# Calculus-antiderivative

posted by Callie

Find f.
f ''(θ) = sin θ + cos θ, f(0) = 5, f '(0) = 3

My steps:
f'(θ)=cosθ-sinθ+C
When f'(0)=3, C=-2, so f'(θ)=cosθ-sinθ-2.

f(θ)=-sinθ-cosθ-2x+D
When f(0)=5, D=6, so f is -sinθ-cosθ-2x+6.

How is that wrong?

1. Reiny

f'Ø) = -cosx + sinx + C
since f'(0) = 3
3 = -cos0 + sin0 + c
3 = -1 + 0 + c ----> c = 4

so f'(x) = -cosx + sinx + 4
f(x) = -sinx - cosx + 4x + d
and f(0) = 5
5 = -sin0 - cos0 + 0 + d
5 = 0 - 1 + d
d = 6

f(x) = -sinx - cosx + 4x + 6

Just realized I used x instead of Ø, no big deal to change

I suggest you take a better look at the derivatives of both sine and cosine

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