Contextual Applications of Derivatives: Get Stronger Answer Key

Limits at Infinity and Asymptotes

  1. x=1x=1
  2. x=−1,x=2
  3. x=0
  4. Yes, there is a vertical asymptote
  5. Yes, there is a vertical asymptote
  6. 0
  7. ∞
  8. −17
  9. –2
  10. –4
  11. Horizontal: none, vertical: x=0
  12. Horizontal: none, vertical: x=±2
  13. Horizontal: none, vertical: none
  14. Horizontal: y=0, vertical: x=±1
  15. Horizontal: y=0, vertical: x=0 and x=−1
  16. Horizontal: y=1, vertical: x=1
  17. Horizontal: none, vertical: none
  18. Answers will vary, for example: y=2xx−1
  19. Answers will vary, for example: y=4xx+1
  20. y=0
  21. ∞
  22. y=3
  23. The function starts in the third quadrant, increases to pass through (−1, 0), increases to a maximum and y intercept at 4, decreases to touch (2, 0), and then increases to (4, 20).
  24. An upward-facing parabola with minimum between x = 0 and x = −1 with y intercept between 0 and 1.
  25. This graph starts at (−2, 4) and decreases in a convex way to (1, 0). Then the graph starts again at (4, 0) and increases in a convex way to (6, 3).
  26. This graph has vertical asymptote at x = 0. The first part of the function occurs in the second and third quadrants and starts in the third quadrant just below (−2π, 0), increases and passes through the x axis at −3π/2, reaches a maximum and then decreases through the x axis at −π/2 before approaching the asymptote. On the other side of the asymptote, the function starts in the first quadrant, decreases quickly to pass through π/2, decreases to a local minimum and then increases through (3π/2, 0) before staying just above (2π, 0).
  27. This graph has vertical asymptotes at x = ±π/2. The graph is symmetric about the y axis, so describing the left hand side will be sufficient. The function starts at (−π, 0) and decreases quickly to the asymptote. Then it starts on the other side of the asymptote in the second quadrant and decreases to the the origin.
  28. This function starts at (−2π, 0), increases to near (−3π/2, 25), decreases through (−π, 0), achieves a local minimum and then increases through the origin. On the other side of the origin, the graph is the same but flipped, that is, it is congruent to the other half by a rotation of 180 degrees.

Applied Optimization Problems

  1. The critical points can be the minima, maxima, or neither.
  2.  
  3. False; y=−x2 has a minimum only
  4.  
  5. h=623 in.
  6. 1
  7. 100 ft by 100 ft
  8. 40 ft by 40 ft
  9. 19.73 ft.
  10. T(θ)=40θ3v+40cosθv
  11.  
  12.  
  13. approximately 34.02 mph
  14.  
  15. 4
  16. 0
  17.  
  18. Maximal: x=5,y=5; minimal: x=0,y=10 and y=0,x=10
  19. Maximal: x=1,y=9; minimal: none
  20. 4π3√3
  21. 6
  22. r=2,h=4
  23. (2,1)
  24. (0.8351,0.6974)
  25. A=20r−2r2−12πr2
  26. C(x)=5x2+32x
  27. P(x)=(50−x)(800+25x−50)

L’Hôpital’s Rule

  1. ∞
  2. 12a
  3. 1nan−1
  4. Cannot apply directly; use logarithms
  5. Cannot apply directly; rewrite as limx→0x3
  6. 6
  7. –2
  8. –1
  9. n
  10. −12
  11. 12
  12. 1
  13. 16
  14. 1
  15. 0
  16. 0
  17. –1
  18. ∞
  19. 0
  20. 1e
  21. 0
  22. 1
  23. 0
  24. tan(1)
  25. 2

Newton’s Method

    1. F(xn)=xn−(xn)3+2xn+13(xn)2+2
    2. F(xn)=xn−exnexn
    3. |c|>0.5 fails, |c|≤0.5 works
    4. c=1f′(xn)
      1. x1=1225,x2=312625
      2. x1=−4,x2=−40
      1. x1=1.291,x2=0.8801
      2. x1=0.7071,x2=1.189
      1. x1=−2625,x2=−1224625
      2. x1=4,x2=18
      1. x1=610,x2=610
      2. x1=2,x2=2
    9. 3.1623 or –3.1623
    10. 0, –1, or 1
    11. 0
    12. 0.5188 or –1.2906
    13. 0
    14. 4.493
    15. 0.159, 3.146
    16. To find candidates for maxima and minima, we need to find the critical points f′(x)=0. Show that to solve for the critical points of a function f(x), Newton’s method is given by xn+1=xn−f′(xn)f′′(xn).
    17. We need f to be twice continuously differentiable.
    18. x=0
    19. x=−1

<lix=5.619

  • x=−1.326
  • There is no solution to the equation.
  • It enters a cycle.
  • 0
  • –0.3513
  • Newton: 11 iterations, secant: 16 iterations
  • Newton: three iterations, secant: six iterations
  • Newton: five iterations, secant: eight iterations
  • E=4.071
  • 4.394%

 

Antiderivatives

  1. F′(x)=15x2+4x+3
  2. F′(x)=2xex+x2ex
  3. F′(x)=ex
  4. F(x)=ex−x3−cos(x)+C
  5. F(x)=x22−x−2cos(2x)+C
  6. F(x)=12x2+4x3+C
  7. F(x)=25(√x)5+C
  8. F(x)=32x2/3+C
  9. F(x)=x+tan(x)+C
  10. F(x)=13sin3(x)+C
  11. F(x)=−12cot(x)−1x+C
  12. F(x)=−secx−4cscx+C
  13. F(x)=−18e−4x−cosx+C
  14. −cosx+C
  15. 3x−2x+C
  16. 83x3/2+45x5/4+C
  17. 14x−2x−12x2+C
  18. f(x)=−12x2+32
  19. f(x)=sinx+tanx+1
  20. f(x)=−16x3−2x+136
  21. Answers may vary; one possible answer is f(x)=e−x
  22. Answers may vary; one possible answer is f(x)=−sinx
  23. 5.867 sec
  24.  
  25. 7.333 sec
  26.  
  27. 13.75 ft/sec2
  28. F(x)=13x3+2x
  29. F(x)=x2−cosx+1
  30. F(x)=−1x+1+1
  31. True
  32. False