Find the critical numbers of the function. (Use n to denote any arbitrary integer values.)
f(θ) = 12 cos θ + 6 sin2θ
My answer was pi/2 and 3pi/2, but those were wrong. I don't understand the 'n' part, either.
Ah, finding critical numbers can be a bit tricky, but fear not, I'm here to help! In this case, we need to find the values of θ where the derivative of the function f(θ) is equal to zero or is undefined. Let's break it down step by step.
To find the critical numbers, we first need to find the derivative of f(θ). Do you need help with that?
To find the critical numbers of the function f(θ) = 12 cos θ + 6 sin^2θ, we need to take the derivative of the function and set it equal to zero. The critical numbers occur when the derivative is zero or undefined.
Taking the derivative of f(θ) with respect to θ, we have:
f'(θ) = -12 sin θ + 12 sin θ cos θ
Setting f'(θ) equal to zero, we have:
-12 sin θ + 12 sin θ cos θ = 0
Factoring out 12sin θ, we get:
12 sin θ (1 - cos θ) = 0
This equation is satisfied when either 12 sin θ = 0 or 1 - cos θ = 0.
For 12 sin θ = 0, θ can be either zero or any multiple of π since sin θ = 0 for those values.
For 1 - cos θ = 0, we solve for θ as follows:
1 - cos θ = 0
cos θ = 1
θ = 2nπ where n is an arbitrary integer
So, the critical numbers are θ = 0, θ = 2nπ, and any multiple of π.
Therefore, the correct critical numbers for the function f(θ) = 12 cos θ + 6 sin^2θ are θ = 0 and θ = 2nπ, where n is an arbitrary integer.
To find the critical numbers of the function f(θ) = 12 cos θ + 6 sin²θ, we need to find the values of θ where the derivative of the function equals zero or is undefined.
First, let's find the derivative of f(θ). The derivative of cos θ is -sin θ, and the derivative of sin²θ can be found using the chain rule:
f'(θ) = -12 sin θ + 6 * 2sinθ * cosθ
= -12 sin θ + 12sinθ * cosθ
To find the critical numbers, we set f'(θ) equal to zero and solve for θ:
-12 sin θ + 12sinθ * cosθ = 0
Factoring out 12sinθ, we have:
12 sin θ (1 - cosθ) = 0
To find the values of θ that make this equation true, we divide both sides by 12 sin θ:
1 - cosθ = 0
Solving for cosθ, we find:
cosθ = 1
Since cosθ equals 1 when θ = 0, we have found one critical number.
To understand the "n" part, it indicates that there may be multiple solutions. In this case, cosθ = 1 has infinitely many solutions since it repeats every 2π. Therefore, we express all the critical numbers using the "n" notation.
For the equation cosθ = 1, the solutions can be written as:
θ = 2πn
So, the critical numbers of the function f(θ) = 12 cos θ + 6 sin²θ are θ = 2πn, where n represents any integer value.
Please note that if you were specifically asked to find the critical numbers within a certain interval or restricted range, you would need to further analyze the function accordingly.
f'(Ø) = -12sinØ + 6cos2Ø
= 0 for max/min
I will use x instead of Ø , easier to type
12sinx = 6cos 2x
2sinx = cos 2x
but cos 2x = 1 - 2sin^2 x
2sinx = 1 - 2sin^2 x
2sin^2 x + 2sinx - 1 = 0
sinx = (-2 ± √12)/4
= (-1 ± √3)/2
sinx = (-1+√3)/2 OR sinx = (-1-√3)/2
the last part is not possible since -1<sinx<1
set your calculator to radians to get
x = .3747 or x = π-.3747 = -2.7669
plug those values into the original function to get the max and min values
see:
http://www.wolframalpha.com/input/?i=plot+f%28%CE%B8%29+%3D+12+cos+%CE%B8+%2B+6+sin2%CE%B8+
since were solving a sin(x) function equation and the period is 2π, adding or subtracting multiples of 2π yields more answers.
switching back to Ø
Ø = .3747 + 2nπ OR Ø = -2.7669 + 2nπ , where n is any integer
will yield
Some people also consider the x-intercepts as "critical values"
12 cos θ + 6 sin2θ = 0
2cosØ + sin 2Ø = 0
2cosØ + 2sinØcosØ = 0
cosØ(1 + sinØ) = 0
cosØ = 0 or sinØ = -1
if cosØ =0 , Ø = π/2 or 3π/2
again, the period of cosØ is 2π
so, if cosØ = 0
Ø = π/2 + 2nπ OR Ø = 3π/2 + 2nπ , where n is an integer
if sinØ = -1 , Ø = 3π/2 yielding the same general solution as above