作业帮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+6B.y= -6x+5C.y= -2x+1D.y= 2x-3E.y= 7x-8
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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+6B.y= -6x+5C.y= -2x+1D.y= 2x-3E.y= 7x-8
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
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+6B.y= -6x+5C.y= -2x+1D.y= 2x-3E.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.
答案A
f'(x) = (x)'(1-2x)^3 + x[(1-2x)^3]'=(1-2x)^3 - 6x(1-2x)^2,所以f'(1)= -7 ，
切线方程为y - (-1) = -7 ( x - 1) 化简得y = -7x + 6 , 所以选(A)
答案：A
这是简单的导数题
先求出f(x)=x(1-2x)^3在点（1,-1)处的导数，为-7，在代用（1，-1）即可
看了以上几个回答,表示虽然不知道它说得是神马,但选A应该是正确答案...