sandy524
find the derivative of f(x) = cos(x) − sin^2(x) on [0,2pi]
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sandy524
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i now know that its -sin(x)(2cos(x)+1)
sandy524
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can yu guys help me find the critical points
myininaya
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Are you sure it ask you to find f'(x) on [a,b]?
myininaya
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If so, you did that. You are done.
sandy524
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yes. may you help me find the critical points of this now
myininaya
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Set f'(x)=0 and solve for x to find the critical numbers.
myininaya
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You only need to solve this on the interval [0,2 pi]
sandy524
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can we do it together please
myininaya
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You are already factored f'(x)
Now set both factors equal to 0.
sandy524
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how?
can we work out the problem
sandy524
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step by step please
myininaya
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Yes. I'm walking you through it.
You have -sin(x)*(1+2cos(x))=0
Set both factors equal to 0.
sandy524
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ok
myininaya
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Just like if you have a*b=0 then either a=0 or b=0 or both=0
sandy524
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this is all i have
sandy524
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im confused about the results
myininaya
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Use your unit circle.
Keep in mind you only have to look between 0 and 2pi
myininaya
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You have -sin(x)=0 or 1+2cos(x)=0
Solve both of these for the trig function.
Solve the first one for sin(x) and solve the second one for cos(x).
Tell me what you get after doing this.
sandy524
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-sinx=pin
sandy524
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\[\pi n\]
myininaya
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Well you are trying to solve -sin(x)=0 for x.
You could divide both sides by -1 since -1 will never be 0.
Try solving this: sin(x)=0
Then we will come back to 1+2cos(x)=0