habitat area is A 0 243 . (c) We want to find A so that kA 1 / 5 = 2 kA 1 / 5 0 . Cancelling the k ’s gives A 1 / 5 = 2 A 1 / 5 0 . Now raise both sides to the fifth power, giving A = 32 A 0 . The new habitat area is 32 A 0 . (Review 1.3.4) 11. (a) Since 1000 = N (0) = N 0 e r · 0 = N 0 we get N 0 = 1000. Since 2000 = N (2) = 1000 e r · 2 , we have 2 = e 2 r . Take natural logarithms, giving 2 r = ln 2 so that r = ln 2 2 . The formula is therefore N ( t ) = 1000 e t ln 2 / 2 . (b) Solving N ( t ) = 3000 gives e t ln 2 / 2 = 3, so that t ln 2 2 = ln 3, and then t = 2 ln 3 ln 2 ≈ 3 . 17. (c) Solving N ( t ) = 4000 gives e t ln 2 / 2 = 4, so that t ln 2 2 = ln 4, and then t = 2 ln 4 ln 2 = 4. (Review 1.3.5) 12. (a) f − 1 ( x ) = √ x − 1 (b) f − 1 ( x ) = − 1 + e x/ 2 (c) f − 1 ( x ) = x 1 / 5 (Review 1.3.6) 13. (a) (i) ln x + ln( x 2 + 1) = ln( x · ( x 2 + 1)). (ii) log ( x 1 / 3 ) − log ( ( x + 1) 1 / 3 ) = 1 3 log x − 1 3 log( x + 1) = 1 3 (log x − log( x + 1)) = 1 3 log x x +1 . (iii) 2 + log 2 x = log 2 4 + log 2 x = log 2 (4 x ). (b) (i) ln 7 ln 2 (ii) ln 6 ln 10 (iii) ln 2 ln x (Review 1.3.7) 14. (a) (i) The amplitude is 2, and the period is 2 π . (ii) The amplitude is 2, and the period is 2 π 3 . (iii) The amplitude is 3, and the period is 2 π π/ 2 = 4. (b) (i) The range is all of R . The maximum domain is all of R except for the points (2 n +1) 2 π where n is an integer (since tangent is undefined at odd multiples of π / 2 ).

  9. Page 9
    Read page 9 as text

    8 Preview and Review (ii) The range is [ − 1 , 1], since − 1 ≤ cos x ≤ 1 for all x . The domain is R . (c) Compress cos x by a factor of 2 in the x -direction and stretch it by a factor of 3 in the y -direction to get 3 cos 2 x . (Review 1.3.8) 15. - 6 - 4 - 2 2 x - 0.5 0.5 y ( a ) - 4 - 2 2 4 x 0.5 1. 1.5 y ( b ) - 4 - 2 2 4 x - 0.5 0.5 y ( c ) - 4 - 2 2 4 x - 0.5 0.5 y ( d ) - 8 - 4 4 8 x - 0.5 0.5 y ( e ) (Review 1.4.1) 16. (a) (i) 2 kg (ii) 30 kg (iii) 100 kg (b) Adult weight Puppy weight = 20 kg 0 . 4 kg = 50 (c) The point representing the cat will lie one tick below the Jack Russell Terrier. (Review 1.4.2) 17. (a) Since both axes are log scales, this is a power relationship; since the slope of the line is 0 . 41, the relationship is D = k A 0 . 41 . (b) Since only the vertical axis is a log scale, this is an exponential relationship; since the slope of the line is − 1, the relationship is N = k 10 − m .

  10. Page 10
    Read page 10 as text

    1.2 Preliminaries 9 (c) Since only the vertical axis is a log scale, this is an exponential relationship; since the slope of the line is 0 . 86, the relationship is N = k 10 0 . 86 t ≈ k 7 . 24 t . (d) Since only the vertical axis is a log scale, this is an exponential relationship; since the slope of the line is − 0 . 09, the relationship is N = k 10 − 0 . 09 t ≈ k 0 . 813 t . (Review 1.4.3) 18. (a) This is the graph of patient 1 shifted two hours to the right, so it is graph 4. (b) This graph should start from zero at 8AM just as for patient 1, but it should have a longer tail since it takes longer to enter the bloodstream. This is graph 3. (c) This graph should have the same shape as the graph of patient A, which is graph 4, but is half as high. This is graph 2. (Review 1.4.4) 1.2 Preliminaries  1.2.1 1. (a) Walking 4 units to the right and to the left from − 1 we get the numbers 3 and − 5, respectively. (b) | x − ( − 1) | = 4, so x + 1 = ± 4, yielding x = 3 or x = − 5. 2. With three numbers there will be three pairwise distances: |− 5 − 2 | = |− 7 | = 7 , | 2 − 7 | = |− 5 | = 5 and |− 5 − 7 | = |− 12 | = 12. 3. (a) 2 x + 4 = ± 6, so either 2 x = 2 giving the solution x = 1 or 2 x = − 10 and the other solution is x = − 5. (b) x − 3 = ± 2, from which we see that x = 5 or x = 1. (c) 2 x − 3 = ± 5, so 2 x = 8 giving x = 4 or 2 x = − 2 and x = − 1. (d) 1 − 5 x = ± 6, so that 5 x = 1 ± 6. Then 5 x = − 5, so x = − 1, or 5 x = 7, so x = 7 5 . 4. (a) The equation implies ± (2 x + 4) = ± (5 x − 2), which reduces to two equations 2 x + 4 = 5 x − 2 and 2 x + 4 = − (5 x − 2), so the solutions are: x = 2 and x = − 2 7 . (b) As in part (a) we get 1 + 2 u = ± (5 − u ), and solving both gives 1 + 2 u = 5 − u with solution u = 4 3 , and 1 + 2 u = − (5 − u ) = u − 5 with solution u = − 6. (c) As in part (a) we get the equations 4 + t 2 = ± ( 3 2 t − 2) that solves to t = 6 and t = − 1. (d) As in part (a) 2 s − 6 = ± (3 − s ); solving both equations gives s = 3 and s = 3. So s = 3 is the only solution. 5. (a) We can rewrite the absolute value as two inequalities − 4 ≤ 5 x − 2 ≤ 4 ⇒ − 4 + 2 ≤ 5 x ≤ 4 + 2 ⇒ − 2 ≤ 5 x ≤ 6 ⇒ − 2 5 ≤ x ≤ 6 5 . (b) Rewriting as a pair of inequalities gives 3 − 4 x < − 8 or 3 − 4 x > 8. Solving the first we get: 4 x > 11, or x > 11 4 , and solving the second we get 4 x < − 5, or x < − 5 4 . So x < − 5 4 or x > 11 4 . (c) Rewriting as a pair of inequalities gives 7 x + 4 ≥ 3, which solves as x ≥ − 1 7 , and 7 x + 4 ≤ − 3 which gives 7 x ≤ − 7 or x ≤ − 1. So x ≤ − 1 or x ≥ − 1 7 .

