Question

56. How many tangent lines to the curve $y=x /(x+1)$ pass through the point (1,2)? At which points do these tangent lines touch the curve?

Answer 1

There are 2 tangent lines that pass through the point

$y=\frac{1}{(-1+\sqrt{3)^2} } (x-1)+2$

and

$y=\frac{1}{(-1-\sqrt{3)^2} } (x-1)+2$

Explanation:

Given:

$y=\frac{x}{x+1}$

The point-slope form of the equation of a line tells us that the form of the tangent lines must be:

$y=m(x-1)+2$ $[1]$

For the lines to be tangent to the curve, we must substitute the first derivative of the curve for $m$:

$\frac{dy}{dx} =\frac{d(x)}{dx}(x+1)-x^\frac{d(x+1)}{dx}$

$\frac{dy}{dx} =\frac{x+1-x}{(x+1)^2}$

$\frac{dy}{dx}= \frac{1}{(x+1)^2}$

$m=\frac{1}{(x+1)^2}$ $[2]$

Substitute equation [2] into equation [1]:

$y=\frac{x-1}{(x+1)^2}+2$ $[1.1]$

Because the line must touch the curve, we may substitute $y=\frac{x}{x+1}:$

$\frac{x}{x+1}=\frac{x-1}{(x+1)^2}+2$

Solve for x:

$x(x+1)=(x-1)+2(x+1)^2$

$x^2+x=x-1+2x^2+4x+2$

$x^2+4x+1$

$x\frac{-4±\sqrt{4^2-4(1)(1)} }{2(1)}$

$x=-2$ ± $\sqrt{3}$

$x=-2$ ± $\sqrt{3}$ $and$ $x=-2-\sqrt{3}$

There are 2 tangent lines.

$y=\frac{1}{(-1+\sqrt{3)^2} } (x-1)+2$

and

$y=\frac{1}{(-1-\sqrt{3)^2} } (x-1)+2$