f" = -36sin(6x)
f' = 6cos(6x)+c
f'(0) = 6+c = -5, so c = -11
f' = 6cos(6x)-11
f = sin(6x)-11x+c
f(0) = c = -3
f = sin(6x)-11x-3
f(π/3) = sin(2π)-11(π/3)-3
= -11π/3 - 3
Suppose f′′(x)=−36sin(6x) and
f′(0)=−5, and f(0)=−3
f(π/3)=?
1 answer