f ''(x) = 6+cosx
f '(x) = 6x + sinx + c
f (x) = 3x^2 - cosx + cx + k
f(0) = 1
0 - cos 0 + 0 + k = 1
-1 +k= 1
k = 2
so f(x) = 3x^2 - cosx + cx + 2
f(7π/2) = 0
3(7π/2)^2 - cos(7π/2) + (7π/2)c +2 = 0
solve for c, plug into the f(x) equation and you got it
Find f.
f ''(x) = 6 + cos x, f(0) = −1,
f(7π/2) = 0
3 answers
asdf
Find f.
f ''(x) = 6 + cos x, f(0) = −1, f(9π/2) = 0
f ''(x) = 6 + cos x, f(0) = −1, f(9π/2) = 0