6x-4cos(3X)Critical Numbers
To find the critical numbers of the function f(x) = 6x - 4cos(3x), we need to find the values of x where the derivative of the function equals zero or is undefined. Here are the steps:
Step 1: Find the derivative of f(x):
f'(x) = 6 - 4(-3sin(3x)) = 6 + 12sin(3x)
Step 2: Set the derivative equal to zero and solve for x:
6 + 12sin(3x) = 0
Step 3: Solve for sin(3x):
12sin(3x) = -6
sin(3x) = -6/12
sin(3x) = -1/2
Step 4: Solve for 3x:
In order to find the solution for sin(3x) = -1/2, we need to determine the angles in the unit circle where the sine function equals -1/2. Those angles are -π/6 and -5π/6.
So, we have two possible solutions for 3x:
3x = -π/6 and/or 3x = -5π/6
Step 5: Solve for x:
Divide both sides of the equation by 3:
x = -π/18 and/or x = -5π/18
Therefore, the critical numbers of the function f(x) = 6x - 4cos(3x) are x = -π/18 and x = -5π/18.
To find the critical numbers of the function 6x - 4cos(3x), we need to follow a few steps:
Step 1: Find the derivative of the function.
The derivative of 6x with respect to x is simply 6.
The derivative of cos(3x) with respect to x can be found using the chain rule. The derivative of cos(u), where u is a function of x, is -sin(u) times the derivative of u with respect to x. In this case, u = 3x, so the derivative of cos(3x) with respect to x is -3sin(3x).
Therefore, the derivative of the whole function 6x - 4cos(3x) is 6 - 4(-3sin(3x)), which simplifies to 6 + 12sin(3x).
Step 2: Set the derivative equal to zero and solve for x.
To find the critical numbers, we set the derivative equal to zero and solve for x:
6 + 12sin(3x) = 0
Step 3: Solve for x.
To solve this equation, we isolate the sin(3x) term by subtracting 6 from both sides:
12sin(3x) = -6
Then, divide both sides by 12:
sin(3x) = -0.5
Step 4: Find the angle(s) whose sine is -0.5.
To find the values of x for which sin(3x) = -0.5, we can use the inverse sine function (also known as arcsine or sin^(-1)).
Using a calculator or a table of trigonometric values, we can find that sin^(-1)(-0.5) is approximately -30 degrees or -π/6 radians.
Step 5: Solve for x.
We want to find the values of x, not 3x. The relationship between them is x = (1/3)θ, where θ is the angle in radians.
For θ = -π/6 radians, x = (1/3)(-π/6) = -π/18.
For θ = -30 degrees, x = (1/3)(-30) = -10.
Therefore, the critical numbers of the function 6x - 4cos(3x) are x = -π/18 and x = -10.
There is no function here. Is there an equal sign somewhere here? A function? What is your question?
presumably you meant
y = 6x - 4cos(3x)
y' = 6 + 12sin(3x)
y'' = 36cos(3x)
Now, I'm sure you know all about 1st and second derivatives and how they relate to critical numbers:
y has a max/min where y' = 0 and y'' not zero
y has an inflection point where y'' = 0
y is increasing where y' >. 0
y is concave up when y'' > 0
No go plug and chug