3. Given the function defined by y = x + sinx for all x such that -π/2<=x<=3π/2
a. Find the coordinate of all maximum and minimum points on the given interval. Justify your answers.
b. Find the coordinates of all points of inflection on the given interval. Justify your answers.
please be explisive!
thanks!
now look at the ends
at x = - pi/2
y = - pi/2 + sin(-pi/2) = -pi/2 -1
minimum I bet
at x = 3 pi/2
y = 3 pi/2 + sin (3 pi/2) = 3 pi/2 -1
maximum
a. To find the maximum and minimum points on the given interval, we need to find the critical points of the function and then determine whether they are maximum or minimum points.
Step 1: Find the derivative of the function:
dy/dx = 1 + cos(x)
Step 2: Set the derivative equal to zero and solve for x:
1 + cos(x) = 0
Solving this equation, we get:
cos(x) = -1
x = π
Step 3: Determine the nature of the critical point:
We can use the second derivative test to determine if x = π is a maximum or a minimum point.
Taking the second derivative of the function:
d²y/dx² = -sin(x)
Substituting x = π into the second derivative, we get -sin(π) = 0.
Since the second derivative is zero at x = π, the second derivative test is inconclusive. Therefore, x = π is not a maximum or minimum point.
Step 4: Test the endpoints of the interval:
At the endpoints of the interval, x = -π/2 and x = 3π/2, the function is y = -π/2 + sin(-π/2) = -π/2 -1 and y = 3π/2 + sin(3π/2) = 3π/2 - 1.
Therefore, the coordinates of the maximum and minimum points on the given interval are (-π/2, -π/2 - 1) and (3π/2, 3π/2 - 1), respectively.
b. To find the inflection points on the given interval, we need to find the points where the concavity of the function changes.
Step 1: Find the second derivative of the function:
d²y/dx² = -sin(x)
Step 2: Set the second derivative equal to zero and solve for x:
-sin(x) = 0
x = 0, π
Step 3: Determine the nature of the points of inflection:
To determine the nature of the inflection points, we can look at the concavity of the function at x = 0 and x = π.
At x = 0, the second derivative is -sin(0) = 0. Since the second derivative is zero at x = 0, we cannot conclude the concavity at this point.
At x = π, the second derivative is -sin(π) = 0. Similar to the previous point, we cannot determine the concavity at x = π.
Therefore, there are no points of inflection on the given interval (-π/2, 3π/2).
To find the coordinates of the maximum and minimum points on the given interval, we need to first find the critical points of the function.
a. Critical points occur when the derivative of the function is equal to zero or does not exist.
First, let's find the derivative of the function:
y' = 1 + cos(x)
To find the critical points, we set the derivative equal to zero:
1 + cos(x) = 0
cos(x) = -1
Since -1 is the value that cosine takes on at π radians, we have:
x = π
However, we need to check if this critical point lies within the given interval. Since -π/2 <= x <= 3π/2, the critical point x = π is within this interval.
Therefore, the coordinate of the only critical point, and therefore the maximum and minimum point, is (π, π + sin(π)).
To justify this answer, we need to analyze the behavior of the function around the critical point using the second derivative test.
The second derivative is:
y'' = -sin(x)
Plugging in x = π into the second derivative, we get:
y''(π) = -sin(π) = 0
Since the second derivative is zero at x = π, we cannot determine the concavity of the function at this point. Therefore, the second derivative test is inconclusive in this case.
b. To find the coordinates of the inflection points, we need to find the points where the concavity of the function changes. This occurs when the second derivative is equal to zero or does not exist.
The second derivative is:
y'' = -sin(x)
To find the inflection points, we set the second derivative equal to zero:
-sin(x) = 0
sin(x) = 0
The values of x where sin(x) = 0 are x = 0, π, 2π, 3π, etc.
Now we need to check if these values lie within the given interval. Since -π/2 <= x <= 3π/2, x = 0, π, and 2π are within this interval. However, x = 3π is not within the given interval.
Therefore, the coordinates of the inflection points are:
(0, 0 + sin(0)), (π, π + sin(π)), and (2π, 2π + sin(2π)).
To justify these answers, we need to analyze the behavior of the function around the inflection points.
The second derivative is:
y'' = -sin(x)
Plugging in the values of x into the second derivative, we get:
y''(0) = -sin(0) = 0
y''(π) = -sin(π) = 0
y''(2π) = -sin(2π) = 0
Since the second derivative is zero at these points, we cannot determine the concavity of the function at these points. Therefore, the second derivative test is inconclusive in these cases.
To summarize:
a. The coordinate of the maximum and minimum point on the given interval is (π, π + sin(π)).
b. The coordinates of the inflection points on the given interval are (0, 0 + sin(0)), (π, π + sin(π)), and (2π, 2π + sin(2π)).
To determine the maximum and minimum points on the given interval for the function y = x + sin(x), we need to find the critical points where the derivative of the function is equal to zero or undefined. Let's go step by step to find these points:
a. Finding critical points:
1. First, we find the derivative of y with respect to x. Since y = x + sin(x), the derivative of y with respect to x will be the derivative of x (which is 1) plus the derivative of sin(x).
dy/dx = 1 + cos(x)
2. Setting the derivative equal to zero to find critical points:
1 + cos(x) = 0
3. Solving the equation for x:
cos(x) = -1
x = π
4. Checking the endpoints of the given interval:
We know that the given interval is -π/2 <= x <= 3π/2. From this, we see that -π/2 is an endpoint of the interval, and plugging it into the derivative equation gives:
1 + cos(-π/2) = 1 - 0 = 1
Similarly, plugging 3π/2 into the derivative equation gives:
1 + cos(3π/2) = 1 + 0 = 1
Since the derivative is not zero at the endpoints, we can ignore them for finding the maximum and minimum points.
5. Now, let's check the second derivative to determine whether the critical point at x = π is a maximum or minimum. We find the second derivative by differentiating the first derivative with respect to x.
d^2y/dx^2 = -sin(x)
6. Substituting x = π into the second derivative:
-sin(π) = 0
The second derivative is zero at x = π, which means we cannot conclude whether the point is a maximum or a minimum based on the second derivative test. So, we'll need to use an alternative method to justify the answer.
Since there are no critical points in the given interval other than x = π, it is the only point we need to check to identify any maximum or minimum.
7. To determine whether x = π is a maximum or minimum point, we analyze the behavior of the function around that point:
- To the left of x = π (x < π): y = x + sin(x) is increasing.
- To the right of x = π (x > π): y = x + sin(x) is decreasing.
Therefore, we can conclude that at x = π, there is a minimum point.
b. Points of inflection:
To find the points of inflection, we need to find the x-values where the concavity of the function changes. This occurs when the second derivative changes sign or is equal to zero.
1. Since the second derivative is -sin(x), it changes sign when sin(x) changes sign, which are at x = -π/2 and x = 3π/2.
2. However, we need to check whether these points are within the given interval. Among these, only x = -π/2 falls within the interval -π/2 <= x <= 3π/2. Thus, x = -π/2 is the point of inflection.
Therefore:
a. The coordinate of the minimum point is (π, sin(π)) or (π, 0). This point is justified by analyzing the behavior of the function around x = π.
b. The coordinate of the point of inflection is (-π/2, sin(-π/2)) or (-π/2, -1). This point is justified by analyzing the sign change in the concavity of the function.
dy/dx = 1 + cos x
0 when cos x = -1
that is when x = pi
d^2y/dx^2 = -sin x
-sin x = 0 when x = pi
so this is not a max or min but a point of inflection