  11. Page 11
    Read page 11 as text

    10 Preview and Review (d) Rewriting gives − 7 < 3 + 2 x < 7 ⇒ − 7 − 3 < 2 x < 7 − 3 ⇒ − 10 < 2 x < 4 ⇒ − 5 < x < 2 . 6. (a) − 6 < 2 x + 3 < 6; subtracting 3 gives − 9 < 2 x < 3, so that − 9 2 < x < 3 2 . (b) Rewriting as a pair of inequalities gives 3 − 4 x ≥ 2 or 3 − 4 x ≤ − 2, so that x ≤ 1 4 or x ≥ 5 4 . (c) − 1 ≤ x + 5 ≤ 1, which is the same as − 6 ≤ x ≤ − 4. (d) Since no absolute value can be negative, there are no values of x that will satisfy the equation.  1.2.2 7. Use the point slope formula y − y 0 = m ( x − x 0 ). In this case we get y − 2 = − 2( x − 3) ⇒ y − 2 = − 2 x + 6 ⇒ 2 x + y − 8 = 0 . 8. Use the point slope formula y − y 0 = m ( x − x 0 ). In this case we get y − ( − 1) = 1 4 ( x − 2) ⇒ y + 1 = 1 4 x − 1 2 ⇒ − x + 4 y + 6 = 0 . 9. Use the point slope formula y − y 0 = m ( x − x 0 ). In this case we get y − ( − 2) = − 3( x − 0) ⇒ y + 2 = − 3 x ⇒ 3 x + y + 2 = 0 . 10. Use the point slope formula y − y 0 = m ( x − x 0 ). In this case we get y − 5 = 1 2 ( x − ( − 3)) ⇒ 2 y − 10 = x + 3 ⇒ − x + 2 y − 13 = 0 . 11. First compute the slope using the two given points: m = y 2 − y 1 x 2 − x 1 = 4 − ( − 3) 1 − ( − 2) = 7 3 . Now use the point slope formula y − y 0 = m ( x − x 0 ) with ( x 0 , y 0 ) = (1 , 4) (we could equally well use the other point, ( − 2 , − 3)): y − 4 = 7 3 ( x − 1) ⇒ 3( y − 4) = 7( x − 1) ⇒ 3 y − 12 = 7 x − 7 ⇒ − 7 x + 3 y − 5 = 0 . 12. First compute the slope using the two given points: m = y 2 − y 1 x 2 − x 1 = − 4 − 4 1 − ( − 1) = − 4 . Now use the point slope formula y − y 0 = m ( x − x 0 ) with ( x 0 , y 0 ) = ( − 1 , 4) (we could equally well use the other point, (1 , − 4)): y − 4 = − 4( x − ( − 1)) ⇒ y − 4 = − 4 x − 4 ⇒ 4 x + y = 0 .

  12. Page 12
    Read page 12 as text

    1.2 Preliminaries 11 13. First compute the slope using the two given points: m = y 2 − y 1 x 2 − x 1 = 1 − 3 2 − 0 = − 1 . Now use the point slope formula y − y 0 = m ( x − x 0 ) with ( x 0 , y 0 ) = (0 , 3) (we could equally well use the other point, (2 , 1)): y − 3 = − 1( x − 0) ⇒ y − 3 = − x ⇒ x + y − 3 = 0 . 14. First compute the slope using the two given points: m = y 2 − y 1 x 2 − x 1 = 5 − ( − 1) 4 − 1 = 6 3 = 2 . Now use the point slope formula y − y 0 = m ( x − x 0 ) with ( x 0 , y 0 ) = (1 , − 1) (we could equally well use the other point, (4 , 5)): y − ( − 1) = 2( x − 1) ⇒ y + 1 = 2 x − 2 ⇒ 2 x − y − 3 = 0 . 15. Horizontal lines are always of the form y = k ; since this line goes through ( 4 , 1 4 ) , its equation is y = 1 4 . The standard form is 4 y − 1 = 0. 16. Horizontal lines are always of the form y = k ; since this line goes through (0 , − 1), its equation is y = − 1. The standard form is y + 1 = 0. 17. Vertical lines are always of the form x = k ; since this line goes through ( − 2 , 0), its equation is x = − 2. The standard form is x + 2 = 0. 18. Vertical lines are always of the form x = k ; since this line goes through (2 , − 3), its equation is x = 2. The standard form is x − 2 = 0. 19. Use the slope-intercept form y = mx + b . We are given m = 3 and b = 2, so the equation is y = 3 x + 2, which in standard form is 3 x − y + 2 = 0. 20. Use the slope-intercept form y = mx + b . We are given m = − 1 and b = 5, so the equation is y = − x + 5, which in standard form is x + y − 5 = 0. 21. Use the slope-intercept form y = mx + b . We are given m = 1 2 and b = 2, so the equation is y = 1 2 x + 2, or 2 y = x + 4, which in standard form is x − 2 y + 4 = 0. 22. Use the slope-intercept form y = mx + b . We are given m = − 1 3 and b = 1 3 , so the equation is y = − 1 3 x + 1 3 , which in standard form is x + 3 y − 1 = 0. 23. Use the point slope formula y − y 0 = m ( x − x 0 ) where ( x 0 , y 0 ) = (1 , 0) is the given point. In this case we get y − 0 = − 2( x − 1) ⇒ y = − 2 x + 2 ⇒ 2 x + y − 2 = 0 . 24. Use the point slope formula y − y 0 = m ( x − x 0 ) where ( x 0 , y 0 ) = ( − 2 , 0) is the given point. In this case we get y − 0 = 1( x − ( − 2)) ⇒ y = x + 2 ⇒ x − y + 2 = 0 . 25. Use the point slope formula y − y 0 = m ( x − x 0 ) where ( x 0 , y 0 ) = ( − 1 2 , 0 ) is the given point. In this case we get y − 0 = − 1 2 ( x − ( − 1 2 )) ⇒ y = − 1 2 x − 1 4 ⇒ 2 x + 4 y + 1 = 0 .

  13. Page 13
    Read page 13 as text

    12 Preview and Review 26. Use the point slope formula y − y 0 = m ( x − x 0 ) where ( x 0 , y 0 ) = ( − 1 2 , 0 ) is the given point. In this case we get y − 0 = 1 5 ( x − ( − 1 2 )) ⇒ y = 1 5 x + 1 10 ⇒ 2 x − 10 y + 1 = 0 . 27. Since the line we want is parallel to x + 2 y − 4 = 0, it must have the same slope. Now, x + 2 y − 4 = 0 implies that 2 y = − x + 4, or y = − 1 2 x + 2, so the line has slope m = − 1 2 . Now use the point slope formula y − y 0 = m ( x − x 0 ) where ( x 0 , y 0 ) = (2 , − 3) is the given point. In this case we get y − ( − 3) = − 1 2 ( x − 2) ⇒ y + 3 = − 1 2 x + 1 ⇒ 2 y + 6 = − x + 2 ⇒ x + 2 y + 4 = 0 . 28. Since the line we want is parallel to x − 2 y + 4 = 0, it must have the same slope. Now, x − 2 y + 4 = 0 implies that 2 y = x + 4, or y = 1 2 x + 2, so the line has slope m = 1 2 . Now use the point slope formula y − y 0 = m ( x − x 0 ) where ( x 0 , y 0 ) = (1 , 2) is the given point. In this case we get y − 2 = 1 2 ( x − 1) ⇒ y − 2 = 1 2 x − 1 2 ⇒ x − 2 y + 3 = 0 . 29. The line passing through (0 , 2) and (3 , 0) has slope m = 0 − 2 3 − 0 = − 2 3 , so the line we want has the same slope. Now use the point slope formula y − y 0 = m ( x − x 0 ) where ( x 0 , y 0 ) = ( − 1 , − 1) is the given point. In this case we get y − ( − 1) = − 2 3 ( x − ( − 1)) ⇒ y + 1 = − 2 3 x − 2 3 ⇒ 2 x + 3 y + 5 = 0 . 30. The line passing through (0 , − 4) and (2 , 1) has slope m = 1 − ( − 4) 2 − 0 = 5 2 , so the line we want has the same slope. Now use the point slope formula y − y 0 = m ( x − x 0 ) where ( x 0 , y 0 ) = (2 , − 1) is the given point. In this case we get y − ( − 1) = 5 2 ( x − 2) ⇒ y + 1 = 5 2 x − 5 ⇒ 5 x − 2 y − 12 = 0 . 31. The line 2 y − 5 x + 7 = 0 can be written as 2 y = 5 x − 7, or y = 5 2 x − 7 2 , so it has slope 5 2 . Since the line we want is perpendicular to it, it must have slope m ⊥ = − 1 m = − 2 5 . It passes through (1 , 4), so using the point slope formula y − y 0 = m ( x − x 0 ) we get y − 4 = − 2 5 ( x − 1) ⇒ 5 y − 20 = − 2 x + 2 ⇒ 2 x + 5 y − 22 = 0 . 32. The line x − 2 y + 3 = 0 can be written as 2 y = x + 3, or y = 1 2 x + 3 2 , so it has slope m 0 = 1 2 . Since the line we want is perpendicular to it, it must have slope m = − 2 (since for perpendicular lines we must have m · m 0 = − 1). It passes through (1 , − 1), so using the point slope formula y − y 0 = m ( x − x 0 ) we get y − ( − 1) = − 2( x − 1) ⇒ y + 1 = − 2 x + 2 ⇒ 2 x + y − 1 = 0 .

