f(x)=cos(2x)-7sin(x)-4 0<x<2*3.14 what are the extreme points of the function above in the x range. thanks in advance.
You need to remember that a function reaches its extremes at values of x that denote the zeroes of derivative.
You need to find at what values of x derivative of f(x) cancels, hence you need to find the equation of derivative and then you need to cancel it such that:
`f'(x) = -2sin(2x) - 7cos x`
`` If `f'(x) = 0 =gt-2sin(2x) - 7cos x = 0`
You need to solve the trigonometric equation in (0,2pi), hence you need to write the formula of sine of double angle such that:
`-4sin x*cos x - 7 cos x = 0`
Factoring out - cos x yields:
`- cos x*(4 sin x + 7) = 0 =gt cos x = 0 =gt x = +- pi/2`
`4 sin x + 7 = 0 =gt sin x = -7/4lt -1` contrary to the fact that the values of sine function are in `[-1,1].`
Hence, the solutions to the equation of derivative, in interval (0,2pi) are `x = -pi/2; x = pi/2` .
Hence, the function reaches its extremes at `x = -pi/2` and `x = pi/2` .