For the function y=2sin(x)-cos^2(x) on [0, 2pi] find the following:

Domain
x and y intercepts
Vertical asymptotes
Horizontal asymptotes
Symmetry
F'(x)
Critical numbers
Increasing f(x)
Decreasing f(x)
Extrema
F"(x)
Possible points of inflection
Concave up
Concave down
Points of inflection

I got some:
Domain: [0, 2pi]
y-intercept: y=-1
Vertical asymptotes: none
Horizontal asymptotes: none
F'(x): 2cosx+2cosxsinx
Critical numbers: x=pi/2, x=pi, x=(3pi)/2
Increasing f(x): (0, pi/2) and ((3pi)/2, 2pi)
Decreasing f(x): (pi/2, pi) and (pi, (3pi)/2). Is that possible if the intervals are right next to each other?
Extrema: absolute max f(0)=-1 and f(2pi)=-1, absolute min/relative min f((3pi)/2)=-2 and f(pi/2)=-2
F"(x): -2sinx+2cos^2(x)-2sin^2(x)

To make it clearer, I can't figure out
x-intercept
Symmetry
Unsure about extrema
Possible points of inflection
Concave up
Concave down
Points of inflection

I appreciate your time very much. Thank you.

3 answers

lets look at your graph first of all
http://www.wolframalpha.com/input/?i=plot+y%3D2sin%28x%29-cos%5E2%28x%29

clearly no vertical or horizontal asymptotes
you had that,
your y-intercept is correct
symmetry: about each vertical line passing through a max or a min

I disagree with your critical numbers,
2cosx + 2cosxsinx = 0
2cosx(1 + sinx) = 0
cosx = 0 or sinx = -1
x = π/2 or x = 3π/2
(you had an extra x = 0, which you can see from the graph and my solution cannot be)

points of inflection:
f '' (x) = -2sinx + 2cosx(cosx) + 2sinx(-sinx)
= -2sinx + 2cos^2 x - 2sin^2 x
= 0 for pts of inflection

-sinx + cos^2 x - sin^2 x = 0
-sinx + (1 - sin^2 x) - sin^2 x = 0
2sin^2 x + sinx - 1 = 0
(2sinx - 1)(sinx + 1) = 0
sinx = 1/2 or sinx = -1
x = π/6 or 5π/6 or 3π/2

looking at the critical values you should be able to discern where the function is increasing and decreasing, and where it is concave up or down

Thank you so much.

Thank you