An equasion of the line tangent to the graph of f(x)=x(1-2x)^3 at the point is?

A) y=-7x+6
B) y=-6x+5
C) y=-2x+1
D) y=2x+3
E) y=7x-8

Question

An equasion of the line tangent to the graph of f(x)=x(1-2x)^3 at the point is?

A) y=-7x+6
B) y=-6x+5
C) y=-2x+1
D) y=2x+3
E) y=7x-8

raden · Answer

what point is it ?

anonymous · Answer

oops sorry. At the point (1,-1)

raden · Answer

first, determine the slope (m) by take the first derivative of f(x)

raden · Answer

use the product rule, and also the chain rule

anonymous · Answer

(1−2x)^3−6x(1−2x)^2 ?

raden · Answer

yup, good job
now to get m, put x=1 to f '
m =  (1−2(1))^3−6(1)(1−2(1))^2 = .... ?

anonymous · Answer

so the slope is -7.
A?

raden · Answer

i got like that too :)
now, use to form tangen line use the formula a straight line equation :
y - y1 = m(x - x1)

here, given (x1,y1) = (1,-1)
and m we got = -7

anonymous · Answer

y+1=-7(x+1)
y=-7x+6
OK thank you!

raden · Answer

yes, that's right
you're welcome