Difference between revisions of "Location of Roots Theorem"

Revision as of 22:33, 15 February 2008

The location of roots theorem is one of the most intutively obvious properties of continuous functions, as it states that if a continuous function attains positive and negative values, it must have a root (i.e. it must pass through 0).

Statement

Let $f:[a,b]\rightarrow\mathbb{R}$ be a continuous function such that $f(a)<0$ and $f(b)>0$ . Then there is some $c\in (a,b)$ such that $f(c)=0$ .

Proof

Let $A=\{x|x\in [a,b],\; f(x)<0\}$

As $a\in A$ , $A$ is non-empty. Also, as $A\subset [a,b]$ , $A$ is bounded

Thus $A$ has a least upper bound, $\begin{align}\sup A& =u\in A.\end{align}$ (Error compiling LaTeX. Unknown error_msg)

If $f(u)<0$ :

As $f$ is continuous at $u$ , $\exists\delta>0$ such that $x\in V_{\delta}(u)\implies f(x)<0$ , which contradicts (1).

Also if $f(u)>0$ :

$f$ is continuous imples $\exists\delta>0$ such that $x\in V_{\delta}(u)\implies f(x)>0$ , which again contradicts (1) by the Gap lemma.

Hence, $f(u)=0$ .

@@ Line 17: / Line 17: @@
 Also if <math>f(u)>0</math>:
-<math>f</math> is continuous imples <math>\exists\delta>0</math> such that <math>x\in V_{\delta}(u)\implies f(x)>0</math>, which again contradicts (1) by the [[Gap Lemma]].
+<math>f</math> is continuous imples <math>\exists\delta>0</math> such that <math>x\in V_{\delta}(u)\implies f(x)>0</math>, which again contradicts (1) by the [[Gap lemma]].
 Hence, <math>f(u)=0</math>.

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Difference between revisions of "Location of Roots Theorem"

Revision as of 22:33, 15 February 2008

Statement

Proof

See Also