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

The key here is to convert the given quotient into a product as follows:

f '' = 9/sqrt(t) = 9/(t)^1/2=9(t)^-1/2 . You can integrate from this form.

The steps from here are:

Step 1 Integrate f '' to get f' (this is an indefinite integral, so there will be a constant of integration to find - designate C₁.)

Step 2 From Step 1, use the given f '(4) =15 to solve for C₁ (let f ' = 15 and t =4)

Step 3 Integrate f ' to get f (there will be another constant of integration to find, designate C₂.)

Step 4 From Step 3, use the given f(4) = 33 to solve for C₂ (let f = 33 and t = 4)

Step 1 Implemented {f '(t) is the integral of f ''(t)} f '(t) = ∫9(t)^-1/2dt = 9{1/(1-1/2)}(t)^1/2 + C₁ = 18(t)^1/2+C₁

Step 2 Implemented f '(4) =15 =(18)(4)^1/2 + C₁ → C₁ =15 - 36 = -21

Step 3 Implemented f(t) = ∫{18(t)^1/2 -21}dt = 18{1/(1+1/2}(t)^3/2 - 21t + C₂ = 12(t)^3/2 -21t +C₂

Step 4 implemented f(4) = 33 = (12)(4)^3/2 - (21)(4) + C₂ → C₂ = 33 - 96 + 84 = 21

note for Step 4: (4)^3/2 = {sqrt(4)}³ = 8

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

Answer 2

f'(t) = ∫f"(t)dt = 9∫ t^-1/2dt = 18√t + C₁

Since f'(4) = 15, 36 + C₁ = 15. So, C₁ = -21

f'(t) = 18√t - 21

f(t) = ∫f'(t)dt = 12t^3/2 - 21t + C₂

Since f(4) = 33, we have 96 - 84 + C₂ = 33. So, C₂ = 21.

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