f(x) = cos(x) - (cos(x))^2 for -pi is less than or equal to x is less than or equal to pi
1) x-intercepts?
2) x and y coordinates of all relative maximum points?
3) intervals on which f is increasing?
then i have to sketch f, so tips on doing so would be appreciated. thank you!
I assume you know how to plot a graph. Choose 10 or more x values between -pi and pi, compute f(x) for each, and graph them with f(x) on the y axis. Draw a smooth line between the points.
The x intercepts are where y = 0. That would be whenever cos x = 0 or 1. You should know where those angles are x = 0 and x = pi/2 are two of them.
Compute the derivative f'(x) and set it equal to zero. The solutions of that equation will be where the function has a relative maximum or minimum.
f'(x) = -sin x + 2 sin x cos x
= sin x (2 cos x -1)
which equals 0 at x = 0, x = pi/2 and cos^1(1/2). Only the last of those is a maximum.
f is increasing where f'(x) is positive. Drawing the graph should help you find that region.
To find the answers to the questions, we'll follow these steps:
1) To find the x-intercepts, we need to find the values of x for which f(x) is equal to zero. In this case, f(x) = cos(x) - (cos(x))^2. So, we set f(x) = 0 and solve for x:
cos(x) - (cos(x))^2 = 0
Factoring out cos(x), we get:
cos(x) * (1 - cos(x)) = 0
So, either cos(x) = 0 or 1 - cos(x) = 0.
For cos(x) = 0, the solutions are x = π/2 and x = 3π/2.
For 1 - cos(x) = 0, the solution is x = 0.
Therefore, the x-intercepts are x = 0, x = π/2, and x = 3π/2.
2) To find the x and y coordinates of the relative maximum points, we need to find the critical points of f(x) where its derivative is equal to zero or undefined. Let's first find the derivative of f(x):
f'(x) = -sin(x) + 2cos(x)sin(x)
Setting f'(x) = 0, we get:
-sin(x) + 2cos(x)sin(x) = 0
Factoring out sin(x), we get:
sin(x) * (2cos(x) - 1) = 0
So, either sin(x) = 0 or 2cos(x) - 1 = 0.
For sin(x) = 0, the solutions are x = 0, x = π, and x = 2π.
For 2cos(x) - 1 = 0, the solution is x = π/3.
Now, we need to evaluate f(x) at these critical points to find the y coordinates.
f(0) = cos(0) - (cos(0))^2 = 1 - 1 = 0
f(π) = cos(π) - (cos(π))^2 = -1 - 1 = -2
f(2π) = cos(2π) - (cos(2π))^2 = 1 - 1 = 0
f(π/3) = cos(π/3) - (cos(π/3))^2 = √3/2 - 3/4 = (4√3 - 3)/4
Therefore, the relative maximum points are (0, 0), (π, -2), (2π, 0), and (π/3, (4√3 - 3)/4).
3) To determine the intervals where f(x) is increasing, we need to find the intervals where f'(x) > 0.
Let's analyze the sign of f'(x) in different intervals:
- For x < 0: sin(x) < 0, cos(x) > 0, so f'(x) > 0.
- For 0 < x < π/3: sin(x) > 0, cos(x) > 1/2, so f'(x) < 0.
- For π/3 < x < π/2: sin(x) > 0, cos(x) < 1/2, so f'(x) > 0.
- For π/2 < x < π: sin(x) > 0, cos(x) < 0, so f'(x) < 0.
- For π < x < 2π: sin(x) > 0, cos(x) > 0, so f'(x) > 0.
Based on the above analysis, f(x) is increasing on the intervals (-∞, 0) and (π/3, π/2) U (π, 2π).
To sketch f(x), you can use the information obtained above. Plot the x-intercepts at x = 0, x = π/2, and x = 3π/2. Mark the relative maximum points at (0, 0), (π, -2), (2π, 0), and (π/3, (4√3 - 3)/4). Connect the points with a smooth curve, keeping in mind the increasing intervals mentioned earlier. Also, take into consideration the symmetry of the cosine function.