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

1 Answer
Mar 11, 2016

The tangent line in #x_0=4# is:

#y=-3/4x+4#

Explanation:

Given:

#y=f(x)#

the tangent line equation at a certain point #P(x_0,y_0)# is:

#y-y_0=m(x-x_0)#

with slope #m=f'(x_0)#

Therefore we have to find the first derivative of #f(x)#

#f(x)=x/(x-2)^2=g(x)/(h(x))#

Using the Quotient Rule and Chain Rule

#f'(x)=(g'(x)*h(x)-h'(x)*g(x))/[h(x)]^2=#
#=(1(x-2)^2-2(x-2)*x)/[(x-2)^2]^2=(x-2)((x-2-2x)/(x-2)^4)=#
#=color(green)cancel((x-2))(-x-2)/(x-2)^(color(green)cancel(4)^3)=#
#=-(x+2)/(x-2)^3#

#:.f'(x)=-(x+2)/(x-2)^3#

#f'(x_0=4)=-(4+2)/(4-2)^3=-cancel(6)^3/cancel(8)^4=-3/4#

#y_0=f(x_0=4)=4/(4-2)^2=4/4=1#

Therefore

#y-y_0=m(x-x_0)<=> y-1=-3/4(x-4)#

#y=-3/4x+3+1=-3/4x+4#

graph{(y-(x/(x-2)^2))(y+3/4x-4)=0 [-7.96, 17.35, -6.01, 6.65]}