Find f. f ‴(x) = cos(x), f(0) = 7, f ′(0) = 7, f ″(0) = 7

Answer 1

f'''(x) = cosx

f" = sinx + 7

f' = -cosx + 7x + 8

f = -sinx + (7/2)x^2 + 8x + 7

Answer 2

Hi Parie,

This question is straightforward using trig derivatives. To find f given its third derivative, we just have to integrate a few times and plug in the initial condition.

∫f '''(x)dx = ∫cos(x)dx

f ''(x) = sinx + C

f ''(0) = 7 = sin(0) + C ==> C = 7

∫sin(x) + 7 dx = -cos(x) + 7x + C

f ' (0) = 7 = -cos(0) + 7(0) + C

7 = -1 + C ==> C = 8

f(x) = ∫(-cos(x) + 7x + 8)dx = -sin(x) + (7/2)x² + 8x + C

f(0) = 7 = -sin(0) + (7/2)0^2 + 8(0) + C ==> C = 7

Thus, f(x) = -sin(x) + (7/2)x² +8x + 7. You can verify this by taking each derivative and plugging in 0, and you should get 7. Hope this helps!