Find the x-coordinates where f '(x) = 0 for f(x) = 2x + sin(4x) in the interval [0, π].
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To find the x-coordinates where f '(x) = 0 for f(x) = 2x + sin(4x) in the interval [0, π], we need to take the derivative of f(x) and set it equal to zero.
First, let's find the derivative of f(x). The derivative of 2x is 2, and the derivative of sin(4x) is 4cos(4x) using the chain rule. Therefore, the derivative of f(x) is:
f '(x) = 2 + 4cos(4x)
Now, we need to set f '(x) = 0 and solve for x. We have:
2 + 4cos(4x) = 0
Subtracting 2 from both sides:
4cos(4x) = -2
Dividing by 4:
cos(4x) = -1/2
Now, we need to find the values of x in the interval [0, π] that satisfy this equation. The cosine function has a value of -1/2 at two points in the interval [0, 2π], which are π/3 and 5π/3. However, we are interested in the interval [0, π], so we will only consider π/3 as a possible solution.
To find the x-coordinates in the interval [0, π], we substitute π/3 into the original function f(x) = 2x + sin(4x):
f(π/3) = 2(π/3) + sin(4(π/3))
= (2π/3) + sin(4π/3)
= (2π/3) + sin(4π/3 - 2π)
= (2π/3) + sin(4π/3 - 6π/3)
= (2π/3) + sin(-2π/3)
= (2π/3) - sin(2π/3)
= (2π/3) - sqrt(3)/2
Therefore, the x-coordinate where f '(x) = 0 in the interval [0, π] is π/3, and the corresponding y-coordinate is (2π/3) - sqrt(3)/2.