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This is a 60 in 60 Challenge: 60 questions in 60 minutes

Some are designated as Calc active. The expectation is that you’ll be able to get a handle on the solution within a minute, not actually do the question. The 60 Focused Questions should create a wealth of other questions . Some of the questions are answered in the next slide.

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This is a 60 in 60 Challenge: 60 questions in 60 minutes

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  1. Some are designated as Calc active. The expectation is that you’ll be able to get a handle on the solution within a minute, not actually do the question. The 60 Focused Questions should create a wealth of other questions. Some of the questions are answered in the next slide. This is a 60 in 60 Challenge:60 questions in 60 minutes

  2. #1

  3. #1 Solution

  4. #2 • Find dy/dx for:

  5. #2 Solution

  6. Doesn’t this allow you to consider all limit questions a) at a point or b) at the infinities as well as any Indeterminate forms (of which this is one) #3

  7. Use L’Hopital – answer is 1/2 #3 solution

  8. The above is non-functional. Could you find dy/dx ? dx/dy? The value(s) of dy/dx @ y=1? The name of any horizontal/vertical tangents, if they exist? #4 Let’s try implicit Diff. again !!!

  9. dy/dx=(4x – y2)/(2xy – 3) dy/dx @ y = 1 would require finding the value(s) for x. When y = 1, then x = 3/2 or -1. Note that for (3/2, 1) the value of dy/dx is undefined (vertical tangency) and for (-1, 1) the value of dy/dx is 1. #4 Solution

  10. #5 There is no mystery to the question and the key to its solution is to use the Change of Base Formula. Never forget that all derivatives can be checked in some way or another.

  11. #5 solution

  12. Euler and the linear approximation model go hand in hand……. • Find the approximated value of f(4) for f(x) = ln(2x+1), starting @ x = 1 w/ a step size of 1.5 #6 (C active)

  13. f(x + dx) ≈ f(x) + f’ (x) dx f(1 + 1.5) ≈ f(1) + f’ (1) (1.5) f(1 + 1.5) ≈ f(1) + f’ (1) (1.5) f(2.5) ≈ ln(3) + (2/3)(1.5) f(2.5) ≈ ln(3) + 1 f(2.5 + 1.5) ≈ f(2.5) + f ‘ (2.5)(1.5) f( 4 ) ≈ [ ln 3 + 1] + (2/6)(1.5) f( 4 ) ≈ [ln 3 + 1] + ½ f( 4 ) ≈ ln 3 + 1.5 #6 solution

  14. Given a graph of the derivative of a function and an initial condition, then sketch the graph of the function………… #7

  15. #7 solution f ‘ (x) f(x)

  16. Calculate the area of the region formed by the function , the x-axis, the y-axis, and the line x = 2π. #8 (C active)

  17. #8 solution

  18. The function g(x) = • Find g(4) • Find g ‘ (4) #9

  19. g(4) = ln 9 – ln 5 by u-substitution g ‘ (4) = #9 solution

  20. If f(x)= then which of the following is FALSE ? • A) f(0) = 0 • B) f is continuous at x for all x ≥ 0. • C) f(1) > 0 • D) f ‘ (1) = • E) f (-1) > 0 #10

  21. Choice (E) #10 soluton

  22. Find A question that demands your understanding of the definition of the derivative. Of course, L’Hopital’s Rule would also be appropriate ! #11

  23. f(x) = 8x8 This limit is the definition of the derivative of the above f(x) function evaluated @ x = ½ The answer is f ‘(1/2) = 64 (1/2) 7 = 1/2 #11 solution

  24. #12

  25. If Then #12 solution

  26. Let’s see what you learned from the last question……… #13

  27. #13 solution

  28. What is the volume of the solid formed when the region bounded by y = ex,y = ln(1/x), x = .1, & y = 0 is rotated about the x-axis? Sketch the region and store the intersection point. This will require two integrations. Be efficient with your use of FnInt. #14 (C active)

  29. #14 Solution NOTE: ln(1/x) is easier to “see” and work with by simplifying to ln (x-1) which is equivalent to - ln(x) NOTE: A boundary line of x = .1 is necessary because if the y-axis is used (w/out the x-axis boundary), then an improper integral results !!

  30. This looks harmless……… Find #15

  31. (ln2)/2 #15 solution

  32. What is the average value of the function 3t3 – t2 over the interval -1 ≤ t ≤ 2 ? #16

  33. 2.75 #16 solution

  34. A region in the plane is bounded by the graph of y = 1/x, the x-axis, the line x = m, and the line x = 2m, m > 0. The area of this region A) is independent of m B) increases as m increases C) decreases as m increases D) decreases as m increases when m < ½; increases as m increases when m > ½ E) increases as m increases when m < ½; decreases as m increases when m > ½ #17

  35. A) #17 solution

  36. The 4th degree Taylor polynomial for f about x = 1 is given by T(x) = 5 - 10(x -1) + 15(x – 1)2 – 20(x – 1)3 + 25(x – 1)4 Calculate f(1) + f‘ (1) + fiv (1) #18

  37. T(x) = 5 - 10(x -1) + 15(x – 1)2 – 20(x – 1)3 + 25(x – 1)4 f(1) = 5 f ‘ (1) = -10 f ‘’ (1) = 30 f ‘’’ (1) = -120 fiv (1) = 600 So, f(1) + f ‘ (1) + fiv (1) = 595 #18 solution

  38. Create the 3rd degree Maclaurin polynomial for f(x) = ln(1 – x) Use this polynomial to approximate ln(3) #19

  39. From previous experience (or memorization) we know ln (1 + x) = • So, ln (1 – x) = • And ln (1 – (-2) ) = ln (3) ≈ • NOTE: ln 3 ≈1.0986, so the 3rd degree polynomial is not a great approximation #19 solution

  40. Which of the following series converge? A) I only B) II only C) III only D) II and III E) I and II #20

  41. Choice E Item I may be broken up into to summations; one from 0 to 1 which is a finite series that must converge and the other from 1 to ∞ which when compared to the p-series with p = 2 must converge. Item II closely resembles a geometric series with r = 2/3. Convergence must occur Item III is a p-series with p = ¾ < 1, so it diverges. #20 solution

  42. Use the definition of the derivative to find the instantaneous rate of change for f(x) = sin(2x) #21

  43. #21 solution

  44. Which of the following is an equation of the normal line to y = sin x + cosx at x = π ? A) y = - x +π – 1 B) y = x – π + 1 C) y = x – π – 1 D) y = x + π + 1 E) y = x + π - 1 #22

  45. Which of the following gives dy/dx for the parametric curve y = 3 sin t, x = 2 cost ? A) -3/2 cot t B) 3/2 cot t C) -2/3 tan t D) 2/3 tan t E) tan t #23

  46. Choice A

  47. Which of the following gives dy/dx if y = sin -1 (2x) ? #24

  48. Choice C • Consider all the inverse trig derivatives, including tan -1 (f (x) ) and the associated antiderivatives that are sure to come up such as: #24 solution

  49. A slope field for the differential equation dy/dx = 42 – y will show A line with slope -1 and y-intercept of 42 A vertical asymptote at x = 42 A horizontal asymptote at y = 42 A family of parabolas opening downward A family of parabolas opening to the left. #25

  50. A family of exponential functions result that by transformational geometry result in horizontal asymptotes at y = 42, regardless of the constant C that you use in the general solution of the differential equation. So, choice C #25 solution

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