Given the function y = 2cos(x) - 1, use Calculus methods to determine the:
a) intervals of increase and decrease
b) local max and mins
c) intervals of concavity
d) inflection points
my answers:
a) decrease: [0,pi) increase: (pi,2pi]
b) local min: (pi,-3)
c) concave down:(-infinity, pi/2] U [3pi/2, infinity)
concave up: (pi/2, 3pi/2)
d) (pi/2,-1)
(3pi/2,-1)
All correct except:
for b) what about the local max's ?
d) disagree with your concavity properties. Look at the initial sketch of your graph
I assume you wanted everything in the domain from 0 to 2π
Then I see concave down for 0 < x < π/2 as well as 3π/2 < x < 2π
concave up for π/2 < x < 3π/2
y = 2 cos(x) - 1
y' = -2 sin x positive slope from pi to 2 pi agree, 0 at 0 and pi
y" = - 2 cos x positive (holds water) from pi/2 to 3 pi/2
agree so far, but sketch a graph
To determine the intervals of increase and decrease, local max and min points, intervals of concavity, and inflection points of the function y = 2cos(x) - 1, we can use calculus methods.
a) Intervals of Increase and Decrease:
To find the intervals of increase and decrease, we need to determine where the derivative of the function is positive (increasing) or negative (decreasing).
1. Find the derivative of the function:
dy/dx = -2sin(x)
2. Set the derivative equal to zero to find critical points:
-2sin(x) = 0
To solve this equation, we set sin(x) = 0, which occurs at x = nπ, where n is an integer.
The critical points are x = 0, x = π, and x = 2π.
3. Test the intervals between the critical points and beyond to determine the intervals of increase and decrease.
For x ≤ 0:
Choose a test point, like x = -1, and substitute it into the derivative:
dy/dx = -2sin(-1) = 2sin(1)
Since sin(1) is positive, the derivative is positive (increasing). Thus, the function is increasing on the interval (-∞, 0].
For 0 < x < π:
Choose a test point, like x = 1, and substitute it into the derivative:
dy/dx = -2sin(1)
Since sin(1) is positive, the derivative is negative (decreasing). Thus, the function is decreasing on the interval (0, π).
For π < x < 2π:
Choose a test point, like x = 3, and substitute it into the derivative:
dy/dx = -2sin(3)
Since sin(3) is negative, the derivative is positive (increasing). Thus, the function is increasing on the interval (π, 2π].
Therefore, the intervals of decrease are [0, π) and the intervals of increase are (π, 2π].
b) Local Max and Min:
To find local maximum and minimum points, we need to examine the critical points and the endpoints of the intervals.
The critical points are x = 0, x = π, and x = 2π.
For x = 0:
Evaluate the function at x = 0:
y = 2cos(0) - 1 = 2(1) - 1 = 1
No other points are closer to 0 than this value for this interval. Therefore, (0, 1) is a local maximum.
For x = π:
Evaluate the function at x = π:
y = 2cos(π) - 1 = 2(-1) - 1 = -3
No other points are closer to π than this value for this interval. Therefore, (π, -3) is a local minimum.
No other critical points have local maximums or minimums.
c) Intervals of Concavity:
To find intervals of concavity, we need to examine the second derivative of the function.
1. Find the second derivative of the original function:
d²y/dx² = -2cos(x)
2. Set the second derivative equal to zero to find possible inflection points:
-2cos(x) = 0
To solve this equation, we set cos(x) = 0, which occurs at x = (n+1/2)π, where n is an integer.
The possible inflection points are x = π/2 and x = 3π/2.
3. Test the intervals between the possible inflection points and beyond to determine the intervals of concavity.
For x < π/2:
Choose a test point, like x = 0, and substitute it into the second derivative:
d²y/dx² = -2cos(0) = -2
Since the second derivative is negative, the function is concave down on the interval (-∞, π/2].
For π/2 < x < 3π/2:
Choose a test point, like x = π, and substitute it into the second derivative:
d²y/dx² = -2cos(π) = 2
Since the second derivative is positive, the function is concave up on the interval (π/2, 3π/2).
For x > 3π/2:
Choose a test point, like x = 2π, and substitute it into the second derivative:
d²y/dx² = -2cos(2π) = -2
Since the second derivative is negative, the function is concave down on the interval (3π/2, ∞).
Therefore, the function is concave down on (-∞, π/2] U [3π/2, ∞) and concave up on (π/2, 3π/2).
d) Inflection Points:
From previous calculations, we found that the possible inflection points are x = π/2 and x = 3π/2.
Evaluate the function at these points:
For x = π/2:
y = 2cos(π/2) - 1 = 2(0) - 1 = -1
(x = π/2, y = -1) is an inflection point.
For x = 3π/2:
y = 2cos(3π/2) - 1 = 2(0) - 1 = -1
(x = 3π/2, y = -1) is an inflection point.
Therefore, the inflection points are (π/2, -1) and (3π/2, -1).
So, to summarize the answers:
a) Intervals of Increase and Decrease: Decrease on [0, π) and Increase on (π, 2π].
b) Local Max and Min: Local max at (0, 1) and local min at (π, -3).
c) Intervals of Concavity: Concave down on (-∞, π/2] U [3π/2, ∞) and concave up on (π/2, 3π/2).
d) Inflection Points: (π/2, -1) and (3π/2, -1).