Find f. f ''(t) = 9/ sqrt t , f(4) = 33, f '(4) = 15

Question

Find f

f ''(t) = 9/ sqrt t

f(4) = 33

f '(4) = 15

Answer 1

Hello Raghad,

f''(t) = 9/sqrt(t) = 9/t^1/2 = 9t^-1/2 means f'(t) = [9t^1/2/(1/2)] + C = 18t^1/2 + C

To find C, plug 4 into f'(t): f'(4) = 18(4)^1/2 + C = 15 -> 18(2) + C = 15 so C = 15 - 18(2) = 15-36 = -21

C = -21

Let's find the antiderivative of f'(t): f(t) = ∫(18t^1/2 -21) = 18t^3/2/(3/2) - 21t + D = 18(2/3)t^3/2 - 21t + D

f(t) = 12t^3/2 - 21t + D

You're told f(4) = 33 so let's use that to find D: f(4) = 33 = 12(4)^3/2 - 21(4) + D = 12(8) - 84 + D = 33

so 96 - 84 + D = 33 ---> 12 + D = 33 ---> D = 21

So: f(t) = 12t^3/2 - 21t + 21