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

Beginner2021-12-14Added 26 answers

The form of the tangent lines must be:

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

Now, substitute the first derivative of the curve for m

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

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

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

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

Now, substitute equation$\left[2\right]$ into equation $\left[1\right]$

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

We have to substitute$y=\frac{x}{x+1}$

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

Solve for x

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

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

${x}^{2}+4x+1$

$x=\frac{-4\pm \sqrt{{4}^{2}-4\left(1\right)\left(1\right)}}{2\left(1\right)}$

$x=-2\pm \sqrt{3}$

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

There are two tangent lines

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

and

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

Now, substitute the first derivative of the curve for m

Now, substitute equation

We have to substitute

Solve for x

There are two tangent lines

and

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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