Asked by Sara
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?
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?
Answers
Answered by
Reiny
You would now sub those two values into the original equation. The larger answer is your maximum the smaller value is your minimum
Answered by
Steve
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.
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.
Answered by
Sara
Thanks!
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