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]}