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2.5

2.5. THE SECOND DERIVATIVE. Solution: (a) f”(x) ≥ 0. (b) f”(x) ≤ 0. (c) f”(x) ≤ 0 for x < 0 f”(x) ≥ 0 for x > 0. Solution:. Solution: For t < t 0 , the rate of growth, dP / dt , is increasing and d 2 P/dt 2 ≥ 0. The rate at t 0 , dP / dt is a maximum.

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2.5

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  1. 2.5 THE SECOND DERIVATIVE

  2. Solution: (a) f”(x) ≥ 0 (b) f”(x) ≤ 0 (c) f”(x) ≤ 0 for x < 0 f”(x) ≥ 0 for x > 0 Solution:

  3. Solution: For t < t0, the rate of growth, dP/dt, is increasing and d2P/dt2 ≥ 0. The rate at t0, dP/dt is a maximum. For t > t0, the rate of growth, dP/dt, is decreasing and d2P/dt2 ≤ 0.

  4. Solution: Since the values of dv/dt are decreasing over the intervals we expect d2v/dt2 ≤ 0 . The fact that dv/dt > 0 tells us the car is speeding up; the fact that d2v/dt2 ≤ 0 Tells us that the rate of increase decreased over this period of time.

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