  14. Page 14
    Read page 14 as text

    1.2 Preliminaries 13 33. The given line, passing through ( − 2 , 1) and (1 , − 2), has slope m = y 2 − y 1 x 2 − x 1 = ( − 2) − 1 1 − ( − 2) = − 3 3 = − 1 . Therefore the slope of the perpendicular line is m ⊥ = − 1 m = 1. It passes through (5 , − 1), so using the point slope formula y − y 0 = m ( x − x 0 ) we get y − ( − 1) = 1( x − 5) ⇒ y + 1 = x − 5 ⇒ x − y − 6 = 0 . 34. The given line, passing through ( − 2 , 0) and (1 , 1), has slope m = y 2 − y 1 x 2 − x 1 = 1 − 0 1 − ( − 2) = 1 3 . Therefore the slope of the perpendicular line is m ⊥ = − 1 m = − 3. It passes through (4 , − 1), so using the point slope formula y − y 0 = m ( x − x 0 ) we get y − ( − 1) = − 3( x − 4) ⇒ y + 1 = − 3 x + 12 ⇒ 3 x + y − 11 = 0 . 35. The line we want is horizontal and passes through (1 , 3). Since all horizontal lines are of the form y = k , here we must have y = 3, or y − 3 = 0. 36. The line we want is horizontal and passes through (1 , 5). Since all horizontal lines are of the form y = k , here we must have y = 5, or y − 5 = 0. 37. The line we want is vertical and passes through ( − 2 , 3). Since all vertical lines are of the form x = k , here we must have x = − 2, or x + 2 = 0. 38. The line we want is vertical and passes through (3 , 1). Since all vertical lines are of the form x = k , here we must have x = 3, or x − 3 = 0. 39. A line perpendicular to a horizontal line is a vertical line. Since all vertical lines are of the form x = k , and the line passes through (1 , − 3), we must have x = 1, or x − 1 = 0. 40. A line perpendicular to a horizontal line is a vertical line. Since all vertical lines are of the form x = k , and the line passes through (1 , 3), we must have x = 1, or x − 1 = 0. 41. A line perpendicular to a vertical line is a horizontal line. Since all horizontal lines are of the form y = k , and the line passes through (7 , 3), we must have y = 3, or y − 3 = 0. 42. A line perpendicular to a vertical line is a horizontal line. Since all horizontal lines are of the form y = k , and the line passes through ( − 2 , 5), we must have y = 5, or y − 5 = 0. 43. (a) We convert each of these into centimeters, remembering that the formula y = 30 . 5 x requires that x be measured in feet: (i) y = 30 . 5 · 6 ft = 183 cm (ii) y = 30 . 5 · 4 12 ft ≈ 30 . 5 · 0 . 1667 ft ≈ 10 . 2 cm (iii) 1 ft , 7 in = 1 + 7 12 ft ≈ 1 . 583 ft, so y ≈ 30 . 5 · 1 . 583 ft ≈ 48 . 3 cm. (iv) y = 30 . 5 · 20 . 5 12 ft ≈ 30 . 5 · 1 . 708 ft ≈ 52 . 1 cm (b) Divide both sides of y = 30 . 5 x by 30 . 5, giving y 30 . 5 = x , or x = y 30 . 5 . Since x is measured in feet and y in centimeters, this gives a way to convert from centimeters to feet.

  15. Page 15
    Read page 15 as text

    14 Preview and Review (c) (i) x = 195 30 . 5 ≈ 6 . 39 ft (ii) x = 12 30 . 5 ≈ 0 . 39 ft (iii) x = 48 30 . 5 ≈ 1 . 57 ft 44. (a) Since 1 pound equals 2 . 20 kg, let y = 2 . 20 x , where x is kilograms and y is pounds. (b) (i) y = 2 . 20 x , so 63 = 2 . 20 x , or x ≈ 28 . 6 kg. (ii) y = 2 . 20 x , so 5 = 2 . 20 x , or x ≈ 2 . 27 kg. (iii) y = 2 . 20 · 2 . 5 = 5 . 5 pounds. (iv) y = 2 . 20 · 76 ≈ 167 pounds. 45. Distance = rate · time; here time = 15 min = 1 4 h = 0 . 25 h, and the distance is 10 mi. The constant of proportionality is miles per hour or “mph”. Using the given values gives 10 mi = speed · 0 . 25 h ⇒ speed = 10 0 . 25 = 40 mph . The speed (constant of proportionality) is 40 mph. 46. Let y be the number of seeds produced, and x be the biomass. Since y is proportional to x , we have y = mx where m is the constant of proportionality. Then 13 = m · 213, so that m = 13 213 ≈ 0 . 061. 47. 1 ft = 0 . 305 m, so 3 . 279 ft = 1 m. Then 1 m 2 = (1 m)(1 m) = (3 . 279 ft)(3 . 279 ft) ≈ 10 . 75 ft 2 . 48. 1 ha = 10 000 m 2 , and 1 acre = 4046 . 86 m 2 . Therefore the area ratio of a hectare to an acre (that is, the number of acres in a hectare) is 1 ha 1 acre = 10 000 m 2 4046 . 86 m 2 ≈ 2 . 471 acre . 49. (a) Since the relationship is linear, if we let y be the number of liters and x the number of ounces, we have x = my where m = 33 . 81. So the relationship is x = 33 . 81 y , or y ≈ 0 . 0296 x . (b) With x = 12 ounces, we get y = 0 . 0296 · 12 ≈ 0 . 355 l. 50. (a) Since the relationship is linear, if we let y be the number of kilometers and x the number of miles, we have y km = mx mi where m = 1 . 609. So the relationship is y = 1 . 609 x . (b) We have 434 = 1 . 609 x , so that x = 434 1 . 609 ≈ 270 mi. 51. (a) Since one cup of flour weighs 120 g, it follows that 2 . 5 cups of flour weighs 2 . 5 · 120 = 300 g. (b) Since one cup of flour weighs 120 g, one gram of flour is 1 120 cups. Thus 225 g of flour is 225 · 1 120 = 1 . 875 cups. (c) Let y be the number of cups and x the weight in grams. Then from part (b), y = 1 120 x . 52. (a) Let x be the temperature in ◦ C and let y be the temperature in ° F. Then we have the points (0 , 32) and (100 , 212). First find the slope between these two pts: m = 212 − 32 100 − 0 = 180 100 = 18 10 = 9 5 The y -intercept is 32. A linear equation that relates Celsius and Fahrenheit then is y = 9 5 x +32, or F = 9 5 C + 32

