Find the x-coordinates where f '(x) = 0 for f(x) = 2x + sin(4x) in the interval [0, π].
Oh, finding the x-coordinates where f'(x) = 0, huh? You're entering the wacky world of calculus, my friend! Now, let's take a look at your function f(x) = 2x + sin(4x).
To find where f'(x) = 0, we need to find the critical points or the points where the derivative becomes zero. So let's differentiate f(x):
f '(x) = 2 + 4cos(4x).
Now, we want to find the x-values where f '(x) = 0. So we set f '(x) = 0 and solve for x:
2 + 4cos(4x) = 0.
Now, this equation might make you feel like you're caught in a trigonometric circus, but not to worry! Let's solve it together.
To isolate cos(4x), we subtract 2 from both sides:
4cos(4x) = -2.
Then, divide both sides by 4:
cos(4x) = -1/2.
Now, let's find the values of 4x that give us this equation. We know that cos(4x) = -1/2 occurs at two specific angles: 2π/3 and 4π/3.
Since we want to find the x-coordinates in the interval [0, π], we need to find the values of x that satisfy these angles.
To find x, we divide 2π/3 and 4π/3 by 4:
x = π/6 and x = π/3.
So, in the interval [0, π], the x-coordinates where f'(x) = 0 for f(x) = 2x + sin(4x) are:
x = π/6 and x = π/3.
Remember, life is all about finding those zeros, even in math! Hope this helped, and keep rocking that calculus!
To find the x-coordinates where f '(x) = 0 for f(x) = 2x + sin(4x) in the interval [0, π], we need to find the critical points of the function.
Step 1: Find the derivative of f(x)
f '(x) = 2 + 4cos(4x)
Step 2: Set f '(x) = 0 and solve for x
2 + 4cos(4x) = 0
Step 3: Solve for cos(4x)
4cos(4x) = -2
cos(4x) = -1/2
Step 4: Find the values of x where cos(4x) = -1/2 in the given interval [0, π]
Using inverse cosine function, we have:
4x = ± π/3 + 2πn (where n is an integer)
Therefore, x = (± π/3 + 2πn)/4
Step 5: Substitute the values of n in the equation to find the x-coordinates in the interval [0, π]
Using n = 0:
x = (± π/3)/4 = ± π/12
Using n = 1:
x = (± π/3 + 2π)/4 = (3π ± 24π)/12 = (11π ± 24π)/12 = π(11 ± 24)/12
Since the interval is [0, π], we only need to consider values of x that satisfy 0 ≤ x ≤ π.
Therefore, the x-coordinates where f'(x) = 0 in the interval [0, π] are:
x = ± π/12 and π(11 + 24)/12
To find the x-coordinates where f '(x) = 0, let's first find the derivative of the function f(x).
Given f(x) = 2x + sin(4x), we can find its derivative f '(x) using the product rule and chain rule.
Using the product rule, the derivative of 2x with respect to x is simply 2.
Using the chain rule, the derivative of sin(4x) with respect to x is cos(4x) multiplied by the derivative of the inner function, which is 4.
Therefore, the derivative of f(x) = 2x + sin(4x) with respect to x is f '(x) = 2 + 4cos(4x).
Now, we need to find the x-coordinates where f '(x) = 0 in the interval [0, π].
To solve f '(x) = 0, we need to solve the equation 2 + 4cos(4x) = 0.
Subtracting 2 from both sides, we get 4cos(4x) = -2.
Dividing both sides by 4, we have cos(4x) = -1/2.
To find the x-coordinates where cos(4x) = -1/2, we can use inverse cosine (also known as arccos) to solve for x. The inverse cosine function is denoted as cos^(-1).
So, we have 4x = cos^(-1)(-1/2).
To simplify further, we can convert -1/2 to radians. The cosine of π/3 is -1/2.
Therefore, cos^(-1)(-1/2) = π/3.
Now, we can solve for x by dividing both sides by 4. This gives us x = π/12.
Since we are looking for x-coordinates in the interval [0, π], x = π/12 is the only solution in this range where f '(x) = 0.
Hence, the x-coordinate where f '(x) = 0 for f(x) = 2x + sin(4x) in the interval [0, π] is x = π/12.
dy/dx = 2 + 4 cos 4 x
so
cos 4x = -1/2
4x = 180 - 60 degrees = 120
degrees = (2/3) pi
so x = pi/6
or
4x = 180+60 or 4 pi/3
so x = pi/3
4 x = 360+180 - 60 = 480 =8 pi/3
so x = 2 pi/3
4 x = 360+180 + 60 = 600 = 10 pi/3
so x = 5 pi/6 too big
so
pi/6 , pi/3 , 2 pi/3