AravindG
find local maxima and local minima for the function f(x)=sin xcos x
0<x<2



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AravindG
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@hartnn

uzumakhi
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for this you have to find f'(x)

AravindG
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ok ....cos x + sin x
then?

uzumakhi
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and then take f'(x)=0

AravindG
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sin x+cos x=0

AravindG
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sinx=cos x

AravindG
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nw what ?

uzumakhi
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cos x + sin x = 0
cos x = sinx

AravindG
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k then ?

AravindG
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why u delete that?

uzumakhi
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find the value of x

AravindG
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x=pi/4 isnt it?

uzumakhi
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no we have to take the value from 0 to 2

AravindG
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then hw ??

uzumakhi
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take x = 135 degrees

uzumakhi
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convert it to radians

AravindG
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2.355

AravindG
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x=3pi/4

hartnn
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if x is between 0 to 2pi, then x= 3pi/4 is correct.

AravindG
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k then hw we find local maxima and local minima ?

hartnn
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local only means 'in that interval' , here 0 to 2pi

AravindG
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so what are maxima and minima here ?

hartnn
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so here u'll have only one value , at x=3pi/4
find f(x) when x=3pi/4

AravindG
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also at x=7pi/4 we have only one value so
f'(X)= ives 2 values for x isnt it?

AravindG
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f'(X)=0 gives 2 values for x isnt it?

hartnn
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oh yes! i missed that.
now its even simpler..... find the second derivate, what u get ?

AravindG
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sinx +cos x

AravindG
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whats the use of taking second derivative ?

hartnn
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to know which point is max. or min. , u know how ?

AravindG
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nop

AravindG
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is it by substituting 3pi/4 and 7pi/4 in f''(X) and getting the sign to tell which is maxima and minima?

AravindG
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k gt it !!!

hartnn
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yes, if f''(x) > 0 , then max or min ?

hartnn
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ok, good :)

AravindG
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3pi/4 is max other is minima !!

AravindG
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can u check up the answer for the next qn also ?

hartnn
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sure ...

AravindG
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i will post separately