  16. Page 16
    Read page 16 as text

    1.2 Preliminaries 15 (b) Solving the equation from part (a) for C gives C = 5 9 ( F − 32) . Then the lower and upper normal body temperatures in humans are 97 . 6 ° F corresponds to 5 9 (97 . 6 − 32) = 5 9 · 65 . 6 ≈ 36 . 44 ◦ C 99 . 6 ° F corresponds to 5 9 (99 . 6 − 32) = 5 9 · 67 . 6 ≈ 37 . 56 ◦ C (c) Setting C = F in the equation from part (a) gives F = 9 5 F + 32 ⇒ − 4 5 F = 32 ⇒ F = − 40 . The two scales read the same at a temperature of − 40 ° . 53. (a) Since the relationship is linear, these two points, (0 . 16 , 0 . 52) and (1 . 0 , 1 . 0), must lie on a line. The slope of that line must be m = y 2 − y 1 x 2 − x 1 = 1 . 0 − 0 . 52 1 . 0 − 0 . 16 = 0 . 48 0 . 84 ≈ 0 . 5714 . The equation of the line is then (using the point slope form with the point (1 . 0 , 1 . 0)) y − 1 . 0 = 0 . 5714( x − 1 . 0) ⇒ y = 0 . 5714 x + 0 . 4286 . (b) Substitute each value of x into the equation from part (a): (i) y = 0 . 5714 · 0 . 5 + 0 . 4286 = 0 . 7143 (ii) y = 0 . 5714 · 0 . 9 + 0 . 4286 = 0 . 9429 (iii) y = 0 . 5714 · 0 + 0 . 4286 = 0 . 4286 (c) Solving the equation in part (a) for x , we get y = 0 . 5714 x + 0 . 4286 ⇒ y − 0 . 4286 = 0 . 5714 x ⇒ x = 1 0 . 5714 y − 0 . 4286 0 . 5714 ⇒ x = 1 . 75 y − 0 . 75 . 54. (a) A plot of two perpendicular lines, with their inclinations labeled, is below. ` 2 has also been translated so we can see what the relative angles of inclination are: θ 2 θ 1 x y l 2 l 1 From the graph, comparing ` 1 with the dashed line, we see that θ 1 = θ 2 + π 2 .

  17. Page 17
    Read page 17 as text

    16 Preview and Review (b) We see that the slopes of the two lines are m 2 = 4 y 4 x = tan θ 2 m 1 = 4 y 4 x = − tan( π − θ 1 ) = tan θ 1 (c) Now using the fact from (a) that − θ 2 = π 2 − θ 1 , we have m 1 = tan θ 1 = 1 cot θ 1 = 1 tan( π 2 − θ 1 ) = 1 tan( − θ 2 ) = cot( − θ 2 ) = − cot θ 2 . (d) m 1 m 2 = − cot θ 2 · tan θ 2 = − 1.  1.2.3 55. Using the general equation r 2 = ( x − x 0 ) 2 + ( y − y 0 ) 2 with r = 2 and ( x 0 , y 0 ) = (1 , − 2) gives 4 = ( x − 1) 2 + ( y − ( − 2)) 2 ⇒ 4 = ( x − 1) 2 + ( y + 2) 2 . 56. Using the general equation r 2 = ( x − x 0 ) 2 + ( y − y 0 ) 2 with r = 4 and ( x 0 , y 0 ) = (2 , 3) gives 4 2 = ( x − 2) 2 + ( y − 3) 2 ⇒ 16 = ( x − 2) 2 + ( y − 3) 2 . 57. (a) Using the general equation r 2 = ( x − x 0 ) 2 + ( y − y 0 ) 2 with r = 4 and ( x 0 , y 0 ) = (2 , 5) gives 4 2 = ( x − 2) 2 + ( y − 5) 2 ⇒ 16 = ( x − 2) 2 + ( y − 5) 2 . (b) When x = 0 the circle is on the y -axis. Setting x = 0 gives 16 = (0 − 2) 2 +( y − 5) 2 ⇒ 16 = 4+( y − 5) 2 ⇒ 12 = ( y − 5) 2 ⇒ ± 2 √ 3 = y − 5 ⇒ y = 5 ± 2 √ 3 . The points of intersection are (0 , 5 ± 2 √ 3). (c) When y = 0 the circle is on the x -axis. Setting y = 0 gives 16 = ( x − 2) 2 + (0 − 5) 2 ⇒ 16 = ( x − 2) 2 + 25 ⇒ − 9 = ( x − 2) 2 . But the right hand side is the square of a real number, which cannot be negative, so this is impossible and the circle does not intersect the x -axis. 58. (a) If we start with a point at (2 , − 5), and think of increasing the radius of a circle centered at that point, it will touch one axis when the radius becomes 2, and will touch the second axis when the radius becomes 5, so 2 ≤ r < 5. (b) As in part (a), it will intersect both axes when r ≥ 5. 59. ( x + 2) 2 + y 2 = 25, so ( x − ( − 2)) 2 + ( y − 0) 2 = 5 2 . This is a circle with center ( − 2 , 0) and radius 5. 60. ( x + 1) 2 + ( y − 3) 2 = 9, so ( x − ( − 1)) 2 + ( y − 3) 2 = 3 2 . This is a circle with center ( − 1 , 3) and radius 3.

  18. Page 18
    Read page 18 as text

    1.2 Preliminaries 17 61. Completing the square, we get 0 = x 2 + y 2 + 6 x + 2 y − 12 0 = ( x 2 + 6 x + 9) + ( y 2 + 2 y + 1) − 12 − 9 − 1 22 = ( x + 3) 2 + ( y + 1) 2 = ( x − ( − 3)) 2 + ( y − ( − 1)) 2 , which is a circle with center ( − 3 , − 1) and radius √ 22. 62. Completing the square, we get 0 = x 2 + y 2 + 2 x − 4 y + 1 0 = ( x 2 + 2 x + 1) + ( y 2 − 4 y + 4) + 1 − 1 − 4 4 = ( x + 1) 2 + ( y − 2) 2 , which is a circle with center ( − 1 , 2) and radius 2.  1.2.4 63. (a) Convert to radian measure: 65 ° · π 180 ° = 13 36 π rad. (b) Convert to degree measure: 11 12 π · 180 ° π = 11 · 180 ° 12 = 165 ° . 64. (a) − 15 ◦ · π 180 ◦ = − π 12 radians or 23 12 π radians. (b) 7 π 4 · 180 ◦ π = 1260 ◦ 4 = 315 ◦ . 65. (a) sin ( − π 4 ) = − √ 2 2 . (b) cos ( 5 π 6 ) = − √ 3 2 . (c) tan ( π 6 ) = 1 √ 3 . 66. (a) sin ( 5 4 π ) = √ 2 2 . (b) cos ( − 11 π 6 ) = √ 3 2 . (c) tan ( π 3 ) = √ 3. 67. (a) From the definition of cosine as adjacent divided by hypotenuse, it is clear that reflecting α across the y -axis negates the cosine; that is, cos( π − α ) = − cos α . Since cos π 3 = 1 2 √ 3, we get cos 2 π 3 = cos ( π − π 3 ) = − cos π 3 = − 1 2 √ 3 . Then, since cos( − α ) = cos α , we also get cos 4 π 3 = cos ( 2 π − 2 π 3 ) = cos ( − 2 π 3 ) = cos 2 π 3 = − 1 2 √ 3 . So the two solutions are α = 2 π 3 , 4 π 3 . (b) From the table in the text, we see that tan α = sin α cos α is equal to 1 √ 3 when α = π 6 . The other solution is α = π + π 6 = 7 π 6 , since for that angle both sine and cosine have the same numerical values as for π 6 , but they are both negative.

