Y=2sinx-cos^2(x)

F"(x)=-sinx+cos^2-sin^2(x)

Find the points of inflection and concavities

well, you remember that

inflection is where y" = 0
concave up when y" > 0

For y" I get 2 times your answer, but that doesn't change its properties.
y' = 2cosx(sinx+1)
y" = -sinx + cos^2x - sin^2x
= -sinx + 1 - 2sin^2x
= -(2sinx-1)(sinx+1)

Note that x = 3pi/2 is not a point of inflection, since y' is also zero there.

Now you can clearly see where y"=0 or is positive/negative. Compare that against the graph of y:

http://www.wolframalpha.com/input/?i=+2sinx-cos^2%28x%29

To find the points of inflection and concavities of the function f(x) = 2sin(x) - cos^2(x), we need to determine where the concavity changes by identifying the points where the second derivative, f''(x), changes sign.

First, let's find the second derivative of f(x). We already have the expression F''(x) = -sin(x) + cos^2(x) - sin^2(x).

To find where f''(x) changes sign, we set f''(x) equal to zero and solve for x.

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

Rearrange the equation:

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

Combine like terms:

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

Now, we have a quadratic equation in terms of sin(x). Let's write sin^2(x) as 1 - cos^2(x).

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

simplify:

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

Rearrange the equation:

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

Now, we have a quadratic equation in terms of cos(x). Let's solve this equation to find the values of cos(x) where f''(x) changes sign.

We can use the quadratic formula to solve for cos(x):

cos(x) = (-b ± √(b^2 - 4ac)) / 2a

In this case, a = 2, b = -1, and c = -1. Plugging these values into the quadratic formula gives us:

cos(x) = (1 ± √(1 + 4*2)) / (2*2)

cos(x) = (1 ± √9) / 4

cos(x) = (1 ± 3) / 4

cos(x) = 1/2 or cos(x) = -1

Now, let's find the corresponding values of x for each solution:

For cos(x) = 1/2:

x = arccos(1/2)

Using the unit circle or a calculator, we find:

x = π/3 or x = 5π/3

For cos(x) = -1:

x = arccos(-1)

Again, using the unit circle or a calculator, we find:

x = π

Now, we have three critical points where f''(x) changes sign: x = π/3, x = 5π/3, and x = π.

To determine the concavities at these points, we need to evaluate the second derivative f''(x) for values of x in the intervals between these critical points:

Let's choose the interval (0, π/3):

Pick a test point, for example, x = π/6:

f''(π/6) = -sin(π/6) + cos^2(π/6) - sin^2(π/6)

Using trigonometric identities, sin(π/6) = 1/2 and cos(π/6) = √3/2:

f''(π/6) = -(1/2) + (√3/2)^2 - (1/2)^2

f''(π/6) = -(1/2) + 3/4 - 1/4

f''(π/6) = 0

Since f''(x) = 0 for x = π/6, the concavity does not change on this interval.

Now, let's choose the interval (π/3, π).

Pick a test point, for example, x = π/2:

f''(π/2) = -sin(π/2) + cos^2(π/2) - sin^2(π/2)

Using trigonometric identities, sin(π/2) = 1 and cos(π/2) = 0:

f''(π/2) = -(1) + (0)^2 - (1)^2

f''(π/2) = -1

Since f''(x) = -1 for x = π/2, the concavity changes from concave up to concave down on this interval.

Finally, let's choose the interval (π, 5π/3):

Pick a test point, for example, x = (7π/6):

f''(7π/6) = -sin(7π/6) + cos^2(7π/6) - sin^2(7π/6)

Using trigonometric identities, sin(7π/6) = -1/2 and cos(7π/6) = -√3/2:

f''(7π/6) = -(-1/2) + (-√3/2)^2 - (-1/2)^2

f''(7π/6) = 1/2 + 3/4 - 1/4

f''(7π/6) = 0

Since f''(x) = 0 for x = 7π/6, the concavity does not change on this interval either.

Therefore, the points of inflection occur at x = π/2 (concave up to concave down) and x = 5π/3 (concave down to concave up), and the concavity of the function f(x) = 2sin(x) - cos^2(x) doesn't change in the intervals (0, π/3) and (π, 5π/3).

In summary, the points of inflection are (π/2, -1) and (5π/3, -1), and the concavity is as follows: concave up on (0, π/2) and (5π/3, 2π) and concave down on (π/2, 5π/3).