SOLUTION: Is my antiderivative correct? Q: Find the function f such that f'(x) = 2sin(x) + sec^2(x) and f(0) = 3 f'(x) = 2sin(x) + sec^2(x) f(x) = -2cos(x) + tan(x) + c f(0) = -2cos(

Algebra ->  Finance -> SOLUTION: Is my antiderivative correct? Q: Find the function f such that f'(x) = 2sin(x) + sec^2(x) and f(0) = 3 f'(x) = 2sin(x) + sec^2(x) f(x) = -2cos(x) + tan(x) + c f(0) = -2cos(      Log On


   

Question 1001596: Is my antiderivative correct?
Q: Find the function f such that f'(x) = 2sin(x) + sec^2(x) and f(0) = 3
f'(x) = 2sin(x) + sec^2(x)
f(x) = -2cos(x) + tan(x) + c
f(0) = -2cos(0) + tan(0) + c = 3
f(0) = +2+0+c=3
f(0) = c = 1
So final solution:
f(x) = -2cos(x) + tan(x) + 1
Is this correct?
Thank you
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
You made an error when you computed "cos(0)". That is equal to 1

So -2cos(0) = -2*1 = -2

You shouldn't have +2

So

f(0) = -2+0+c=3
-2+c = 3
c = 3+2
c = 5

Making the equation to be f(x) = -2cos(x) + tan(x) + 5