  19. Page 19
    Read page 19 as text

    18 Preview and Review 68. (a) From the definition of cosine as adjacent divided by hypotenuse, it is clear that reflecting α across the y -axis negates the cosine; that is, cos( π − α ) = − cos α . Since cos π 4 = 1 2 √ 2, we get cos 3 π 4 = cos ( π − π 4 ) = − cos π 4 = − 1 2 √ 2 . Then, since cos( − α ) = cos α , we also get cos 5 π 4 = cos ( 2 π − 3 π 4 ) = cos ( − 3 π 4 ) = cos 3 π 4 = − 1 2 √ 2 . So the two solutions are α = 3 π 4 , 5 π 4 . (b) Since sec α = 1 cos α , we are looking for solutions to cos α = 1 √ 2 = 1 2 √ 2. From the table in the text, one solution is α = π 4 ; since cos( − α ) = cos α , α = − π 4 is another solution. Adding 2 π to get it into the required range gives α = 7 π 4 . The two solutions are α = π 4 , 7 π 4 . 69. Using the facts that tan θ = sin θ cos θ and that sec θ = 1 cos θ , we get sin 2 θ + cos 2 θ = 1 sin 2 θ + cos 2 θ cos 2 θ = 1 cos 2 θ (Divide both sides by cos 2 θ ) sin 2 θ cos 2 θ + cos 2 θ cos 2 θ = 1 cos 2 θ (Distribute) tan 2 θ + 1 = sec 2 θ (Make substitution) 70. Using the facts that cot θ = cos θ sin θ and that csc θ = 1 sin θ , we get sin 2 θ + cos 2 θ = 1 sin 2 θ + cos 2 θ sin 2 θ = 1 sin 2 θ (Divide both sides by sin 2 θ ) sin 2 θ sin 2 θ + cos 2 θ sin 2 θ = 1 sin 2 θ (Distribute) 1 + cot 2 θ = csc 2 θ (Make substitution) 71. Using the identity sin 2 θ +cos 2 θ = 1, the equation cos 2 θ − 2 = 2 sin θ is the same as (1 − sin 2 θ ) − 2 = 2 sin θ , or sin 2 θ + 2 sin θ + 1 = (sin θ + 1) 2 = 0. Thus sin θ = − 1, so that θ = 3 π 2 . 72. Use the identity 1 + tan 2 x = sec 2 x to rewrite the equation: sec 2 x = √ 3 tan x + 1 ⇒ 1 + tan 2 x = √ 3 tan x + 1 ⇒ tan x (tan x − √ 3) = 0 . The solutions to this equation are the solutions to tan x = 0 or tan x = √ 3 for 0 ≤ x < π , which are x = 0 (tan x = 0) , x = π 3 (tan x = √ 3) .

  20. Page 20
    Read page 20 as text

    1.2 Preliminaries 19  1.2.5 73. (a) 2 4 8 − 2 / 3 = 2 4 ( 2 3 ) − 2 / 3 = 2 4 2 3( − 2 / 3) = 2 4 2 − 2 = 2 2 = 4. (b) 3 3 3 − 1 / 2 3 1 / 2 = 3 3 3 − 1 / 2 3 − 1 / 2 = 3 3 − 1 / 2 − 1 / 2 = 3 2 = 9. (c) 5 k (25) k − 1 5 2 − k = 5 k ( 5 2 ) k − 1 5 − (2 − k ) = 5 k 5 2( k − 1) 5 k − 2 = 5 k +(2 k − 2)+( k − 2) = 5 4 k − 4 . 74. (a) (2 4 · 2 − 3 / 2 ) 2 = (2 8 / 2 · 2 − 3 / 2 ) 2 = (2 8 / 2 − 3 / 2 ) 2 = (2 5 / 2 ) 2 = 2 5 = 32. (b) ( 6 5 / 2 · 6 2 / 3 6 1 / 3 ) 3 = ( 6 5 / 2+2 / 3 6 1 / 3 ) 3 = ( 6 15 / 6+4 / 6 6 2 / 6 ) 3 = ( 6 19 / 6 6 2 / 6 ) 3 = ( 6 (19 / 6) − (2 / 6) ) 3 = ( 6 17 / 6 ) 3 = 6 (17 / 6) · 3 = 6 17 / 2 . (c) ( 3 − 2 k +3 3 4+ k ) 3 = ( 3 − 2 k +3 − 4 − k ) 3 = ( 3 − 3 k − 1 ) 3 = 3 ( − 3 k − 1)(3) = 3 − 9 k − 3 . 75. (a) log 4 x = − 2, so that x = 4 − 2 = 1 4 2 = 1 16 . (b) log 1 / 3 x = − 3, so that x = ( 1 3 ) − 3 = 1 ( 1 3 ) 3 = 1 1 27 = 27. (c) log 10 x = − 2, so that x = 10 − 2 = 1 10 2 = 1 100 . 76. (a) log 2 x = − 3, so that x = 2 − 3 = 1 8 . (b) log 1 / 4 x = − 1 2 , so that x = ( 1 4 ) − 1 / 2 = 1 ( 1 / 4 ) 1 / 2 = 4 1 / 2 = 2. (c) log 3 x = 0, so that x = 3 0 = 1. 77. (a) log 1 / 2 32 = x , so that 32 = ( 1 2 ) x = 1 x 2 x = 2 − x . Therefore 2 − x = 32 = 2 5 , so that x = − 5. (b) log 1 / 3 81 = x , so that 81 = ( 1 3 ) x = 1 x 3 x = 3 − x . Therefore 3 − x = 81 = 3 4 , so that x = − 4. (c) log 10 0 . 001 = x , so that 10 x = 0 . 001 = 10 − 3 . It follows that x = − 3. 78. (a) log 3 81 = x , so that 81 = 3 4 = 3 x . It follows that x = 4. (b) log 5 1 25 = x , so that 1 25 = 5 − 2 = 5 x . It follows that x = − 2. (c) log 10 1000 = x , so that 1000 = 10 3 = 10 x . It follows that x = 3. 79. (a) − ln ( 1 3 ) = ln [ ( 1 3 ) − 1 ] = ln 3. (b) log 4 ( x 2 − 4) = log 4 [( x − 2)( x + 2)] = log 4 ( x − 2) + log 4 ( x + 2). (c) log 2 4 3 x − 1 = (3 x − 1) log 2 (2 2 ) = (3 x − 1)(2 · log 2 2) = (3 x − 1)(2 · 1) = 6 x − 2. 80. (a) − log 3 1 4 = log 3 1 1 / 4 = log 3 4. (b) log ( x 3 − x x − 1 ) = log ( x ( x 2 − 1) x − 1 ) = log ( x ( x − 1)( x +1) x − 1 ) = log( x ( x + 1)) = log( x 2 + x ). (c) ln ( e x − 2 ) = ( x − 2) ln e = x − 2.

