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=(1)/((-1+√(3)^2) ) (x-1)+2`

and

`y=(1)/((-1-√(3)^2) ) (x-1)+2`

Step-by-step explanation:

Given:

`y=(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`:

`(dy)/(dx) =(d(x))/(dx)(x+1)-x^(d(x+1))/(dx) \\ \\`

`(dy)/(dx) =(x+1-x)/((x+1)^2)`

`(dy)/(dx)= (1)/((x+1)^2)`

`m=(1)/((x+1)^2)`
`[2]`

Substitute equation [2] into equation [1]:

`y=(x-1)/((x+1)^2)+2`
`[1.1]`

Because the line must touch the curve, we may substitute
`y=(x)/(x+1):`

`(x)/(x+1)=(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(-4±√(4^2-4(1)(1)) )/(2(1))`

`x=-2` ±
`√(3)`

`x=-2` ±
`√(3)`
`and`
`x=-2-√(3)`

There are 2 tangent lines.

`y=(1)/((-1+√(3)^2) ) (x-1)+2`

and

`y=(1)/((-1-√(3)^2) ) (x-1)+2`