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.
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
Sure, let's go through the remaining parts step-by-step:
x-intercepts:
To find the x-intercepts, we need to solve the equation y = 2sin(x) - cos^2(x) = 0. This means that 2sin(x) - cos^2(x) = 0. Rearranging the equation, we have cos^2(x) = 2sin(x). Since cos^2(x) is always positive and sin(x) is always between -1 and 1, the equation cos^2(x) = 2sin(x) can only be true if sin(x) is either 1 or -1. Therefore, the x-intercepts occur when sin(x) = 1 or sin(x) = -1. The solutions to these equations are x = pi/2 and x = 3pi/2 respectively.
Symmetry:
To determine the symmetry of the function, we can check if it is an even or odd function. An even function is symmetric about the y-axis, implying that f(-x) = f(x). On the other hand, an odd function is symmetric about the origin, indicating that f(-x) = -f(x). To determine if a function is even or odd, we need to check if replacing x with -x in the given function y = 2sin(x) - cos^2(x) results in the same function or its negative. Let's test this:
f(-x) = 2sin(-x) - cos^2(-x) = -2sin(x) - cos^2(x)
Since f(-x) = -f(x), the function is odd.
Extrema:
To find the extrema of a function, we need to locate the points at which the function reaches a maximum or minimum value. We have already found the critical points, which are x = pi/2, x = pi, and x = 3pi/2. To determine the nature of these critical points (whether they are maxima or minima), we can use the Second Derivative Test. The second derivative of the function is f''(x) = -2sin(x) + 2cos^2(x) - 2sin^2(x). Evaluate the second derivative at the critical points and check the sign of f''(x).
f''(pi/2) = -2sin(pi/2) + 2cos^2(pi/2) - 2sin^2(pi/2) = -2 + 2(0) - 2(1) = -4
Since f''(pi/2) is negative, it suggests that there is a relative maximum at x = pi/2.
By applying the same process, you can determine the nature of the remaining critical points.
Possible Points of Inflection:
To find the possible points of inflection, we need to locate the points where the concavity of the function changes. A point of inflection occurs when the second derivative changes sign.
Set the second derivative f''(x) = 0 and solve for x to find the potential points of inflection. The equation to solve is:
-2sin(x) + 2cos^2(x) - 2sin^2(x) = 0
You can use the unit circle to find the values of x that satisfy this equation.
Concave Up and Concave Down:
To determine where the function is concave up or down, we need to examine the concavity of the function. The concavity changes at points of inflection. You can use the second derivative f''(x) to determine the sign of the second derivative for different intervals.
If f''(x) > 0, the function is concave up, and if f''(x) < 0, the function is concave down. Use the critical points and the points of inflection to divide the x-axis into intervals, and then analyze the sign of f''(x) in each interval.
Points of Inflection:
Points of inflection occur where the concavity changes, which means at those points, the function changes from being concave up to concave down or vice versa. To find the points of inflection, we need to find the x-values where the second derivative changes sign. These values correspond to the points of inflection.
I hope this explanation helps clarify the remaining parts of the function analysis. If you have any further questions, feel free to ask!