  21. Page 21
    Read page 21 as text

    20 Preview and Review 81. (a) Since e 3 x − 1 = 2, we have (3 x − 1) ln e = ln 2, so that 3 x − 1 = ln 2 and therefore x = 1+ln 2 3 . (b) Since e − 2 x = 10, we have − 2 x ln e = ln 10, so that − 2 x = ln 10 and therefore x = − ln 10 2 . (c) Since e x 2 − 1 = 10, we have ( x 2 − 1) ln e = ln 10, so that x 2 − 1 = ln 10 and therefore x = ± √ 1 + ln 10. 82. (a) Since 625 = 5 4 , we have 5 x = 5 4 so that x = 4. (b) Since 256 = 4 4 , we have 4 4 x = 4 4 , so that 4 x = 4 and thus x = 1. (c) Since 0 . 0001 = 10 − 4 , we have 10 2 x = 10 − 4 , so that 2 x = − 4 and thus x = − 2. 83. (a) Since ln( x − 3) = 5, we have e ln( x − 3) = e 5 , so that x − 3 = e 5 . Therefore x = e 5 + 3. (b) Since ln( x + 2) + ln( x − 2) = ln (( x + 2)( x − 2)) ln( x 2 − 4), we have ln( x 2 − 4) = 1, so that e ln( x 2 − 4) = e 1 = e . Therefore x 2 − 4 = e so that x = ± √ e + 4. However, x = − √ e + 4 is invalid, since then x − 2 < 0 so that ln( x − 2) is undefined. The only solution is x = √ e + 4. (c) We have log 3 x 2 − log 3 2 x = log 3 x 2 2 x = log 3 x 2 , so that log 3 x 2 = 2. Therefore 3 log 3 ( x/ 2) = 3 2 = 9, so that x 2 = 9 and then x = 18. 84. (a) log 3 (2 x − 1) = 2 ⇒ 2 x − 1 = 3 2 = 9, so that 2 x = 10 and then x = 5. (b) ln(2 − 3 x ) = 0 ⇒ 2 − 3 x = e 0 = 1, so that 3 x = 1 and then x = 1 3 . (c) log( x ) − log( x + 1) = log x x +1 . Therefore log x x +1 = log 2 3 , so that x x +1 = 2 3 . It follows that x = 2.  1.2.6 85. (3 − 2 i ) − ( − 5 + 2 i ) = 3 − 2 i + 5 − 2 i = 8 − 4 i 86. (6 + i ) − 4 i = 6 + i − 4 i = 6 − 3 i 87. (4 − 2 i ) + (9 + 4 i ) = 4 + 9 − 2 i + 4 i = 13 + 2 i 88. (6 − 4 i ) + (2 + 5 i ) = 6 + 2 − 4 i + 5 i = 8 + i 89. 4(5 + 3 i ) = 20 + 12 i 90. (2 − 3 i )(3 + 2 i ) = 6 + 4 i − 9 i − 6 i 2 = 6 − 5 i + 6 = 12 − 5 i since i 2 = − 1. 91. (6 − i )(6 + i ) = 36 − 6 i + 6 i − i 2 = 36 − ( − 1) = 37. 92. ( − 4 − 3 i )(4 + 3 i ) = − 4 · 4 − 4 · 3 i − 3 i · 4 − 9 i 2 = − 16 − 12 i − 12 i − 9( − 1) = − 16 − 24 i + 9 = − 7 − 24 i . 93. Since conjugation replaces i by − i , we get z = 1 + 2 i = 1 − 2 i . 94. z + u = (1 + 2 i ) + (2 − 3 i ) = 1 + 2 i + 2 − 3 i = 3 − i . 95. z + v = (1 + 2 i ) + (1 − 5 i ) = 1 + 2 i + 1 − 5 i = 2 − 3 i = 2 + 3 i . 96. v − w = (1 − 5 i ) − (1 + i ) = 1 − 5 i − 1 − i = − 6 i = 6 i . 97. vw = (1 − 5 i )(1 + i ) = 6 − 4 i = 6 + 4 i . 98. uz = (2 − 3 i )(1 + 2 i ) = 8 + i = 8 − i .

  22. Page 22
    Read page 22 as text

    1.2 Preliminaries 21 99. Using the quadratic formula with a = 2, b = − 3, c = 2, we get x 1 , 2 = − b ± √ b 2 − 4 ac 2 a = − ( − 3) ± √ ( − 3) 2 − 4 · 2 · 2 2 · 2 = 3 ± √ − 7 4 = 3 ± √ 7 i 4 . 100. Using the quadratic formula with a = 1, b = 1, c = 1, we get x 1 , 2 = − b ± √ b 2 − 4 ac 2 a = − 1 ± √ 1 2 − 4 · 1 · 1 2 · 1 = − 1 ± √ − 3 2 = − 1 ± √ 3 i 2 . 101. Using the quadratic formula with a = − 1, b = 1, c = 2, we get x 1 , 2 = − b ± √ b 2 − 4 ac 2 a = − 1 ± √ 1 2 − 4 · ( − 1) · 2 2 · ( − 1) = − 1 ± √ 9 − 2 = − 1 ± 3 − 2 , so that the two roots are x 1 = − 1+3 − 2 = − 1 and x 2 = − 1 − 3 − 2 = 2. 102. Using the quadratic formula with a = 1, b = 2, c = 3, we get x 1 , 2 = − b ± √ b 2 − 4 ac 2 a = − 2 ± √ 2 2 − 4 · 1 · 3 2 · 1 = − 2 ± √ − 8 2 = − 2 ± 2 √ − 2 2 = − 1 ± √ 2 i. 103. Using the quadratic formula with a = 1, b = 1, c = 6, we get x 1 , 2 = − b ± √ b 2 − 4 ac 2 a = − 1 ± √ 1 2 − 4 · 1 · 6 2 · 1 = − 1 ± √ − 23 2 = − 1 ± √ 23 i 2 . 104. Using the quadratic formula with a = − 2, b = 4, c = − 3, we get x 1 , 2 = − b ± √ b 2 − 4 ac 2 a = − 4 ± √ 4 2 − 4 · ( − 2) · ( − 3) 2 · ( − 2) = − 4 ± √ − 8 4 = − 4 ± 2 i √ 2 4 = − 1 ± √ 2 2 i. 105. The discriminant is b 2 − 4 ac = ( − 4) 2 − 4 · 3 · ( − 7) = 16 + 84 = 100 , so there are two real roots, which are x 1 , 2 = 4 ± √ 100 6 = 4 ± 10 6 = − 1 , 7 3 . 106. The discriminant is b 2 − 4 ac = ( − 1) 2 − 4 · 1 · ( − 1) = 1 + 4 = 5 , so there are two real roots, which are x 1 , 2 = 1 ± √ 5 2 . 107. The discriminant is b 2 − 4 ac = ( − 1) 2 − 4 · 3 · ( − 4) = 49 , so there are two real roots, which are x 1 , 2 = 1 ± √ 49 6 = 1 ± 7 6 = − 1 , 4 3 .

  23. Page 23
    Read page 23 as text

    22 Preview and Review 108. The discriminant is b 2 − 4 ac = ( − 1) 2 − 4 · 4 · 1 = − 15 , so there are two complex roots, which are complex conjugates. They are x 1 , 2 = 1 ± √ − 15 8 = 1 ± √ 15 i 8 . 109. The discriminant is b 2 − 4 ac = ( − 5) 2 − 4 · 3 · 6 = − 47 , so there are two complex roots, which are complex conjugates. They are x 1 , 2 = 5 ± √ − 47 6 = 5 ± √ 47 i 6 . 110. The discriminant is b 2 − 4 ac = ( − 1) 2 − 4 · ( − 3) · ( − 4) = − 47 , so there are two complex roots, which are complex conjugates. They are x 1 , 2 = 1 ± √ − 47 − 6 = − 1 ± √ 47 i 6 . 111. If z = a + bi , then z = a − bi , so that z + z = ( a + bi ) + ( a − bi ) = a + bi + a − bi = 2 a z − z = ( a + bi ) − ( a − bi ) = a + bi − a + bi = 2 bi. 112. If z = a + bi , then z = a − bi . Since taking complex conjugates negates the imaginary portion of the number we also have ( z ) = a − bi = a + bi = z. 113. If z = a + bi , then ( z ) = a − bi = a + bi = z. 114. Let z = a + bi and w = c + di . Then z + w = ( a + bi ) + ( c + di ) = ( a + c ) + ( b + d ) i = ( a + c ) − ( b + d ) i = ( a − bi ) + ( c − di ) = z + w. 115. Let z = a + bi and w = c + di . Then zw = ( a + bi )( c + di ) = ( ac − bd ) + ( ad + bc ) i = ( ac − bd ) − ( ad + bc ) i = ( a − bi )( c − di ) = z · w.

