Question

My question: An equation of the line tangent to the graph of f(x) = x(1 - 2x)3 at the point (1, –1) is??

My answer: y = –2x + 1

Answer 1

take the derivative using the product rule
f ' (x) = x(3)(1-2x)^2 (-2) + (1-2x)^3
sub in the x of the point(1,-1)
f ' (1) = 1(3)(1)(-2) + (-1)
= -7 which will be the slope of your tangent

Now you have the slope of -7 and a point (1,-1)
Use your favourite method of finding the equation of the line.

Answer 2

Okay, thanks

Answer 3

Final answer would be y=-7x+6 one way to figure it out is by finding you slope which is -7. You will then want to use the equation y=mx+b. We have the coordinates, (1,-1). The slope, -7 is m. 1 is x and -1 is y. Plug it into the equation which is -1=-7(1)+b. Now solve for b. -7(1)=-7
-1=-7+b. Plus 7 on both sides. -1+7= 6.
Y=mx+b, Y=-7x+6 =)

Answer 4

我日您爹