Intersection between $y=\frac{x^2+1}{x+1}$ and $y=x+2$ on $y$-axis

Question

I stumbled upon this problem: Considering the function $f:(-1;\infty)\rightarrow R$, $$f(x)=\frac{x^2+1}{x+1}$$ Find the sum of the values on positive $y$-axis where the tangent of the function is perpendicular to the line $y=x+2$

After letting $(0,a)$ be the point on positive $y$-axis, with the equation of the tangent $y-{{y}_{0}}=f'({{x}_{0}})(x-{{x}_{0}})$ gives slope to be $-1$ So the tangent is $y-a=-1(x-0)=-x$

Is this correct? Can you help me furthermore with this problem?

Answer 1

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

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

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

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

set this to $-1$ as tangent of $f(x)$ is perpendicular to $y=x+2$

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

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

$2x^2+4x =0$

$2x(x+2)=0$

$\therefore x=0,-2$

So at $x=0,-2$ the tangents drawn will be perpendicular to the line $y-x+2$

On the OY axis lies $x=0,y=1$, So at the point $(0,1)$ the tangent of the function $f(x)$ is perpendicular to $y=x+2$

EDIT 1:

graph for reference

Answer 2

The searched equation of the Tangent line has the equation $$y=-x+n$$ to compute $n$ we Need to solve the equation $$-2x^2+x(n-1)+n-1=0$$ where the discriminant should equal to Zero. Doing so we get $$(n-1)^2+8(n-1)=0$$ so $$(n-1)(n+7)=0$$ It is $$(-x+n)(x+1)=x^2+1$$ This equation must have only one Solutions, so the discriminant must be Zero.

Answer 3

First we are searching for all $x\in \Bbb R$, where $$f'(x)=-1\ ,$$ which is the condition to have in $X(x,f(x))$, point on the graph of $f$, a tangent perpendicular to some given line with slope $+1$.

The derivative of $$f(x)=\frac{x^2+1}{x+1}=\frac{x^2-1+2}{x+1} =(x-1)+\frac{2}{x+1}\text{ is } f'(x)=1-\frac 2{(x+1)^2}\ . $$ The equation $f'(x)=-1$ is $-\frac 2{(x+1)^2}=-2$, equivalently $(x+1)^2=1$, so it has exactly two solutions, $0,-2$.

Initial post:

The corresponding tangents through $(0,f(0))=(0,1)$ and $(-2,f(-2))=(-2,-5)$ have slope $-1$, equations $y=-x+1$, respectively $y=-x-7$, they hit $Oy$, (equation $x=0$,) at $$1,\ -7\ ,$$ the sum of these values is $-6$.

Edited, thanks go to Sonkun: The point $-2$ is not in the domain of $f$, so $(0,1)$ is the only point to be considered, it is already on $Oy$, so the required sum of $y$-valuess on $Oy$ is $1$.