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2014 AP CALCULUS AB FRQs

2014 AP CALCULUS AB FRQs. Average rate of change = lbs/day. On the fifteenth day the amount of grass clippings is decreasing at the rate of 0.1635 lbs /day. Average amount of clippings: Solve days. Tangent line at is Solve days. y = 4. f (x).

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2014 AP CALCULUS AB FRQs

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  1. 2014 AP CALCULUS AB FRQs

  2. Average rate of change = lbs/day

  3. On the fifteenth day the amount of grass clippings is decreasing at the rate of 0.1635 lbs/day

  4. Average amount of clippings: Solvedays

  5. Tangent line at is Solve days

  6. y= 4 f(x) Intersections of the two graphs: and

  7. y= 4 f(x) Length of the rectangle: y – f(x)

  8. y= 4 f(x)

  9. y= 4 f(x)

  10. y= 4 f(x)

  11. , therefore if g(x) increases . This happens on the intervals and (−3, 2) If g(x) i decreases. This happens on and (0, 4) g(x)is increasing and concave down on and (0, 2)

  12. The slope of the graph of f at is −2, so

  13. Average acceleration of train A: S meters/minute2

  14. Since is differentiable, it is also continuous so the Intermediate Value Theorem applies to for . Therefore, there must be at least a value such that meters/minute

  15. meters

  16. Train B meters meters meters z = distance between the two trains y Train A x At minutes,

  17. f has a local minimum when its derivative, f’, switches from negative to positive. This occurs at x = 1.

  18. f is twice differentiable so its derivative, f’, is both continuous and differentiable. Therefore the Mean Value Theorem can be used on f’ . There must be a value for such that Since

  19. The problem can also be done using Rolle’s Theorem: f is twice differentiable so its derivative, f’, is both continuous and differentiable. And since Rolle’s Theorem can be used on f’. Therefore, there must be a value for such that

  20. Equation of the tangent line:

  21. Using (0, 1):

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