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

 ProHead From: USPosts: 29Votes: 58 Let f for th funcation defined by f(x) = sin^2-sin x, for o Anonymous From: -Posts: -Votes: - F'x = 2sinx*cosx-cosx  Anonymous From: -Posts: -Votes: - F'(x)=2*sinx*cosx-cosx f'(0)=-1  Anonymous From: -Posts: -Votes: - F'(x)=2*sinx*cosx-cosx f'(0)=-1  MisterJay From: USPosts: 9Votes: 27 If u were asking the derivative of the function, then the ans is f'(x)= -2cosxsin(sinx)  Anonymous From: -Posts: -Votes: - Solution f'(x)=2*sinx*cosx-cosx f'(0)=-1  Anonymous From: -Posts: -Votes: - F'(x)=dx/dx==2*sinx*cosx-cosx = sin 2x - cos x ∴ f'(0)=-1  Anonymous From: -Posts: -Votes: - 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  BigNewton From: USPosts: 5Votes: 15 F'(x)=2*sinx*cosx-cosx f'(0)=-1  Enter 