What is the equation of the line tangent to #f(x)=x ^3-3x^2 # at #x=4#?

1 Answer
Feb 14, 2016

#y=24x-80#

Explanation:

First we must find the slope of the tangent line at #x=4#.

#[1]" "f'(x_0)=lim_(x->x_0)(f(x)-f(x_0))/(x-x_0)#

#[2]" "f'(4)=lim_(x->4)(f(x)-f(4))/(x-4)#

#[3]" "f'(4)=lim_(x->4)((x^3-3x^2)-(4^3-3(4)^2))/(x-4)#

#[4]" "f'(4)=lim_(x->4)(x^3-3x^2-16)/(x-4)#

#[5]" "f'(4)=lim_(x->4)(cancel((x-4))(x^2+x+4))/cancel(x-4)#

#[6]" "f'(4)=lim_(x->4)(x^2+x+4)#

#[7]" "f'(4)=(4)^2+4+4#

#[8]" "f'(4)=24#

#color(blue)(m=24)#

Now we must get the point of #f(x)# at #x=4#.

#[1]" "y=x^3-3x^2#

#[2]" "y=4^3-3(4)^2#

#[3]" "y=64-48#

#[4]" "y=16#

#color(red)(P(4,16))#

We can find the equation of the tangent line using the point-slope form.

#[1]" "y-y_0=m(x-x_0)#

#[2]" "y-16=24(x-4)#

#[3]" "y-16=24x-96#

#[4]" "y=24x-96+16#

#[5]" "color(pink)(y=24x-80)#