Question

need help on attachment

Answer 1

\[f(x)=\cos x-\sin^2 x\]
\[f'(x)=-\sin x-2\sin x(-\cos x)=2\sin x \cos x-\sin x\]\[=2\sin x(\cos x-1)\]

Answer 2

(chain rule on sin^2x)

Answer 3

Thanks, what will all the critical points of f in (0,2pi) and the absolute maximum value of
f on [0, 2pi].

Answer 4

critical points are when f'(x)=0 or is undefined
0=2sinx(cosx−1)
sinx=0 -------->x={0,pi,2pi}
cosx-1=0
cosx=1 -------->x={0,pi,2pi}
plug these points into your original equation for f(x), whichever gives the highest value is your absolute maximum. I think you can do that yourself, right?

Answer 5

the derivative of sin is positive cosine not negative cosine