Find all points on the graph of the function
f(x) = 2 cos x + cos^2 x
at which the tangent line is horizontal. (Use n as your arbitrary integer.)
smaller y-value (x,y)=
larger y-value (x,y)=
Oh, equations and tangents, you're really putting me to the test! But don't worry, I'm always up for a challenge. Let's find those points where the tangent line is horizontal!
To find where the tangent line is horizontal, we need to find where the derivative of the function is equal to zero. So, let's take the derivative of f(x) = 2cos(x) + cos^2(x).
f'(x) = -2sin(x) - 2cos(x)sin(x)
Now, set this derivative equal to zero and solve for x:
-2sin(x) - 2cos(x)sin(x) = 0
Sin(x)(-2 - 2cos(x)) = 0
Now, either sin(x) = 0 or -2 - 2cos(x) = 0.
1) If sin(x) = 0, then x = nπ, where n is an arbitrary integer.
2) If -2 - 2cos(x) = 0, then cos(x) = -1. Since the range of cosine is -1 to 1, we have two possible values for x: x = π + 2nπ and x = -π + 2nπ, where n is an arbitrary integer.
So, the points where the tangent line is horizontal are:
1) (x,y) = (nπ, f(nπ)), where n is an arbitrary integer
2) (x,y) = (π + 2nπ, f(π + 2nπ)) and (x,y) = (-π + 2nπ, f(-π + 2nπ)), where n is an arbitrary integer
I hope this clarifies things! Let me know if you need any more help, or if you need a joke to lighten the mood!
To find the points on the graph of the function where the tangent line is horizontal, we need to find the values of x for which the derivative of the function is equal to zero.
Step 1: Find the derivative of the function f(x) = 2 cos(x) + cos^2(x).
f'(x) = -2sin(x) - 2cos(x)sin(x)
Step 2: Set the derivative equal to zero and solve for x.
-2sin(x) - 2cos(x)sin(x) = 0
Factoring out sin(x), we get:
sin(x)(-2 - 2cos(x)) = 0
Setting each factor equal to zero, we have two cases:
Case 1: sin(x) = 0
This occurs when x = nπ, where n is an arbitrary integer.
Case 2: -2 - 2cos(x) = 0
Adding 2 to both sides and dividing by -2, we get:
2cos(x) = -2
cos(x) = -1
This occurs when x = (2n + 1)π, where n is an arbitrary integer.
Step 3: Plug in the values of x into the original function to find the corresponding y-values.
For x = nπ, the corresponding y-value is f(nπ) = 2cos(nπ) + cos^2(nπ)
The value of cos(nπ) alternates between 1 and -1 depending on whether n is even or odd. We can substitute this into the equation.
If n is even: cos(nπ) = 1, so f(nπ) = 2(1) + cos^2(nπ) = 2 + 1 = 3.
If n is odd: cos(nπ) = -1, so f(nπ) = 2(-1) + cos^2(nπ) = -2 + 1 = -1.
Therefore, for x = nπ, the corresponding y-values are either 3 or -1, depending on whether n is even or odd.
For x = (2n + 1)π, the corresponding y-value is f((2n + 1)π) = 2cos((2n + 1)π) + cos^2((2n + 1)π)
Since cos((2n + 1)π) = -1, we can substitute this into the equation:
f((2n + 1)π) = 2(-1) + cos^2((2n + 1)π) = -2 + 1 = -1.
Therefore, for x = (2n + 1)π, the corresponding y-value is -1.
Now, let's find the points with the smallest and largest y-values.
Smaller y-value (x, y):
For x = nπ, the y-value is 3.
Therefore, the point with the smallest y-value is (nπ, 3).
Larger y-value (x, y):
For x = (2n + 1)π, the y-value is -1.
Therefore, the point with the larger y-value is ((2n + 1)π, -1).
In summary,
Smaller y-value (x,y) = (nπ, 3)
Larger y-value (x,y) = ((2n + 1)π, -1)
where n is an arbitrary integer.
To find the points on the graph where the tangent line is horizontal, we need to find the values of x where the derivative of the function f(x) with respect to x is equal to zero.
First, let's find the derivative of the function f(x) = 2 cos x + cos^2 x with respect to x.
Taking the derivative of the first term, 2 cos x, gives us -2 sin x, as the derivative of cosine is negative sine.
Taking the derivative of the second term, cos^2 x, requires the chain rule. The derivative of cos^2 x is -2cos x * sin x, as the derivative of cosine is negative sine and we multiply it by the derivative of the inside function, which is sin x.
Now we can set the derivative equal to zero and solve for x:
-2 sin x - 2 cos x * sin x = 0
Factoring out sin x, we have:
sin x (-2 - 2 cos x) = 0
Now, we have two cases to consider:
Case 1: sin x = 0
This occurs when x is an integer multiple of π, that is, x = n * π, where n is an arbitrary integer.
Case 2: -2 - 2 cos x = 0
To solve this equation, we can rearrange it to get:
2 cos x = -2
Dividing both sides by 2, we have:
cos x = -1
This occurs when x is an odd multiple of π, that is, x = (2n + 1) * π/2, where n is an arbitrary integer.
Now, let's find the corresponding y-values for the points where the tangent line is horizontal.
For the smaller y-value, substitute the x-values from case 1 (x = n * π) into the function f(x) = 2 cos x + cos^2 x:
f(n * π) = 2 cos (n * π) + cos^2 (n * π)
Since cos (n * π) = 1 when n is even and cos (n * π) = -1 when n is odd, we can simplify the expression:
For even n: f(n * π) = 2 + 1 = 3
For odd n: f(n * π) = 2 * (-1) + 1 = -1
Therefore, the smaller y-value is 3 for even n and -1 for odd n.
For the larger y-value, substitute the x-values from case 2 (x = (2n + 1) * π/2) into the function f(x) = 2 cos x + cos^2 x:
f((2n + 1) * π/2) = 2 cos ((2n + 1) * π/2) + cos^2 ((2n + 1) * π/2)
Since cos ((2n + 1) * π/2) = 0, we can simplify the expression:
f((2n + 1) * π/2) = 2 * 0 + 0^2 = 0
Therefore, the larger y-value is 0 for all values of n.
In summary:
Smaller y-value (x, y) = (n * π, 3) for even n
Smaller y-value (x, y) = (n * π, -1) for odd n
Larger y-value (x, y) = ((2n + 1) * π/2, 0) for all values of n.
Since the derivative is the slope of the tangent line, we need
-2sinx - 2cosx*sinx = 0
-2sinx(1+cosx) = 0
sinx = 0 means x = nπ
cosx = -1 means x = (2n+1)π
so, f(x) has a horizontal tangent at x=nπ