an equation of the line tangent to the graph of f(x)=x(1-2x)
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an equation of the line tangent to the graph of f(x)=x(1-2x)^3 at the point (1,-1) is?
A.y= -7x+6
B.y= -6x+5
C.y= -2x+1
D.y= 2x-3
E.y= 7x-8
A.y= -7x+6
B.y= -6x+5
C.y= -2x+1
D.y= 2x-3
E.y= 7x-8
先翻译成中文:
求函数f(x)=x(1-2x)^3在点(1,-1)处的切线方程.
f'(x)=(1-2x)^3+3x(1-2x)^2*(-2)=(1-2x)^2*[(1-8x)]
当x=1,f'(x)=-7
所以切线方程为y+1=f'(x)(x-1)=-7(x-1),化简得y= -7x+6
所以选A.
求函数f(x)=x(1-2x)^3在点(1,-1)处的切线方程.
f'(x)=(1-2x)^3+3x(1-2x)^2*(-2)=(1-2x)^2*[(1-8x)]
当x=1,f'(x)=-7
所以切线方程为y+1=f'(x)(x-1)=-7(x-1),化简得y= -7x+6
所以选A.
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