find the points(in exact form) at which the tangent is horizontal on the given interval:
y=cos(2x)+2cos(x) , [0,2pi)
To find the points at which the tangent is horizontal, we need to find where the derivative of the function equals zero.
First, we find the derivative of the function:
y' = -2sin(2x) - 2sin(x)
Next, we set the derivative equal to zero and solve for x:
-2sin(2x) - 2sin(x) = 0
Factoring out a -2sin(x), we get:
-2sin(x)(cos(x) + cos(2x)) = 0
Setting each factor equal to zero, we have:
-2sin(x) = 0 cos(x) + cos(2x) = 0
From the first equation, sin(x) = 0. This occurs when x = 0, pi, and 2pi.
From the second equation, cos(x) + cos(2x) = 0. This equation is a bit more complicated to solve, but we can use the double-angle identities to simplify it. Using the identity cos(2x) = 2cos^2(x) - 1, we can rewrite the equation as:
cos(x) + 2cos^2(x) - 1 = 0
Rearranging the terms, we have:
2cos^2(x) + cos(x) - 1 = 0
Factoring the quadratic equation, we get:
(2cos(x) - 1)(cos(x) + 1) = 0
Setting each factor equal to zero, we have two possible solutions:
2cos(x) - 1 = 0 cos(x) + 1 = 0
Solving the first equation, we get:
2cos(x) = 1
cos(x) = 1/2
This occurs when x = pi/3 and x = 5pi/3.
Solving the second equation, we get:
cos(x) = -1
This occurs when x = pi.
Therefore, the points on the interval [0, 2pi) at which the tangent is horizontal are x = 0, x = pi/3, x = pi, and x = 5pi/3. To find the corresponding y-values, we substitute these x-values back into the original function:
For x = 0: y = cos(2(0)) + 2cos(0) = 1 + 2 = 3
For x = pi/3: y = cos(2(pi/3)) + 2cos(pi/3) = 1/2 + 1 = 3/2
For x = pi: y = cos(2(pi)) + 2cos(pi) = 1 + (-2) = -1
For x = 5pi/3: y = cos(2(5pi/3)) + 2cos(5pi/3) = 1/2 + 1 = 3/2
Therefore, the points at which the tangent is horizontal on the interval [0, 2pi) are (0, 3), (pi/3, 3/2), (pi, -1), and (5pi/3, 3/2).