Calculus...Simple!?

Question

Given the function f defined by f(x) = cos x - cox^2 x, for x is greater than or equal to -pie and x is less than or equal to pie
(a) Find the x intercepts of the graph of f.
(b) Find the x and y coordinates of all relative maximum points of f. How do i justify this?
(c) Find the intervals of which the graph of f is increasing.
(d) Using information found in parts (a), (b), and (c), sketch the graph of f.

Answer 1

f(x) = cos(x) - cos^2(x), for -pi <= x <= pi

(a) To find the x-intercepts of the graph, set f(x) = 0, and solve for x.

0 = cos(x) - cos^2(x). Factor out cos(x), to obtain,
0 = cos(x) [1 - cos(x)]

And now, equate each factor to 0, and obtain solutions between -pi and pi.

cos(x) = 0 implies x = {-pi/2, pi/2}

1 - cos(x) = 0, so
1 = cos(x), and cos(x) = 1 at the point 0.

Therefore, the x-intercepts are at x = {-pi/2, pi/2, 0}.

(b) To find the x and y intercepts, take the first derivative, and then make it 0.

f(x) = cos(x) - cos^2(x).
f'(x) = -sin(x) - 2cos(x) [-sin(x)]. Simplifying, we get
f'(x) = -sin(x) + 2cos(x)sin(x). Now, making f'(x) = 0,

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

And now we equate each factor to 0

-sin(x) = 0 means sin(x) = 0, which occurs at the points 0 and pi; however, given our restriction, we only accept 0, so x = 0.

1 - 2cos(x) = 0 implies
1 = 2cos(x), or
cos(x) = 1/2. This occurs at x = {pi/3, -pi/3}

Therefore, at the coordinates x = {0, pi/3, -pi/3}, we have our critical values.

To determine if we have a min/max, we find the indicated value at the second derivative; if it is positive, it is concave up (and therefore a minimum. If it is negative, it is concave down (and therefore a maximum.

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

f''(-pi/3) = -cos(-pi/3) - 2sin^2(-pi/3) + 2cos^2(-pi/3)
f''(-pi/3) = -(1/2) - 2[-sqrt(3)/2]^2 + 2[1/2]^2
f''(-pi/3) = -1/2 - 2[3/4] + 2(1/4)
f''(-pi/3) = -1/2 - 3/2 + 1/2 = -3/2, which is negative.
Therefore, we have a minimum at x = -pi/3.

f''(0) = -cos(0) - 2sin^2(0) + 2cos^2(0)
f''(0) = -1 - 0 + 2 = 1, which is positive.
Therefore, we have a maximum at x = 0

f''(pi/3) = -cos(pi/3) - 2sin^2(pi/3) + 2cos^2(pi/3)
f''(pi/3) = -[1/2] - 2[sqrt(3)/2]^2 + 2[1/2]^2
f''(pi/3) = -1/2 - 2[3/4] + 2[1/4]
f''(pi/3) = -1/2 - 3/2 + 1/2 = -3/2, which is negative.
Therefore, we have a minimum at x = pi/3.

This means we are only interested in x = 0, since it is the only relative maximum point.
When x = 0, f(0) = cos(0) - cos^2(0) = 1 - 1 = 0
So x = 0, y = 0 is the relative maximum point of f.

(c) As we just solved, at x = -pi/3 is a local minimum, at x = 0 is a local maximum, and at x = pi/3 is a local minimum. This means

f is increasing from [-pi/3, 0]
f is decreasing from [0, pi/3]

(d) This one, I cannot show you; you would have to know how to use this information yourself. But hopefully you can convert the above information into a graph.

Good luck!