Agostarawz

2022-07-02

How to prove this function has Darboux's property?

$f(x)=\{\begin{array}{ll}\mathrm{cos}(\frac{1}{x})& \text{if}x\ne 0\\ 0& \text{if}x=0\end{array}$

How do I prove this function has Darboux's property? I know it has it because it has antiderivatives, but how do I prove it otherwise, with intervals maybe?

$f(x)=\{\begin{array}{ll}\mathrm{cos}(\frac{1}{x})& \text{if}x\ne 0\\ 0& \text{if}x=0\end{array}$

How do I prove this function has Darboux's property? I know it has it because it has antiderivatives, but how do I prove it otherwise, with intervals maybe?

Ordettyreomqu

Beginner2022-07-03Added 22 answers

Step 1

Take, $a,b\in \mathbb{R}$ with $a<b$. You want to prove that, if y lies between f(a) and f(b), then there is some $c\in [a,b]$ such that $f(c)=y$. If $b<0$ or $a>0$, this is clear, by continuity. If $a<0<b$, take some $n\in \mathbb{N}$ such that $\frac{1}{2\pi n}<b$.

Step 2

Then $f\left([\frac{1}{2\pi n+\pi},\frac{1}{2\pi n}]\right)=[-1,1]$, and therefore there is come $c\in [\frac{1}{2\pi n+\pi},\frac{1}{2\pi n}]$ such that $f(c)=y$. The remaining cases are similar.

Take, $a,b\in \mathbb{R}$ with $a<b$. You want to prove that, if y lies between f(a) and f(b), then there is some $c\in [a,b]$ such that $f(c)=y$. If $b<0$ or $a>0$, this is clear, by continuity. If $a<0<b$, take some $n\in \mathbb{N}$ such that $\frac{1}{2\pi n}<b$.

Step 2

Then $f\left([\frac{1}{2\pi n+\pi},\frac{1}{2\pi n}]\right)=[-1,1]$, and therefore there is come $c\in [\frac{1}{2\pi n+\pi},\frac{1}{2\pi n}]$ such that $f(c)=y$. The remaining cases are similar.

grenivkah3z

Beginner2022-07-04Added 6 answers

Step 1

It's obvious that it has Darboux's property on any interval [a, b] with $0<a<b\vee a<b<0$.

If $a\le 0<b$, then $f(a),f(b)\in [-1,1]$ anyways and there are some $0<{x}_{1}<{x}_{2}<b$ such that $f({x}_{1})=-1$ and $f({x}_{2})=1$. By the previous remark, every value between -1 and 1 (so a fortiori every value between $min\{f(a),f(b)\}$ and $max\{f(a),f(b)\}$) is attained in the interval $[{x}_{1},{x}_{2}]\subseteq (a,b)$.

Step 2

If $a<0\le b$, then $f(a),f(b)\in [-1,1]$ anyways and there are some $a<{x}_{1}<{x}_{2}<0$ such that $f({x}_{1})=-1$ and $f({x}_{2})=1$. For the same reason as the previous case, every value between -1 and 1 is attained in the interval $[{x}_{1},{x}_{2}]\subseteq (a,b)$.

It's obvious that it has Darboux's property on any interval [a, b] with $0<a<b\vee a<b<0$.

If $a\le 0<b$, then $f(a),f(b)\in [-1,1]$ anyways and there are some $0<{x}_{1}<{x}_{2}<b$ such that $f({x}_{1})=-1$ and $f({x}_{2})=1$. By the previous remark, every value between -1 and 1 (so a fortiori every value between $min\{f(a),f(b)\}$ and $max\{f(a),f(b)\}$) is attained in the interval $[{x}_{1},{x}_{2}]\subseteq (a,b)$.

Step 2

If $a<0\le b$, then $f(a),f(b)\in [-1,1]$ anyways and there are some $a<{x}_{1}<{x}_{2}<0$ such that $f({x}_{1})=-1$ and $f({x}_{2})=1$. For the same reason as the previous case, every value between -1 and 1 is attained in the interval $[{x}_{1},{x}_{2}]\subseteq (a,b)$.

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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