f(x) = 6(cos(x))^2 - 12sin(x)
0 ¡Ü x ¡Ü 2(PI)
I have to find where the interval is increasing, decreasing, local mins and maxs, inflection points, and concave up and down. I understand how to do these type of problems. I am just getting messed up with the sin, cos, and pi stuff. So far I have
f'(x)= -12cos(x)sin(x)-12cos(x) and i have the critical numbers being -pi/2 and pi/2. and i am lost from there...
that should be less than or equal to for both. 0 (<=) x (<=) 2pi
To find where the interval is increasing or decreasing, local minima and maxima, inflection points, concave up and down, we need to analyze the first and second derivatives of the function.
First, let's start with finding the critical points and intervals of increasing and decreasing:
1. To find the critical points, set f'(x) = 0 and solve for x.
Given f'(x) = -12cos(x)sin(x)-12cos(x), we need to solve the equation:
-12cos(x)sin(x)-12cos(x) = 0
factoring out -12cos(x):
-12cos(x)(sin(x) + 1) = 0
Setting each factor equal to zero:
-12cos(x) = 0 => cos(x) = 0
sin(x) + 1 = 0 => sin(x) = -1
Finding the solutions:
For cos(x) = 0, we have x = (n + 1/2)π, where n is an integer.
So, x = (n + 1/2)π will be our critical points.
In the given interval, 0 ≤ x ≤ 2π, we have two critical points: x = π/2 and x = 3π/2.
2. To determine where the function is increasing or decreasing, we need to evaluate the sign of the first derivative for different intervals.
Choose test points within each interval:
Interval 1: 0 ≤ x < π/2 (Test point x₀ = 0)
Interval 2: π/2 < x < 3π/2 (Test point x₁ = π)
Interval 3: 3π/2 < x ≤ 2π (Test point x₂ = 2π)
Evaluate the sign of f'(x) at these test points:
f'(x₀) = -12cos(0)sin(0) - 12cos(0) = 0 - 12(1) = -12
f'(x₁) = -12cos(π)sin(π) - 12cos(π) = 0 - 12(-1) = 12
f'(x₂) = -12cos(2π)sin(2π) - 12cos(2π) = 0 - 12(0) = 0
From this, we can determine the intervals:
Interval 1: f'(x) < 0 (Decreasing)
Interval 2: f'(x) > 0 (Increasing)
Interval 3: f'(x) = 0 (Not increasing or decreasing)
Next, let's find the local minima and maxima:
To find the local minima and maxima, we need to check the behavior around the critical points.
1. We have two critical points: x = π/2 and x = 3π/2.
2. Evaluate the second derivative, f''(x), at these critical points.
f''(π/2) = -12cos(π/2) = -12(0) = 0
f''(3π/2) = -12cos(3π/2) = -12(0) = 0
When the second derivative is zero, it does not determine the concavity.
Additional tests are needed to determine whether they are maxima, minima, or neither.
3. Apply the second derivative test:
Choose test points within each interval:
Interval 1: 0 < x < π/2 (Test point x₃ = π/4)
Interval 2: π/2 < x < 3π/2 (Test point x₄ = 5π/4)
Evaluate the sign of f''(x) at these test points:
f''(x₃) = -12cos(π/4) = -6√2 < 0
f''(x₄) = -12cos(5π/4) = 6√2 > 0
From this, we can conclude:
Local maximum at x = π/2
Local minimum at x = 3π/2
Now, let's find the inflection points and determine concavity:
1. To find the inflection points, we need to determine where the second derivative changes sign or equals zero.
f''(x) = -12cos(x)
2. Solve f''(x) = 0:
-12cos(x) = 0 => cos(x) = 0
The solutions for cos(x) = 0 are x = (n + 1/2)π, where n is an integer.
In the given interval, 0 ≤ x ≤ 2π, we have two solutions: x = π/2 and x = 3π/2.
These are the potential inflection points.
3. Determine the concavity by evaluating the sign of f''(x) on each side of the potential inflection points.
Interval 1: 0 < x < π/2 (Test point x₅ = π/4)
Interval 2: π/2 < x < 3π/2 (Test point x₆ = 5π/4)
Evaluate the sign of f''(x) at these test points:
f''(x₅) = -12cos(π/4) = -6√2 < 0 (Concave down)
f''(x₆) = -12cos(5π/4) = 6√2 > 0 (Concave up)
From this, we can conclude:
Inflection point at x = π/2 (Concave down to concave up)
Inflection point at x = 3π/2 (Concave up to concave down)
In summary:
- The interval 0 ≤ x ≤ 2π has three intervals of increasing and decreasing:
- Increasing: 0 ≤ x < π/2, π/2 < x < 3π/2
- Decreasing: 3π/2 < x ≤ 2π
- Local minimum: x = 3π/2
- Local maximum: x = π/2
- Inflection point: x = π/2 (Concave down to concave up), x = 3π/2 (Concave up to concave down)