# Calculus

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?

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

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2. 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.

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3. Thanks!

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