Intermediate Value Theorem: Difference between revisions

Latest revision as of 17:08, 1 February 2021

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

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 $f(c)=k.$

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

@@ Line 2: / Line 2: @@
 ==Statement==
-Let <math>f:[a,b]\rightarrow\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==
-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>
-By [[Location of roots theorem]], <math>\exists c\in (a,b)</math> such that <math>g(c)=0</math>
+Note that <math>g(a)<0</math> and <math>g(b)>0</math>
-or <math>f(c)=k</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>
 <p align=right>QED</p>