Find intervals on which the following functions are increasing and decreasing.

1. f(x) = sin 2x : x is greater than or equal to 0 and less than or equal to pi

2. y = sin x + cos x [0,2pi]

wouldn't those be the intervals where the derivative is + ?

A function is increasing when the derivative is positive. If you are not studying calculus, then you will have to plot and think about the curves. The sine function rises until the angle (2x in this case) reaches pi/2, and then rises again between 3 pi/2 and 2 pi. That means 0< x< pi/4 and

3 pi/8 < x < pi are the reagions where sin 2x rises. You need to also consider the second period of oscillation when 2x goes from 2 pi to 4 pi. This will give you two more intervals for x.

For the second problem, use calculus or consider using the identity
sin x + cos x =
sqrt2 * sin(x + pi/4)
which reaches a maximum at x = pi/4, as well as 5 pi/4

To find the intervals on which a function is increasing and decreasing, we need to analyze the behavior of the derivative of the function. If the derivative is positive, the function is increasing, and if the derivative is negative, the function is decreasing.

Let's start with the first function:
1. f(x) = sin 2x : x is greater than or equal to 0 and less than or equal to pi

To find the derivative of f(x), we can apply the chain rule. The derivative of sin 2x is calculated by multiplying the derivative of sin(x) with the derivative of 2x. The derivative of sin(x) is cos(x) and the derivative of 2x is 2, resulting in 2cos 2x.

Now, we can analyze the sign of the derivative, 2cos 2x, to determine the intervals of increase and decrease.

For the function f(x) = sin 2x, we need to find when 2cos 2x > 0 and when 2cos 2x < 0.

To find when 2cos 2x > 0:
2cos 2x > 0
cos 2x > 0

The cosine function is positive in the first and second quadrants. Since x is greater than or equal to 0 and less than or equal to pi, the range of x fits within the first and second quadrants. Therefore, cos 2x > 0 for all values of x in the interval [0, pi].

To find when 2cos 2x < 0:
2cos 2x < 0
cos 2x < 0

The cosine function is negative in the second and third quadrants. Again, the range of x lies within the first and second quadrants, so we need to find when 2x falls in the third quadrant. This occurs when x is between pi/2 and pi. Therefore, cos 2x < 0 for all values of x in the interval (pi/2, pi].

In conclusion, f(x) = sin 2x is increasing for all values of x in the interval [0, pi/2] and decreasing for all values of x in the interval (pi/2, pi].

Now let's move on to the second function:

2. y = sin(x) + cos(x) [0, 2pi]

To find the intervals of increase and decrease for this function, we need to find when the derivative dy/dx > 0 (increasing) and when dy/dx < 0 (decreasing).

To find the derivative of y, we differentiate sin(x) and cos(x) separately. The derivative of sin(x) is cos(x) and the derivative of cos(x) is -sin(x). Therefore, the derivative of y is dy/dx = cos(x) - sin(x).

To find when dy/dx > 0:
cos(x) - sin(x) > 0

Both cosine and sine functions are positive in the first and fourth quadrants. Therefore, we have two cases to consider:
1. From 0 to pi/4: In this interval, both cosine and sine are positive. Hence, cos(x) - sin(x) > 0.
2. From 5pi/4 to 2pi: In this interval, both cosine and sine are negative. Hence, cos(x) - sin(x) > 0.

To find when dy/dx < 0:
cos(x) - sin(x) < 0

Again, we have two cases:
1. From pi/4 to 5pi/4: In this interval, cosine is positive, but sine is negative. Hence, cos(x) - sin(x) < 0.
2. From pi to 2pi: In this interval, both cosine and sine are negative. Hence, cos(x) - sin(x) < 0.

In conclusion, the function y = sin(x) + cos(x) is increasing in the intervals [0, pi/4) and (5pi/4, 2pi]. It is decreasing in the intervals (pi/4, pi) and (pi, 5pi/4].

Remember that increasing and decreasing are defined relative to the x-values within the given interval.