Consider the function on the interval
(0, 2π).
f(x) = x + 2 sin x
relative maximum (x, y) =
relative minimum (x, y) =
From when I worked out using the first derivative test, I ended up with (2π, 0) for the max and (0,0) for the min. Apparently this isn't correct.
Too bad you didn't show your work...
f' = 1 + 2cosx
f'=0 when cosx = -1/2
x = 2?/3, 4?/3
f" = -2sinx
f"(2?/3) < 0, so that is a maximum
f"(4?/3) > 0, so that is a minimum
See the graph at
http://www.wolframalpha.com/input/?i=x%2B2sinx,+for+x+%3D+0..2pi
Ah, the joys of calculus! Let's take a look at your function, f(x) = x + 2sin(x), and find those relative extrema together.
To find the relative extrema, we'll need to take the first derivative of the function, f'(x), and set it equal to zero. So, let's do some derivative shenanigans here:
f(x) = x + 2sin(x)
f'(x) = 1 + 2cos(x)
Now, we set f'(x) = 0:
1 + 2cos(x) = 0
Oh no, we have a cosine in the equation, and that can take any value between -1 and 1. So, to find the exact values of x that make f'(x) equal to zero, we need to consider the cosine function. In this case, cos(x) = -1/2 will give us an x value.
So, solving cos(x) = -1/2, we find two solutions in the interval (0, 2π): x = 2π/3 and x = 4π/3.
Now that we have these critical points, we can plug them back into the original function f(x) to find the corresponding y values.
For x = 2π/3:
f(2π/3) = 2π/3 + 2sin(2π/3) ≈ 3.528
For x = 4π/3:
f(4π/3) = 4π/3 + 2sin(4π/3) ≈ -0.072
So, the relative maximum of the function f(x) = x + 2sin(x) on the interval (0, 2π) is approximately (2π/3, 3.528), and the relative minimum is approximately (4π/3, -0.072).
To find the relative maximum and minimum of the function f(x) = x + 2sin(x) on the interval (0, 2π), you can follow these steps:
Step 1: Calculate the derivative of the function.
Taking the derivative of f(x) with respect to x will help us determine where the function is increasing or decreasing.
f'(x) = 1 + 2cos(x)
Step 2: Find the critical points.
To find the critical points, we set the derivative equal to 0 and solve for x.
1 + 2cos(x) = 0
Step 3: Solve for x.
Subtracting 1 from both sides of the equation:
2cos(x) = -1
Dividing both sides by 2:
cos(x) = -1/2
Now we need to find the values of x between 0 and 2π (exclusive) where the cosine function equals -1/2. These values correspond to the critical points.
Step 4: Determine the values of x.
Using knowledge of the unit circle or trigonometric identities, cos(x) = -1/2 occurs at two angles: 2π/3 and 4π/3.
However, these angles fall outside the interval (0, 2π). We need to find the critical points in this interval.
Let's consider the first quadrant (0 to π/2):
cos(x) = -1/2 does not have any solutions in this range.
Now, let's consider the second quadrant (π/2 to π):
cos(x) = -1/2 has one solution in this range, which is π - π/3 = 2π/3.
So, the critical point x = 2π/3 falls within the interval (0, 2π).
Finally, the critical points for the derivative f'(x) = 0 are x = 2π/3.
Step 5: Analyze the intervals.
With the critical points identified, we can analyze the intervals between them to determine where the function is increasing or decreasing.
Interval (0, 2π/3):
For values of x in this interval, f'(x) > 0 (positive), indicating that the function f(x) = x + 2sin(x) is increasing.
Interval (2π/3, 2π):
For values of x in this interval, f'(x) < 0 (negative), indicating that the function f(x) = x + 2sin(x) is decreasing.
Step 6: Determine the relative maximum and minimum.
Since the function is increasing on the interval (0, 2π/3) and decreasing on the interval (2π/3, 2π), we can conclude:
Relative maximum: The function has no relative maximum within the interval (0, 2π).
Relative minimum: The relative minimum occurs at x = 2π/3, where the value of the function f(x) = x + 2sin(x) is at its lowest point within the interval (0, 2π).
Therefore, the relative maximum does not exist, and the relative minimum can be identified as the point (2π/3, f(2π/3)) where f(2π/3) can be calculated.
To find the relative maximum and minimum of a function, we can start by finding its critical points. Critical points are the values of x where the first derivative of the function is either zero or undefined. Once we have the critical points, we can use the second derivative test or evaluate the function at these points to determine whether they correspond to a relative maximum or minimum.
Let's start by finding the first derivative of the function f(x) = x + 2sin(x):
f'(x) = 1 + 2cos(x)
Now, we need to find the critical points by solving the equation f'(x) = 0:
1 + 2cos(x) = 0
Subtracting 1 from both sides:
2cos(x) = -1
Dividing by 2:
cos(x) = -1/2
To find the solutions for this equation between (0, 2π), we can refer to the unit circle or trigonometric values. The solutions are x = 2π/3 and x = 4π/3.
Now that we have the critical points, let's apply the second derivative test to determine whether they correspond to a relative maximum or minimum. The second derivative is calculated as follows:
f''(x) = -2sin(x)
Evaluating f''(x) at the critical points:
f''(2π/3) = -2sin(2π/3) = -2√3/2 = -√3
f''(4π/3) = -2sin(4π/3) = -2(-√3/2) = √3
When f''(x) is negative, it indicates that the function is concave down, suggesting a relative maximum. On the other hand, when f''(x) is positive, it indicates that the function is concave up, suggesting a relative minimum.
Considering the values of f''(x) at the critical points, we can conclude:
- At x = 2π/3, f''(x) = -√3 (< 0), suggesting a relative maximum.
- At x = 4π/3, f''(x) = √3 (> 0), suggesting a relative minimum.
Therefore, the relative maximum is at (2π/3, f(2π/3)) and the relative minimum is at (4π/3, f(4π/3)). To evaluate the function at these points, plug in the x-values into the original function f(x) = x + 2sin(x).
So the correct answers are:
- Relative maximum: (2π/3, 2π/3 + 2sin(2π/3))
- Relative minimum: (4π/3, 4π/3 + 2sin(4π/3))