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 $c\in \left[f\left({x}_{1}\right),f\left({x}_{2}\right)\right]$ there exists an $x\in U$ such that $f\left(x\right)=c$. 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 $\phi :\left[0,1\right]\to U$ 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 $\varphi \left(0\right)={x}_{1}$ and $\varphi \left(1\right)={x}_{2}$. Its existence is guaranteed by path-connectedness. The composition $f\circ \varphi$ is continuous, maps $\left[0,1\right]$ to $\mathbf{R}$, and satisfies $f\circ \varphi \left(0\right)=f\left({x}_{1}\right)$ and $f\circ \varphi \left(1\right)=f\left({x}_{2}\right)$. There must exist $t\in \left[0,1\right]$ with $f\circ \varphi \left(t\right)=c$: what can you say about $\varphi \left(t\right)$?

Sonia Ayers

Expert

2022-07-11Added 3 answers

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

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