antennense

Answered

2022-07-09

Let $f$ be a function such that $f:U\subseteq {\mathbb{R}}^{n}\to \mathbb{R}$. $U$ is open in ${\mathbb{R}}^{n}$ and path connected and $f$ is continuous. Let ${x}_{1},{x}_{2}\in U$. Proof that for all $c\in [f({x}_{1}),f({x}_{2})]$ there exists an $x\in U$ such that $f(x)=c$. I'm supposed to use one-dimensional intermediate value theorem to proof this. There is a hint stating that I should look out for a function $\phi :[0,1]\to U$ such that we use a "useful" composition of $f$ and $\phi $. I really don't know how to do this proof I would appreciate help a lot!

Answer & Explanation

Alexis Fields

Expert

2022-07-10Added 14 answers

You could let $\varphi $ be a continuous function satisfying $\varphi (0)={x}_{1}$ and $\varphi (1)={x}_{2}$. Its existence is guaranteed by path-connectedness. The composition $f\circ \varphi $ is continuous, maps $[0,1]$ to $\mathbf{R}$, and satisfies $f\circ \varphi (0)=f({x}_{1})$ and $f\circ \varphi (1)=f({x}_{2})$. There must exist $t\in [0,1]$ with $f\circ \varphi (t)=c$: what can you say about $\varphi (t)$?

Sonia Ayers

Expert

2022-07-11Added 3 answers

you need to find a point $x\in U$ satisfying $f(x)=c$. So far there is a point $t\in [0,1]$, guess i`m right

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