For what values of x does the graph of f have a horizontal tangent? (Use n as your integer variable. Enter your answers as a comma-separated list.)
f(x) = x − 2 sin x
horizontal tangent means slope = 0
f'(x) = 1 - 2cosx
1-2cosx=0
cosx = 1/2
x = 2kpi +/- pi/3
To find the values of x for which the graph of f(x) = x - 2sin(x) has a horizontal tangent, we need to determine when the derivative of f(x) equals zero.
Step 1: Find the derivative of f(x):
f'(x) = 1 - 2cos(x)
Step 2: Set the derivative equal to zero and solve for x:
1 - 2cos(x) = 0
Step 3: Solve for cos(x):
2cos(x) = 1
cos(x) = 1/2
Step 4: Find the values of x:
We know that cosine is equal to 1/2 at π/3 and 5π/3:
For π/3, x = π/3 + 2nπ, where n is an integer.
For 5π/3, x = 5π/3 + 2nπ, where n is an integer.
In summary, the values of x for which the graph of f(x) = x - 2sin(x) has a horizontal tangent are x = π/3 + 2nπ and x = 5π/3 + 2nπ, where n is an integer.
To find the values of x for which the graph of f(x) = x - 2sin(x) has a horizontal tangent, we need to find the points where the derivative of f(x) is equal to zero.
The derivative of f(x) can be found by applying the derivative rules:
f'(x) = 1 - 2cos(x)
To find the values of x where the derivative is equal to zero, we need to solve the equation f'(x) = 0:
1 - 2cos(x) = 0
To isolate the cosine term, we can rearrange the equation:
2cos(x) = 1
Now, we divide both sides of the equation by 2:
cos(x) = 1/2
To find the values of x where the cosine of x is equal to 1/2, we can look at the unit circle or use the inverse cosine function. Since the inverse cosine function gives values between 0 and π, we need to find the solutions within that range.
The solutions to cos(x) = 1/2 are x = π/3 + 2nπ and x = 5π/3 + 2nπ, where n is an integer.
Therefore, the values of x for which the graph of f has a horizontal tangent are x = π/3 + 2nπ and x = 5π/3 + 2nπ, where n is any integer.
In other words, for any integer n, the values of x that give a horizontal tangent are: π/3 + 2nπ and 5π/3 + 2nπ.