Verified answer

given

......f(x)=cos^2(x)-cos(x)

......= cos(x)[ cos(x) - 1 ]

a) zeros:

......cosx = 0

......x = π/2, 3π/2

or

......cosx - 1 = 0

......cosx = 1

......x = 0, π, 2π

x-intercept coordinates:

(0,0), (π/2, 0), (π, 0), (3π/2, 0), (2π, 0)

b) intervals where f(x) is increasing, decreasing

......y = cos^2(x)-cos(x)

......dy/dx = 2cos(x)sin(x) + sin(x)

......= sin(x)[ 2cos(x) + 1 ]

find the zeroes of this function, and test one value in each interval between the zeroes

sin(x) = 0

x = 0, π, 2π

or

2cos(x) + 1 = 0

cos(x) = -1/2

x = 2π/3, 4π/3

critical values (aka zeroes of f'(x)):

x = 0, 2π/3, π, 4π/3, 2π

plug one value between each interval into f'(x), if it gives a positive value then f(x) is increasing, if negative, f(x) is decreasing

.....f'(x) = sin(x)[ 2cos(x) + 1 ]

.....f'(π/2) > 0, increasing on (0, 2π/3)

.....f'(3π/4) < 0, decreasing on (2π/3, π)

.....f'(5π/4) > 0, increasing on (π, 4π/3)

.....f'(3π/2) < 0, decreasing on (4π/3, 2π)

c) extrema are the points where f(x) goes from increasing to decreasing, or vice versa

if it's increasing before, and decreasing after, then it is a local maximum

if its decreasing before and increasing after, it's a local minimum

local maxima: 2π/3, 4π/3

local minimum: π

f(x) = cos^2 x - cos x

a)

x-intercepts: set y=0

cos^2 x - cos x = 0

cos x ( cos x -1) = 0

cos x = 0

x=pi/2, 3pi/2

cos x = 1

x=0, 2pi

x-intercepts are (0,0),(pi/2,0),(3pi/2,0),(2pi,0)

b)

f'(x) = 2 cos x (-sin x) - (-sin x) = 0

- 2 sin x cos x + sin x = 0

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

sin x = 0

x = 0, pi

-2 cos x + 1 = 0

-2 cos x = -1

cos x = 1/2

x = pi/3, 5pi/3

x=0, pi/3,pi, 5pi/3,2pi

Consider the intervals (0,pi/3),(pi/3,pi), (pi,5pi/3),(5pi/3,2pi)

choose one point from each interval and examine the sign of f'(x)

(0,pi/3) : choose pi/6;

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

f'(pi/6) = -0.366 < 0, so f(x) is decreasing on (0, pi/3)

(pi/3,pi) : choose pi/2

f'(pi/2) = 1 > 0, so f(x) is increasing on (pi/3,pi)

(pi,5pi/3) : choose 3pi/2

f'(3pi/2) = -1 < 0 , so f(x) is decreasing on (pi,5pi/3)

(5pi/3,2pi) : choose 7pi/4

f'(7pi/4) = 0.29 > 0 so f(x) is increasing on (pi,5pi/3)

c)

x = 0, pi/3, pi, 5pi/3, 2pi

f''(x) = -2 cos^2(x) + 2 sin^2(x) + cos(x)

f''(0) = -1 < 0 , relative maximum : f(0) =0

f''(pi/3) = 1.5 > 0, relative minimum : f(pi/3) = -1/4

f''(pi) = -3 , relative maximum: f(pi) = 2

f'''(5pi/3) = 1.5, relative minimum : f(5pi/3) =-1/4

f(2pi) = 0