(Q.3) Suppose that f(t) = t^2 + 5t - 4. Simplify your Answer for (a) and (b).

Question

​(a) What is the average rate of change of f(t) over the interval 5 to 6​? _____ (Answer).

​(b) What is the​ (instantaneous) rate of change of f(t) when t = 5? _________ (Answer).

Answer 1

f(t) = t^2 + 5t -4, which gives position or distance as a function of time t

f'(t) = 2t +5 gives velocity as a function of time, it's the derivative of the distance function

f'(5) = 2(5)+5 = 15 instantaneous velocity at time t=5 is 15 units per second

position at time t=5 = 46

position at time t=6 = 62

f'(6) = 17

It moved 62-46 = 16 units in one second, average velocity = 16 = [f'(5)+f'(6)]/2 = (15+17)/2 = 32/2 = 16