Joan Thompson

2021-12-13

How many tangent lines to the curve $y=\frac{x}{x+1}$ pass through the point (1,2)

Ana Robertson

The form of the tangent lines must be:
$y=m\left(x-1\right)+2$ $\left[1\right]$
Now, substitute the first derivative of the curve for m
$\frac{dy}{dx}=\frac{\frac{d\left(x\right)}{dx}\left(x+1\right)-x\frac{d\left(x+1\right)}{dx}}{{\left(x+1\right)}^{2}}$
$\frac{dy}{dx}=\frac{x+1-x}{{\left(x+1\right)}^{2}}$
$\frac{dy}{dx}=\frac{1}{{\left(x+1\right)}^{2}}$
$m=\frac{1}{{\left(x+1\right)}^{2}}$ $\left[2\right]$
Now, substitute equation $\left[2\right]$ into equation $\left[1\right]$
$y=\frac{x-1}{{\left(x+1\right)}^{2}}+2$ $\left[1.1\right]$
We have to substitute $y=\frac{x}{x+1}$
$\frac{x}{x+1}=\frac{x-1}{{\left(x+1\right)}^{2}}+2$
Solve for x
$x\left(x+1\right)=\left(x-1\right)+2{\left(x+1\right)}^{2}$
${x}^{2}+x=x-1+2{x}^{2}+4x+2$
${x}^{2}+4x+1$
$x=\frac{-4±\sqrt{{4}^{2}-4\left(1\right)\left(1\right)}}{2\left(1\right)}$
$x=-2±\sqrt{3}$
$x=-2+\sqrt{3}$ and $x=-2-\sqrt{3}$
There are two tangent lines
$y=\frac{1}{{\left(-1+\sqrt{3}\right)}^{2}}\left(x-1\right)+2$
and
$y=\frac{1}{{\left(-1-\sqrt{3}\right)}^{2}}\left(x-1\right)+2$

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