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.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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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` .

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