Show that if $f : \mathbb{R} \rightarrow \mathbb{R}$ is continuous, with $f(1)=-1$, $f(3)=28$, there exist $x_1, x_2 \in (1,3)$ such that
$f(x_1)f(x_2)+x_1^3x_2=0$
For the above we have that there exists $x_0 \in (1,3) : f(x_0)=0$, since $f$ is continuous in $[1,3]$ and $f(1)\cdot f(3)<0$ (Bolzano's theorem).
Any ideas on how to continue?
Let $\varphi(x,y)=f(x)f(y)+x^{3}y$, $(x,y)\in[1,3]\times[1,3]$, then $\varphi(1,3)=-25<0$. Now find some $x_{0}\in(1,3)$ such that $f(x_{0})=0$, then $\varphi(x_{0},1)=x_{0}^{3}>0$.
Consider the line $L$ joining $(x_{0},1)$ and $(1,3)$, that is, $L(x)=-\dfrac{2}{x_{0}-1}x+=\dfrac{3x_{0}-1}{x_{0}-1}$, then the map $\xi:[1,x_{0}]\rightarrow[1,3]$ defined by $\xi(x)=\varphi(x,L(x))$ is continuous and $\xi(1)=\varphi(1,3)<0$, $\xi(x_{0})=\varphi(x_{0},1)>0$, so some $x_{1}\in(1,x_{0})$ is such that $\xi(x_{1})=0$. Now $1<x_{2}:=L(x_{1})<3$, and we have $\varphi(x_{1},L(x_{1}))=\varphi(x_{1},x_{2})=f(x_{1})f(x_{2})+x_{1}^{3}x_{2}=0$.
Choosing $x_1=1$ the equation turns into $g(x_2)=0$ where $g(x)=-f(x)+x$. It holds that $g(1)=2$ and $g(3)=-25$. Consequently, there exists $x_2\in(1,3)$ such that $g(x_2)=0$. Done.
