Find the x-coordinates where f '(x) = 0 for f(x) = 2x + sin(2x) in the interval [0, 2π].
cos(3π) is also -1, so x = 3π/2
f(x) = 2x + sin(2x)
f ' (x) = 2 + 2cos(2x)
= 0
2cos(2x) = -2
cos 2x = -1
I know cos π = -1
so 2x = π
x = π/2
To find the x-coordinates where f '(x) = 0 for the function f(x) = 2x + sin(2x) in the interval [0, 2π], we need to find the critical points of the function. The critical points occur where the derivative of the function equals zero.
Step 1: Find the derivative of the function.
Let's find the derivative of f(x) = 2x + sin(2x).
f '(x) = d/dx [2x + sin(2x)]
Using the sum rule for derivatives, we differentiate each term separately.
f '(x) = d/dx [2x] + d/dx [sin(2x)]
The derivative of 2x is simply 2, as the derivative of x is 1.
f '(x) = 2 + d/dx [sin(2x)]
Using the chain rule, we differentiate the sine function.
f '(x) = 2 + cos(2x) * d/dx [2x]
The derivative of 2x is 2, as the derivative of x is 1.
f '(x) = 2 + 2cos(2x)
Step 2: Set the derivative equal to zero and solve for x.
To find the x-coordinates where f '(x) = 0, we set 2 + 2cos(2x) = 0.
2cos(2x) = -2
cos(2x) = -1
The cosine function equals -1 at π radians.
Step 3: Solve for x within the given interval.
Since the interval is [0, 2π], we need to find all values of x between 0 and 2π that satisfy cos(2x) = -1.
At π/2 radians, cos(2(π/2)) = cos(π) = -1.
At 3π/2 radians, cos(2(3π/2)) = cos(3π) = -1.
So, we have two solutions within the given interval: x = π/2 and x = 3π/2.
Therefore, the x-coordinates where f '(x) = 0 for the function f(x) = 2x + sin(2x) in the interval [0, 2π] are x = π/2 and x = 3π/2.
To find the x-coordinates where f'(x) = 0 for f(x) = 2x + sin(2x) in the interval [0, 2π], we need to first find the derivative of f(x) and then solve for x when the derivative equals zero.
Let's find the derivative of f(x):
f(x) = 2x + sin(2x)
To find the derivative, we can use the sum rule and chain rule:
f'(x) = d/dx(2x) + d/dx(sin(2x))
= 2 + (cos(2x))(d/dx(2x))
= 2 + 2(cos(2x))
Now, let's solve the equation f'(x) = 0:
2 + 2(cos(2x)) = 0
Subtract 2 from both sides:
2(cos(2x)) = -2
Divide both sides by 2:
cos(2x) = -1
In the interval [0, 2π], we need to find all x-values that make cos(2x) equal to -1.
Recall that the cosine function is equal to -1 when the argument is π radians plus a multiple of 2π. Therefore, we can write:
2x = π + 2nπ, where n is an integer
To find x, divide both sides by 2:
x = π/2 + nπ, where n is an integer
Now we need to find the x-values in the interval [0, 2π]. Plug in n = 0, 1, 2, and so on, until we exceed the upper limit of 2π.
For n = 0, we have:
x = π/2 + 0π = π/2
For n = 1, we have:
x = π/2 + 1π = 3π/2
For n = 2, we have:
x = π/2 + 2π = 5π/2
Since 5π/2 exceeds the upper limit of 2π, we stop here.
Therefore, the x-coordinates where f'(x) = 0 in the interval [0, 2π] are x = π/2 and x = 3π/2.