Question

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

Answer 1

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]}