# Find the value of `f(0)` when `f''(t) = 2(9t + 4)` and `f'(1) = 3` , `f(1) = 5` . Find the value of f(0) when f'(t) = 4 cos(2t) , f(pi/4) =4 .

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To solve for f(0), we need to determine the function, f(t). To do so, integrate f "(t). The integral of this is equal to f '(t) .

`f ''(t) = 2(9t + 4)= 18t + 8`

`f ' (t) = int f ''(t) dt = int (18t + 8) dt = int 18t dt + int 8dt`

`f'(t) = 9t^2 + 8t + C`

To determine the value of C, substitute the given f '(1) = 3.

`3 = 9(1)^2+8(1)+C`

`3=17+C`

`-14=C`

So, `f '(t) = 9t^2+8t-14` .

Then, integrate f'(t) to get f(t).

`f (t) = int f'(t) d(t) = int (9t^2+8t-14) dt = int 9t^2dt + int 8t dt -int 14dt`

`f(t) =3t^3 +4t^2-14t + C`

Substitute f(1)=5 to determine the value of C.

`5 = 3(1)^3 + 4(1)^2-14(1) + C`

`5=-7+C`

`12=C`

Hence,` f(t) = 3t^3+4t^2-14t+12` .

Now that we know the function f(t), we may now be able to solve for f(0).

Substitute t=0 to the function f(t).

`f(0) = 3(0)^3 + 4(0)^2-14(0)+12=12`

**Therefore, `f(0) = 12.` **