Question

Given f''(x)= -4sin(2x) and f'(0)= -4 and f(0)= -2. Find f(pi/3)=

Answer 1

`\begin{gathered} f^(\prime)(x)=\int -4\sin (2x)dx=2\cos (2x)+C \\ f^(\prime)(0)=2+C=-4\rightarrow C=-6 \\ f(x)=\int 2\cos (2x)-6dx=\sin (2x)-6x+C \\ f(0)=C=-2 \\ f(x)=\sin (2x)-6x-2\rightarrow \\ f((\pi)/(3))=\frac{\sqrt[]{3}}{2}-2\pi-2 \end{gathered}`