Let F For Th Funcation Defined By F(x) = Sin^2-sin

ProHead

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Let f for th funcation defined by f(x) = sin^2-sin x, for o

 

Answer No.1

Anonymous

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F'x = 2sinx*cosx-cosx

Answer No.2

Anonymous

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F'(x)=2*sinx*cosx-cosx f'(0)=-1

Answer No.3

Anonymous

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F'(x)=2*sinx*cosx-cosx f'(0)=-1

Answer No.4

MisterJay

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If u were asking the derivative of the function, then the ans is f'(x)= -2cosxsin(sinx)

Answer No.5

Anonymous

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Solution f'(x)=2*sinx*cosx-cosx f'(0)=-1

Answer No.6

Anonymous

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F'(x)=dx/dx==2*sinx*cosx-cosx = sin 2x - cos x ∴ f'(0)=-1

Answer No.7

Anonymous

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Sin^2x-sinx= 0 sinx= 0 and sin x=1 so x=0 x=pi and x=pi/2 f(x)= 2six cosx -cos x = cos x(2sin x-1)=0 cos x= 0 x= pi/2 and 3pi/2 sinx=1/2 x= pi/6 and x= 5pi/6 The values of the function at those points are 0,0,2-1/4,-1/4 The abs max is 2 at x=3pi/2 and the abs min -1/4 at x=pi/6 and x=5pi/6 sign of f(x) cosx ++++++pi/2-----------------3pi/2 2sinx-1 ------pi/6+++++++++++5pi/6-----3pi/2 pi/6<=x<= pi/2 and 5pi/6<=x<=3pi/2 f(x) is increasing

Answer No.8

BigNewton

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F'(x)=2*sinx*cosx-cosx f'(0)=-1

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