Difference between revisions of "Location of Roots Theorem"
(New page: '''Location of roots theorem''' is one of the most intutively obvious properties of continuos functions, as it states that if a continuos function attains positive and negative values, it ...) |
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| − | ''' | + | The '''location of roots theorem''' is one of the most intutively obvious properties of [[continuous function]]s, 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== | ==Statement== | ||
| − | Let <math>f:[a,b]\rightarrow\mathbb{R} | + | Let <math>f:[a,b]\rightarrow\mathbb{R}</math> be a continuous function such that <math>f(a)<0</math> and <math>f(b)>0</math>. Then there is some <math>c\in (a,b)</math> such that <math>f(c)=0</math>. |
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| − | Then <math> | ||
==Proof== | ==Proof== | ||
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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 | ||
| − | Thus <math>A</math> has a [[ | + | Thus <math>A</math> has a [[least upper bound]], <math>\begin{align}\sup A& =u\in A.\end{align}</math> |
If <math>f(u)<0</math>: | If <math>f(u)<0</math>: | ||
| − | As <math>f</math> is | + | As <math>f</math> is continuous at <math>u</math>, <math>\exists\delta>0</math> such that <math>x\in V_{\delta}(u)\implies f(x)<0</math>, which contradicts (1). |
Also if <math>f(u)>0</math>: | Also if <math>f(u)>0</math>: | ||
| − | <math>f</math> is | + | <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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| + | Hence, <math>f(u)=0</math>. | ||
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==See Also== | ==See Also== | ||
*[[Bolzano's intermediate value theorem]] | *[[Bolzano's intermediate value theorem]] | ||
*[[Continuity]] | *[[Continuity]] | ||
Revision as of 16:57, 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 ![$f:[a,b]\rightarrow\mathbb{R}$](http://latex.artofproblemsolving.com/6/7/c/67c320754a36aaec178e61f04bfcbac41c0bf675.png)
be a continuous function such that 
and 
. Then there is some 
such that 
.
Proof
Let ![$A=\{x|x\in [a,b],\; f(x)<0\}$](http://latex.artofproblemsolving.com/3/e/b/3eb80aa39bfe22273871b801405e939933a91974.png)
As 
, 
is non-empty. Also, as ![$A\subset [a,b]$](http://latex.artofproblemsolving.com/1/d/a/1daac1893d5bbe6f0e79a571e2c9f250b1a8e2fd.png)
, is bounded
Thus has a least upper bound, $\begin{align}\sup A& =u\in A.\end{align}$ (Error compiling LaTeX. Unknown error_msg)
If 
:
As 
is continuous at 
, 
such that 
, which contradicts (1).
Also if 
:
is continuous imples such that 
, which again contradicts (1) by the Gap Lemma.
Hence, 
.
