Find all relative extrema and points of inflection of the function:
f(x) = sin (x/2)
0 =< x =< 4pi
=< is supposed to be less than or equal to.
I can find the extrema, but the points of inflection has me stumped. The inflection point is (2pi,0) but shouldn't there be inflection points on x=0 and x=4pi as well?
To find the relative extrema of the function, we need to find the points where the derivative is equal to zero or does not exist. Let's start by finding the derivative of the given function:
f(x) = sin(x/2)
Using the chain rule, we have:
f'(x) = (1/2) * cos(x/2)
To find the points where the derivative is equal to zero, we need to solve the equation f'(x) = 0. Let's solve it:
(1/2) * cos(x/2) = 0
cos(x/2) = 0
For cosine to be equal to zero, the argument (x/2) must be equal to (2k + 1) * (π/2), where k is an integer. So we have:
x/2 = (2k + 1) * (π/2)
Solving for x, we get:
x = (2k + 1) * π, for k ∈ Z
The values of x that satisfy the equation are x = π, 3π, 5π, and so on. However, we are only interested in the values of x within the given interval 0 ≤ x ≤ 4π.
In the given interval, the values of x that satisfy the equation are x = π, 3π. These are the points where the function has a relative extremum.
Now, let's find the points of inflection. A point of inflection occurs where the concavity of the function changes. We can determine the concavity by looking at the second derivative.
To find the second derivative (f''(x)), we differentiate the first derivative (f'(x)):
f''(x) = (-1/4) * sin(x/2)
Setting f''(x) = 0, we have (-1/4) * sin(x/2) = 0. Solving this equation gives us sin(x/2) = 0.
For sine to be equal to zero, the argument (x/2) must be equal to k * π, where k is an integer. So we have:
x/2 = k * π
Solving for x, we get:
x = 2kπ, for k ∈ Z
The values of x that satisfy the equation are x = 0, 2π, 4π, and so on. However, we are only interested in the values of x within the given interval 0 ≤ x ≤ 4π.
In the given interval, the value of x that satisfies the equation is x = 0. This is the point of inflection for the function.
Therefore, the relative extrema of the function f(x) = sin(x/2) in the interval 0 ≤ x ≤ 4π are x = π and x = 3π, and the point of inflection is x = 0.
To find the relative extrema and points of inflection of a function, you'll need to follow these steps:
Step 1: Determine the first derivative of the function.
Step 2: Set the first derivative equal to zero and solve for x to find the critical points.
Step 3: Determine the second derivative of the function.
Step 4: Plug the critical points from Step 2 into the second derivative to determine the concavity of the function.
Step 5: Find any points where the concavity changes to identify the points of inflection.
Let's go through these steps for the function f(x) = sin(x/2) on the interval 0 ≤ x ≤ 4π:
Step 1: Determine the first derivative of the function.
The first derivative is found by differentiating the function, using the chain rule:
f'(x) = (1/2) * cos(x/2)
Step 2: Set the first derivative equal to zero and solve for x to find the critical points.
Setting f'(x) = 0:
(1/2) * cos(x/2) = 0
cos(x/2) = 0
Since cos(x/2) is zero at x = π/2, 3π/2, 5π/2, and 7π/2, these are the critical points.
Step 3: Determine the second derivative of the function.
The second derivative is found by differentiating the first derivative:
f''(x) = (-1/4) * sin(x/2)
Step 4: Plug the critical points from Step 2 into the second derivative to determine the concavity of the function.
For the values x = π/2, the second derivative is f''(π/2) = (-1/4) * sin((π/2)/2) = (-1/4) * sin(π/4) = -sqrt(2)/8 < 0. This indicates concave-down behavior.
For the values x = 3π/2, the second derivative is f''(3π/2) = (-1/4) * sin((3π/2)/2) = (-1/4) * sin(3π/4) = -sqrt(2)/8 < 0. This also indicates concave-down behavior.
For the values x = 5π/2, the second derivative is f''(5π/2) = (-1/4) * sin((5π/2)/2) = (-1/4) * sin(5π/4) = sqrt(2)/8 > 0. This indicates concave-up behavior.
For the values x = 7π/2, the second derivative is f''(7π/2) = (-1/4) * sin((7π/2)/2) = (-1/4) * sin(7π/4) = sqrt(2)/8 > 0. This also indicates concave-up behavior.
Step 5: Find any points where the concavity changes to identify the points of inflection.
From the previous step, we see that the concavity changes at x = 5π/2. Therefore, the point (5π/2, f(5π/2)) is an inflection point.
Regarding your question, there are no inflection points at x = 0 or x = 4π because the concavity does not change at those points. The point (2π, f(2π)) is indeed an inflection point.
To summarize:
- Relative extrema: None (because the function is periodic with no maximum or minimum within the given interval)
- Points of inflection: (2π, f(2π)) and (5π/2, f(5π/2))
The basic sine curve has points of inflection at every x-intercept of the curve,
y' = (1/2) cos x/2
y' = -(1/4) sin x/2
-(1/4)sin x/2 = 0
gives me x/2 = 0 or x/2 = π or x/2 = 2π etc
or
x = 0 or x = 2π or x = 4π etc
then if you domain is 0 ≤ x ≤ 4π
you are correct, there should have been points of inflection at
(0,0), (2π,0) and (4π,0)