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Difference between revisions of "Location of Roots Theorem"

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As <math>a\in A</math>, <math>A</math> is non-empty. Also, as <math>A\subset [a,b]</math>, <math>A</math> is bounded
 
As <math>a\in A</math>, <math>A</math> is non-empty. Also, as <math>A\subset [a,b]</math>, <math>A</math> is bounded
  
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Thus <math>A</math> has a [[least upper bound]], <math>\begin{align}\sup A& =u\in A.\end{align}</math>
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Thus <math>A</math> has a [[least upper bound]], <math>\sup A = u \in A.</math>
  
 
If <math>f(u)<0</math>:
 
If <math>f(u)<0</math>:
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Also if <math>f(u)>0</math>:
 
Also if <math>f(u)>0</math>:
  
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<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]].
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<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>.
 
Hence, <math>f(u)=0</math>.
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*[[Bolzano's intermediate value theorem]]
 
*[[Bolzano's intermediate value theorem]]
 
*[[Continuity]]
 
*[[Continuity]]
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[[Category:Theorems]]

Latest revision as of 14:06, 5 June 2018

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, $\sup A = u \in A.$

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, $f(u)=0$.

See Also

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