Intermediate value theorem: Difference between revisions
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Any [[continuous function]] on an [[interval]] satisfies the [[intermediate value property]]. | Any [[continuous function]] on an [[interval]] satisfies the [[intermediate value property]]. | ||
==Caveats== | |||
===The statement need not be true for a discontinuous function=== | |||
It is possible for a function having a discontinuity to violate the intermediate value theorem. Below is an example, of the function <math>f(x) := (x + 1)\operatorname{sgn}(x)</math> where <math>\operatorname{sgn}</math> is the [[signum function]] and we define it to be zero at 0. | |||
[[File:Intermediatevaluetheoremfailsfordiscontinuousfunction.png|500px]] | |||
Here, we consider the domain <math>[a,b] = [-1,3]</math>, with <math>f(-1) = 0</math> and <math>f(3) = 4</math>, but there is no <math>c \in [-1,3]</math> satisfying <math>f(c) = 1/2</math>, even though <math>1/2 \in [0,4]</math>. | |||
===The function needs to be defined throughout the domain=== | |||
Consider the function <math>f(x) := 1/x</math>. We have <math>f(-1) = -1</math> and <math>f(1) = 1</math>. However, there is no <math>c \in [-1,1]</math> such that <math>f(c) = 1/2</math>. The reason the theorem fails is that <math>f</math> is not defined at the point 0, and hence it is not defined on the domain <math>[-1,1]</math>. | |||
[[File:1byxviolatesintermediatevaluetheorem.png|500px]] |
Latest revision as of 17:05, 15 October 2011
Statement
Full version
Suppose is a continuous function and a closed interval is contained in the domain of (in particular, the restriction of to the interval is continuous). Then, for any between the values and (see note below), there exists such that .
Note: When we say is between and , we mean if and we mean that if .
Short version
Any continuous function on an interval satisfies the intermediate value property.
Caveats
The statement need not be true for a discontinuous function
It is possible for a function having a discontinuity to violate the intermediate value theorem. Below is an example, of the function where is the signum function and we define it to be zero at 0.
Here, we consider the domain , with and , but there is no satisfying , even though .
The function needs to be defined throughout the domain
Consider the function . We have and . However, there is no such that . The reason the theorem fails is that is not defined at the point 0, and hence it is not defined on the domain .