Let f(t)= 1/t for t>0. For what value of t is f'(t) equal to the average rate of change of f on the closed…

Let f(t)= 1/t for t>0. For what value of t is f'(t) equal to the average rate of change of f on the closed interval [a,b]?

Answer 1

Let at the value t = c, f'(t) equal to the average rate of change of f on the closed interval [a,b].

By the Mean Value theorem
f'(c) = {f(b) – f(a)} / (b – a)
f'(c) = {(1/b) – 1/a)} / (b – a) [assuming a >0, b > 0]
f'(c) = {(a – b)/ab} / (b – a)
f'(c) = -1/ab

Now, f(t) = 1/t [for t > 0]
f'(t) = -1/(t^2)

Now, f'(t) = f'(c)
-1/(t^2) = -1/ab
t = √(ab)

Hence, b) is correct.

Answer 2

Where was this problem taken from

Leave a Comment