  24. Page 24
    Read page 24 as text

    1.3 Elementary Functions 23 1.3 Elementary Functions  1.3.1 1. The range is y ≥ 0. - 2 - 1 1 2 x 1 2 3 4 y 2. The range is 0 ≤ y ≤ 4. 1 2 x 1 2 3 4 y 3. The range is 0 ≤ y < 4 (note that y = 4 is excluded since the domain does not include x = − 2). - 2 - 1 x 1 2 3 4 y

  25. Page 25
    Read page 25 as text

    24 Preview and Review 4. The range is 0 ≤ x < ( 3 2 ) 2 , or 0 ≤ x < 9 4 . - 1 2 1 2 1 3 2 x 1 2 y 5. (a) Factoring the numerator, we get x 2 − 1 x − 1 = ( x + 1)( x − 1) x − 1 = x + 1 if x 6 = 1 (so that the function is defined) . (b) No the functions are not equal; they have different domains. The function f is not defined at the point x = 1 while g is defined for all of R . 6. (a) For x ≥ 2 the quantity inside the absolute value is non-negative, so it is equal to 2( x − 2) . For x < 2, the quantity inside the absolute value is negative, and so the absolute value will be equal to − 2( x − 2) = 2(2 − x ). This shows the equality. (Note that the case x = 2 is given in both branches on the right. However, the value is in any case 0, so the two definitions agree.) (b) Indeed f and g are the same function because they are defined on the same domain and f ( x ) = g ( x ) for all x in the domain. This can be seen as follows: for 0 ≤ x ≤ 2 , g ( x ) = 2 | x − 2 | = − 2( x − 2) = 4 − 2 x , and for 2 ≤ x ≤ 3 , g ( x ) = 2 | x − 2 | = 2( x − 2) = 2 x − 4. 7. The function is odd, as the graph below indicates: - 2 - 1 1 2 x - 6 - 4 - 2 2 4 6 y To see this algebraically, notice that f ( − x ) = 3( − x ) = − 3 x = − f ( x ) .

  26. Page 26
    Read page 26 as text

    1.3 Elementary Functions 25 8. The function is even, as the graph below indicates: - 2 - 1 1 2 x 4 8 12 y To see this algebraically, notice that f ( − x ) = 3( − x ) 2 = 3 x 2 = f ( x ). 9. The function is even, as the graph below indicates: - 2 - 1 1 2 x 2 4 6 y To see this algebraically, notice that f ( − x ) = | 3( − x ) | = |− 3 x | = | 3 x | = f ( x ). 10. The function is neither even nor odd, as the graph below implies: - 2 - 1 1 2 x - 5 - 3 - 1 1 3 y To see this algebraically, notice that for example f ( − 1) = 2( − 1) − 1 = − 3, while f (1) = 2 · 1 − 1 = 1, so that f ( − 1) 6 = f (1) and f ( − 1) 6 = − f (1). 11. The function is even, as the graph below implies. To see this algebraically, notice that f ( − x ) = − |− x | = − | x | = f ( x ). - 2 - 1 1 2 x - 2 - 1 y

  27. Page 27
    Read page 27 as text

    26 Preview and Review 12. The function is odd, as the graph below implies: - 2 - 1 1 2 x - 24 - 16 - 8 8 16 24 y To see this algebraically, notice that f ( − x ) = 3( − x ) 3 = − 3 x 3 = − f ( x ) . 13. (a) ( f ◦ g )( x ) = f ( g ( x )) = f (3 + x ) = (3 + x ) 2 . (b) ( g ◦ f )( x ) = g ( f ( x )) = g ( x 2 ) = 3 + x 2 . 14. (a) ( f ◦ g )( x ) = f ( g ( x )) = f (1 − 2 x ) = √ 1 − 2 x . (b) ( g ◦ f )( x ) = g ( f ( x )) = g ( √ x ) = 1 − 2 √ x . 15. (a) ( f ◦ g )( x ) = f ( g ( x )) = f ( √ x ) = 1 − √ x . The domain of this function is { x ∈ R : x ≥ 0 } , since otherwise √ x is undefined. (b) ( g ◦ f )( x ) = g ( f ( x )) = g (1 − x ) = √ 1 − x . The domain of this function is { x ∈ R : x ≤ 1 } , since otherwise √ 1 − x is undefined. 16. (a) ( f ◦ g )( x ) = f ( g ( x )) = f (2 x 2 ) = 1 2 x 2 +1 . The domain of this function is R , since the denominator is never zero and is therefore defined everywhere, and since g ( x ) is defined for all x . (b) ( g ◦ f )( x ) = g ( f ( x )) = g ( 1 x +1 ) = 2 · ( 1 x +1 ) 2 = 2 ( x +1) 2 . The domain of this function is { x ∈ R : x 6 = − 1 } , since the denominator vanishes for x = − 1. 17. (a) ( f ◦ g )( x ) = f ( g ( x )) = f ( √ x ) = 1 √ x . The domain of this function is { x ∈ R : x > 0 } , since √ x is undefined for x < 0, and the fraction is undefined for x = 0. (b) ( g ◦ f )( x ) = g ( f ( x )) = g ( 1 x ) = √ 1 x = 1 √ x . This is the same function as in part (a), so its domain is also { x ∈ R : x > 0 } . 18. ( f ◦ g )( x ) = f ( g ( x )) = f ( √ x + 1) = ( √ x + 1) 4 = ( x + 1) 2 . In order for the composition to be defined, we must have x ≥ 3 (since this is the domain of g ), and also √ x + 1 ≥ 3 (since this is the domain of f ). These two taken together give x ≥ 8. Since ( x + 1) 2 is defined for x ≥ 8, the domain is in fact x ≥ 8. 19. ( f ◦ g )( x ) = f ( g ( x )) = f ( √ x ) = ( √ x ) 2 = x ( g ◦ f )( x ) = g ( f ( x )) = g ( x 2 ) = √ x 2 = | x | . In addition, both compositions have the same domain, namely { x ∈ R : x ≥ 0 } . As a result, | x | = x on that domain, so that f ◦ g = g ◦ f .

