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I recently wrote a test in my precalculus class, and came across a problem which I thought I did correctly, but everyone else taking the course seems to have gotten a different solution. Because of the discrepancy, I was wondering if anyone would be willing to provide their own solution of the problem.

For what their worth, my thoughts on the problem: I thought that the rocket would be above $19.5$ metres after about $0.5$ seconds, and would never again fall below it, and I thought that part c) was a trick, since it never reaches a highest point, and never falls to the ground. As for the instantaneous rate of change, that I know is given by the derivative, which would be $H'(t) = 16t + 32$.)

And here is the problem:

A rocket fired straight up from the ground at a velocity of $32 \text{ ms$^{-1}$}$, and the height above the ground is given by $H(t) = 8t^{2} + 32t$.

a) During what interval will the rocket be at least $19.5$ metres above the ground.

b) Calculate the average rate of change when the rocket reaches its highest point, to the time it falls to the ground.

c)Calculate the instantaneous rate of change at $t = 3$ s.

d)Explain why your answer to part b) and c) and positive and negative, respectively.

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    Not really relevant, but how are derivatives _precalculus_?2012-12-15

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