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](/uploads/image/z/1226391-15-1.jpg?t=an+equation+of+the+line+tangent+to+the+graph+of+f%28x%29%3Dx%281-2x%29%5E3+at+the+point+%281%2C-1%29+is%3FA.y%3D+-7x%2B6B.y%3D+-6x%2B5C.y%3D+-2x%2B1D.y%3D+2x-3E.y%3D+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
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...