Find all points of extrema on the interval [0, 2pi] if y=x+sinx
(-1.-1+(3pi/2))
(pi, pi)
(-1,0)
((3pi/2), 0)
none of these
I got pi as my critical number but I'm confused with the intervals.
dy/dx = 1 + cosx
= 0
cosx = -1
look at your cosine curve.
There is only one place where it has a value of -1 and it is at 180° or π radians for its entire period from 0 to 2π
so when x = π , back in the original equation:
y = π + sinπ
= π + 0 = π
so the point is (π,π)
Well, speaking of confusion, let me just put on my clown wig and try to clear things up for you!
To find the extrema of the function y = x + sin(x) on the interval [0, 2pi], we need to first find the critical points. These are the values of x where either the derivative is zero or it does not exist.
The derivative of y with respect to x is 1 + cos(x). Setting this equal to zero, we get:
1 + cos(x) = 0
Subtracting 1 from both sides, we have:
cos(x) = -1
Now, on the interval [0, 2pi], the only value of x that satisfies cos(x) = -1 is x = pi.
So, as you correctly found, pi is a critical number. Good job!
Now, let's examine the intervals and see if we can determine the extrema.
On the interval [0, pi], the derivative 1 + cos(x) is positive for all x in this interval since cos(x) is always greater than -1. So, there are no extrema in this interval.
On the interval [pi, 2pi], the derivative 1 + cos(x) is negative for all x in this interval since cos(x) is always less than -1. So, there are no extrema in this interval either.
Therefore, the function y = x + sin(x) has no extrema on the interval [0, 2pi].
So, the correct answer is none of these options. Keep up the good work and if you have any more questions, don't hesitate to ask!
To find the points of extrema for the function y = x + sin(x) on the interval [0, 2π], we need to calculate the first derivative of the function and find where it equals zero or does not exist.
The derivative of y = x + sin(x) can be found using the sum rule and the chain rule. The derivative is given by:
y' = 1 + cos(x)
To find where the derivative equals zero, we solve the equation:
1 + cos(x) = 0
Subtracting 1 from both sides, we get:
cos(x) = -1
The cosine function is equal to -1 at π radians. So, one critical number is x = π.
Next, we need to examine the intervals [0, π] and [π, 2π].
In the interval [0, π], the derivative y' = 1 + cos(x) is always positive since the cosine function is always greater than -1. This means there are no critical points in this interval.
In the interval [π, 2π], the derivative y' = 1 + cos(x) is always negative since the cosine function is always less than -1. Again, there are no critical points in this interval.
Therefore, the only point of extremum on the interval [0, 2π] is x = π.
So, the correct answer is: (π, π)
To find the extrema of a function, we need to determine where the function's derivative equals zero or is undefined. Let's first find the derivative of the function y = x + sin(x):
y' = 1 + cos(x).
To find the critical numbers, we set the derivative equal to zero:
1 + cos(x) = 0.
Subtracting 1 from both sides gives:
cos(x) = -1.
The solutions to the equation cos(x) = -1 occur when x is in the set {π}. So, π is the only critical number on the interval [0, 2π].
Now, we need to determine whether these critical numbers correspond to a maximum or minimum. To do that, we can analyze the sign of the derivative in the intervals [0, π] and between π and 2π.
In the interval [0, π], we can test any value between 0 and π, for example, x = 1. Substituting x = 1 into the derivative, we get:
1 + cos(1) > 0.
Therefore, the function is increasing in the interval [0, π] and does not have any extrema in this interval.
In the interval [π, 2π], we can test any value between π and 2π, for example, x = 4. Substituting x = 4 into the derivative, we get:
1 + cos(4) < 0.
Therefore, the function is decreasing in the interval [π, 2π], indicating a possible maximum.
Since the derivative changes sign from positive to negative at x = π, π is a potential maximum point on the interval [0, 2π]. However, we still need to check the endpoints of the interval.
For x = 0, the function is y = 0 + sin(0) = 0.
For x = 2π, the function is y = 2π + sin(2π) = 2π.
Comparing y = 0 and y = 2π, we find that the extrema occur at x = π, where y = π.
Therefore, the point of maximum on the interval [0, 2π] is (π, π).
None of the given options (-1, -1 + (3π/2)), (π, π), (-1, 0), (3π/2, 0) match the correct point of maximum (π, π).