Question

an equation of the line tangent to the graph of f(x)=x(1-2x)^3 at the point (1,-1) is ..

Answer 1

Do you know what a tangent line is?

Answer 2

yes

Answer 3

i just need some help in finding the equation

Answer 4

Take the derivative first.

Answer 5

Do you know how? :)

Answer 6

lol uhm sort of.. hehe. uhm so far i've got 3x(1-2x)^2(-2)

Answer 7

\[ f(x)=x(1-2x)^3\]
\[\frac{ dy }{ dx }=\frac{ d }{ dx }[x(1-2x)^3]\]
Use the product rule \(x*(1-2x)^3\) hence x is being multiplied by (1-2x)^3, and then use the chain rule when you take the derivative of (1-2x)^3.
\[\frac{ dy }{ dx }=1*(1-2x)^3 + 3x(1-2x)^{2}(-2)]\]
\[\frac{ dy }{ dx }=1*(1-2x)^3 -6x(1-2x)^{2}]\]

Answer 8

now, the equation of the tangent line is already given, to find the slope you plug in the given x from (1,-1) which is 1
f'(1)=? will be the slope
then use the point slope formula
\(y-y_1=m(x-x_1)\)
where m is the slope you have solved for when f'(1)=?

Answer 9

wait but why did u change the x into a 1. Isn't the product rule uv'+vu' so the first multiplied by the derivative of the 2nd?

Answer 10

because the derivative of x is 1
\[\frac{ d }{ dx }[1x^1]=1*1x^0=1\]
member the power rule?

Answer 11

power rule is one of the short cuts for taking the derivative, where you have to multiply the exponent by the coefficient and subtract one from the exponent
\[\frac{ dy }{ dx }=nx^{n-1}\]

Answer 12

okay thanks

Answer 13

wait so i plug in one into the original equation?

Answer 14

One thing i forgot the mention, after you solvd for he slope, you plug the slope to m and the given point (1,-1) to (x1,y1) in the point slope formula

Answer 15

plug 1 to the derivative to get the slope

Answer 16

okay so I got 7 for the slope

Answer 17

plug 2 to the derivative to get the slope