Determine the absolute maximum and minimum of f(x)= 2 cosx+ sin 2x

2.598 and -2.598.

Step-by-step explanation:

f(x) = 2 cos x + sin 2x

f'(x) = -2 sin x + 2 cos 2x = 0   for turning points.

cos 2x =  1 - 2 sin^2 x so we have

-2 sin x + 2 - 4 sin^2 x = 0

4sin^2 x + 2 sin x - 2 = 0

2(2 sin^2 x + sin x - 1) = 0

2(2sinx - 1)(sinx + 1) = 0

sin x  = 0.5, -1   when  f(x) is at a turning point.

x = π/6,  -π/2, 5pi/6

The second derivative is  2 cos x + 2 * -2 sin 2x

= 2 cos x - 4 sin 2x

When x = π/6, this is negative , when x = -π/2 it is positive

so x = π/6 gives a maximum f(x) and x = -π/2 gives 0 so this is a point of inflection

When x = π/6 , f(x) = 2.598

When x = 5pi/6,  f(x) = -2.598.

i think the correct option is the second option.

the answer is d on edge

step-by-step explanation: