Question

Does anyone know how to solve the second half of this question?

Answer 1

Answer:

see explanation

Step-by-step explanation:

given

f'(x) = 12cos x - 4sin x, then

f'($\frac{4\pi }{3}$) = 12cos($\frac{4\pi }{3}$) - 4sin($\frac{4\pi }{3}$)

[ cos($\frac{4\pi }{3}$) = -cos($\frac{\pi }{3}$), sin($\frac{4\pi }{3}$ = - sin($\frac{\pi }{3}$) ]

= -12cos($\frac{\pi }{3}$) + 4sin($\frac{\pi }{3}$)

= - 12 × $\frac{1}{2}$ + 4 × $\frac{\sqrt{3} }{2}$

= - 6 + 2$\sqrt{3}$ ≈ - 2.536 ( 3 dec. places )