`f(x) = cos^2(2x)` Consider the function on the interval (0, 2pi). Find the open intervals on which the function is increasing or decreasing, apply first derivative test to identify all...

`f(x) = cos^2(2x)` Consider the function on the interval (0, 2pi). Find the open intervals on which the function is increasing or decreasing, apply first derivative test to identify all relative extrema.

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Textbook Question

Chapter 3, 3.3 - Problem 45 - Calculus of a Single Variable (10th Edition, Ron Larson).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to find the open intervals on which the function is increasing or decreasing, hence, you need to find where the derivative is positive or negative, so, you need to evaluate the first derivative of the function, using chain rule, such that:

`f'(x) = (cos^2(2x))' => f'(x) = 2 cos(2x)*(- sin(2x))*(2x)'`

`f'(x) = -2*sin(4x)`

You need to solve for x the equation f'(x) = 0:

`-2*sin(4x) = 0 => sin 4x = 0 => 4x = pi => x = pi/4`

For `x = pi/6 < pi/4, sin (4*pi/6) = sin (2pi/3) = 2 sin (pi/3) cos(pi/3) = sqrt3/2 > 0`

For `x = pi/3 > pi/4 => sin (4*pi/3) = sin (2*2pi/3) = 2 sin (2pi/3) cos(2pi/3) = sqrt3*(1/4 - 3/4) = -sqrt3/2< 0`

You need to notice that f'(x)<0 on intervals `(pi/4,pi)U(5pi/4,2pi)` and f'(x)>0 on `(0,pi/4)U(pi,5pi/4).`

Hence, the function increases for `x in(0,pi/4)U(pi,5pi/4` ) and the function decreases for `x in (pi/4,pi)U(5pi/4,2pi).`

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