determine the absolute extreme values of the function f(x)=sinx-cosx+6 on the interval 0<=x<=2pi.

This is what i did:
1.) i found the derivative of the function which is f'(x)=cosx+sinx
2.) I set f'(x)=0 and got sinx/cosx=-1 which is tanx=-1 and then x=3π/4 , 7π/4

What would I do after this?

You would now sub those two values into the original equation. The larger answer is your maximum the smaller value is your minimum

well, you have calculated where the extrema occur, so just plug in the values for x and calculate f(x).

or, you can easily see the values because

sinx-cosx = √2(sinx/√2 - cosx/√2)
= √2(sinx cos π/4 - sin π/4 cosx)
= √2 sin(x-π/4)

the max and min are 6±√2

naturally, these occur when x-π/4 is a multiple of π/2, as you correctly calculated.

Thanks!

After finding the critical points of the function by setting the derivative equal to zero, you need to evaluate the function at these critical points as well as at the endpoints of the given interval to determine the absolute extreme values.

Let's start by evaluating the function at the critical points and endpoints:

1. Evaluate f(x) at x = 0:
f(0) = sin(0) - cos(0) + 6 = 0 - 1 + 6 = 5

2. Evaluate f(x) at x = 3π/4:
f(3π/4) = sin(3π/4) - cos(3π/4) + 6 = (√2/2) - (-√2/2) + 6 = √2 + √2 + 6 = 2√2 + 6

3. Evaluate f(x) at x = 7π/4:
f(7π/4) = sin(7π/4) - cos(7π/4) + 6 = (-√2/2) - (√2/2) + 6 = -√2 - √2 + 6 = -2√2 + 6

4. Evaluate f(x) at x = 2π:
f(2π) = sin(2π) - cos(2π) + 6 = 0 - 1 + 6 = 5

Now, compare all these values and identify the maximum and minimum values of f(x) on the given interval:

- The maximum value of f(x) is 2√2 + 6, which occurs at x = 3π/4.
- The minimum value of f(x) is -2√2 + 6, which occurs at x = 7π/4.

Therefore, the absolute maximum value of the function is 2√2 + 6, and the absolute minimum value is -2√2 + 6.