  28. Page 28
    Read page 28 as text

    1.3 Elementary Functions 27 20. (a) ( f ◦ g )( x ) = f ( g ( x )) = f ( x + 1) = ( x + 1) 4 ( g ◦ f )( x ) = g ( f ( x )) = g ( x 4 ) = x 4 + 1 . Since these expressions are not equal, ( f ◦ g )( x ) 6 = ( g ◦ f )( x ). (b) ( f ◦ g )( x ) = f ( g ( x )) = f ( √ x ) = ( √ x ) 4 = x 2 ( g ◦ f )( x ) = g ( f ( x )) = g ( x 4 ) = √ x 4 = x 2 . While these expressions are equal, note that ( f ◦ g )( x ) is defined only for x ≥ 0, while ( g ◦ f )( x ) is defined for any x ∈ R . Since the domains of f ◦ g and g ◦ f are unequal, the functions are unequal. (c) ( f ◦ g )( x ) = f ( g ( x )) = f ( 1 x ) = ( 1 x ) 4 = 1 x 4 ( g ◦ f )( x ) = g ( f ( x )) = g ( x 4 ) = 1 x 4 . Since these expressions are equal, ( f ◦ g )( x ) = ( g ◦ f )( x ). (d) ( f ◦ g )( x ) = f ( g ( x )) = f ( − x ) = ( − x ) 4 ( g ◦ f )( x ) = g ( f ( x )) = g ( x 4 ) = − x 4 . Since ( − x ) 4 = x 4 6 = − x 4 , these expressions are not equal, so ( f ◦ g )( x ) 6 = ( g ◦ f )( x ). (e) ( f ◦ g )( x ) = f ( g ( x )) = f ( | x | ) = | x | 4 ( g ◦ f )( x ) = g ( f ( x )) = g ( x 4 ) = ∣ ∣ x 4 ∣ ∣ . Since | x | 4 = x 4 = ∣ ∣ x 4 ∣ ∣ , these expressions are equal, so ( f ◦ g )( x ) = ( g ◦ f )( x ).  1.3.2 21. x 2 x 3 0.5 1. 1.5 x 1 2 3 y f ( x ) > g ( x ) for 0 < x < 1, and f ( x ) < g ( x ) for x > 1.

  29. Page 29
    Read page 29 as text

    28 Preview and Review 22. x 3 x 5 0.5 1. 1.5 x 1 2 3 4 5 y f ( x ) > g ( x ) for 0 < x < 1, and f ( x ) < g ( x ) for x > 1. 23. 0.5 1. 1.5 x 1 2 3 4 5 y x x 2 x 3 x 4 The curves intersect at x = 0 and again at x = 1. 24. (a) x x 2 0.5 1. 1.5 2. x 1 2 3 4 y (b) f ( x ) ≥ g ( x ) for 0 ≤ x ≤ 1, and f ( x ) ≤ g ( x ) for x = 0 and x ≥ 1. 25. (a) x 2 x 3 0.5 1. 1.5 x 1 2 3 y (b) Multiply both sides of x ≤ 1 by x to get x 2 ≤ x . Note that we do not have to worry about reversing the sign of the inequality since we also know x ≥ 0. (c) Multiply both sides of x ≥ 1 by x to get x 2 ≥ x . Again note that we do not have to worry about reversing the sign of the inequality since we must have x > 0. 26. First suppose 0 ≤ x ≤ 1. Multiplying by x , we have 0 ≤ x 2 ≤ x ≤ 1. Multiplying by x again, we have 0 ≤ x 3 ≤ x 2 ≤ x ≤ 1. Repeating this process n − 1 times, we get the chain of inequalities 0 ≤ x n ≤ x n − 1 · · · ≤ x m ≤ x m − 1 · · · ≤ x ≤ 1 Thus x n ≤ x m if n ≥ m .

  30. Page 30
    Read page 30 as text

    1.3 Elementary Functions 29 Next suppose x ≥ 1. Multiplying by x , we have x 2 ≥ x ≥ 1. Multiplying by x again, we have x 3 ≥ x 2 ≥ x ≥ 1. Repeating this process n − 1 times, we get the chain of inequalities x n ≥ x n − 1 · · · ≥ x m ≥ x m − 1 · · · ≥ x ≥ 1 Thus x n ≥ x m if n ≥ m . 27. (a) Let f ( x ) = y = x 4 ; then f ( − x ) = ( − x ) 4 = x 4 = f ( x ), so that f is even. (b) Let f ( x ) = y = x 3 ; then f ( − x ) = ( − x ) 3 = − x 3 = − f ( x ), so that f is odd. 28. (a) If n is even, write n = 2 k where k is an integer. Then letting f ( x ) = y = x n = x 2 k we get f ( − x ) = ( − x ) n = ( − x ) 2 k = ( ( − x ) 2 ) k = ( x 2 ) k = x 2 k = f ( x ) , so that f is even. (b) If n is odd, write n = 2 k + 1 where k is an integer. Then letting f ( x ) = y = x n = x 2 k +1 we get f ( − x ) = ( − x ) n = ( − x ) 2 k +1 = ( − x ) 2 k · ( − x ) = − x ( ( − x ) 2 ) k = − x ( x 2 ) k = − x · x 2 k = − x 2 k +1 = − f ( x ) , so that f is odd. 29. (a) I ( p ) is a probability, and all probabilities must be between 0 and 1. A graph of I is below: - 1 1 2 p 1 2 3 4 5 I From the graph we see that I ( p ) > 0 everywhere. To see when I ( p ) = 1, solve I ( p ) = 2 p 2 − 2 p + 1 = 1, giving 2 p ( p − 1) = 0, so that p = 0 or p = 1. Thus for 0 ≤ p ≤ 1, the value of I ( p ) ∈ [0 , 1]; it follows that the domain of I is p ∈ [0 , 1]. (b) See part (a). (c) The minimum value of I ( p ) occurs for p = 1 2 , where it is 2 · ( 1 2 ) 2 − 2 · 1 2 + 1 = 1 2 , so the range I ([0 , 1]) = [ 1 2 , 1 ] . 30. (a) The reaction rate R is proportional to the amount of the two reactants that are left; that is, R ∝ AB . Denote by k the constant of proportionality. If x is the concentration of AB , then we have used up x units of both A and B , so that R ( x ) = k ([ A ] − x )([ B ] − x ) = k (3 − x )(1 − x ) since there were three units of A and one unit of B to start with. (b) We cannot use up more of either component than there was to start with; since there is only one unit of B , and we use a nonnegative amount of B , the domain is x ∈ [0 , 1]. Note that for x ≤ 1 that R ( x ) ≤ 3 k , and R ( x ) ≥ 0 everywhere, so that the range of R ( x ) is [0 , 3 k ].

  31. Page 31
    Read page 31 as text

    30 Preview and Review (c) Substituting [ A ] = 6 and [ B ] = 3 in the equation in part (a) gives R ( x ) = k (6 − x )(3 − x ). Here, by the same argument as in part (b), the domain of R ( x ) is x ∈ [0 , 3], and its range is [0 , 18 k ], since the maximum reaction rate occurs when x = 0, at which time it is 18 k . 31. The beetle has a constant speed of 1 m / h. From rate · time = distance , we see that in one hour the beetle goes (1 m / h) · (1 h) = 1 m . Similarly, In two hours the beetle goes (1 m / h) · (2 h) = 2 m In three hours the beetle goes (1 m / h) · (3 h) = 3 m Let the distance in meters be d and the time in hours be t ; then from the distance-rate-time equation we get a polynomial equation of degree 1 in t : d = rate · t ⇒ d = 1 · t ⇒ d = t. 32. The fungal disease spreads radially at the speed of 3 ft / day. To calculate the area of the affected part of the orchard, we need to know its radius. Using the fact that speed · time = distance, we get for the radius (3 ft / day)(2 day) = 6 ft on day two. Thus, the area of infection on the 2 nd day is A = π · 6 2 ≈ 113 . 1 ft 2 . To find the radius on the fourth and eighth days, use the same idea: (3 ft / day)(4 day) = 12 ft on day four (3 ft / day)(8 day) = 24 ft on day eight Therefore the area on these days is A = π · 12 2 = 144 π ≈ 452 . 4 ft 2 on day four A = π · 24 2 = 576 π ≈ 1809 . 6 ft 2 on day eight An equation that expresses the area as a function of time: we know that the circular area is A = πr 2 where r = 3 t ; substituting for r gives a polynomial equation of degree 2 in t : A = πr 2 = π (3 t ) 2 = 9 πt 2 ,  1.3.3 33. The domain is ( −∞ , 1) ∪ (1 , ∞ ), and the range is ( −∞ , 0) ∪ (0 , ∞ )

994 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.