Art of Problem Solving

Intermediate Value Theorem: Difference between revisions

New page: '''Bolzano's intermediate value theorem''' is one of the very interesting properties of continous functions. ==Statement== Let <math>f:[a,b]\righarrow\mathbb{R}</math> Let <math>f</math>...
 
m Added grammar and punctuation
 
(3 intermediate revisions by 2 users not shown)
Line 1: Line 1:
'''Bolzano's intermediate value theorem''' is one of the very interesting properties of continous functions.
The '''Intermediate Value Theorem''' is one of the very interesting properties of continous functions.


==Statement==
==Statement==
Let <math>f:[a,b]\righarrow\mathbb{R}</math>
Take a function <math>f</math> and interval <math>[a,b]</math> such that the following hold:


Let <math>f</math> be continous on <math>[a,b]</math>
<math>f:[a,b]\rightarrow\mathbb{R},</math>


Let <math>f(a)<k<f(b)</math>
<math>f</math> is continuous on <math>[a,b],</math>


Then, <math>\exists c\in (a,b)</math> such that <math>f(c)=k</math>
<math>f(a)<k<f(b).</math>
 
Then, <math>\exists c\in (a,b)</math> such that <math>f(c)=k.</math>


==Proof==
==Proof==
Consider <math>g:[a,b]\rightarrow\mathbb{R}</math> such that <math>g(x)=f(x)-k</math>
Consider <math>g:[a,b]\rightarrow\mathbb{R}</math> such that <math>g(x)=f(x)-k.</math>


note that <math>g(a)<0</math> and <math>g(b)>0</math>
Note that <math>g(a)<0</math> and <math>g(b)>0</math>


By [[Location of roots theorem]], <math>\exists c\in (a,b)</math> such that <math>g(c)=0</math>
By the [[Location of roots theorem]], <math>\exists c\in (a,b)</math> such that <math>g(c)=0</math> or <math>f(c)=k.</math>
 
or <math>f(c)=k</math>
<p align=right>QED</p>
<p align=right>QED</p>


Line 23: Line 23:
*[[Continuity]]
*[[Continuity]]
*[[Location of roots theorem]]
*[[Location of roots theorem]]
[[Category:Analysis]]
[[Category:Theorems]]

Latest revision as of 17:08, 1 February 2021

The Intermediate Value Theorem is one of the very interesting properties of continous functions.

Statement

Take a function $f$ and interval $[a,b]$ such that the following hold:

$f:[a,b]\rightarrow\mathbb{R},$

$f$ is continuous on $[a,b],$

$f(a)<k<f(b).$

Then, $\exists c\in (a,b)$ such that $f(c)=k.$

Proof

Consider $g:[a,b]\rightarrow\mathbb{R}$ such that $g(x)=f(x)-k.$

Note that $g(a)<0$ and $g(b)>0$

By the Location of roots theorem, $\exists c\in (a,b)$ such that $g(c)=0$ or $f(c)=k.$

QED

See Also