Find all points on the graph of the function f(x) = 2 cos(x) + (cos(x))2 at which the tangent line is horizontal. Consider the domain x = [0,2π).

I have pi/2 and 3pi/2 for x values. But when I plug them I get zero. Is this correct as y values or am I solving it wrong? I took the derivative and set it equal to zero. But I believe I am wrong. Help urgent!

well, from the graph at

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

I'd say you were wrong.

f = 2cosx + cos^2(x)
f' = -2sinx - 2sinx cosx
= -2sinx(1+cosx)
f'=0 where sinx=0 or cosx = -1
x = 0 or pi

To find the points on the graph of the function where the tangent line is horizontal, you need to find the x-values where the derivative of the function is equal to zero.

First, let's find the derivative of the function f(x) = 2cos(x) + (cos(x))^2. We can find the derivative using the chain rule.

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

Next, we set the derivative equal to zero and solve for x:

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

Factor out sin(x):

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

Now, we have two possibilities for the derivative to be zero:

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

For the first case, sin(x) = 0 has solutions at x = 0, π, and 2π.

For the second case, -2 - 2cos(x) = 0, we can solve for cos(x):

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

This equation is satisfied when x = π.

So, the possible x-values where the tangent line is horizontal are x = 0, π, and 2π.

Now, we need to find the y-values corresponding to these x-values. Plug these x-values back into the original function f(x) = 2cos(x) + (cos(x))^2.

For x = 0:
f(0) = 2cos(0) + (cos(0))^2 = 2 + 1 = 3

For x = π:
f(π) = 2cos(π) + (cos(π))^2 = -2 + 1 = -1

For x = 2π:
f(2π) = 2cos(2π) + (cos(2π))^2 = 2 + 1 = 3

So, the points on the graph of the function where the tangent line is horizontal are (0, 3) and (2π, 3).