What does the intermediate value theorem mean?

2 Answers
Jan 5, 2016

Answer:

It means that a if a continuous function (on an interval #A#) takes 2 distincts values #f(a)# and #f(b)# (#a,b in A# of course), then it will take all the values between #f(a)# and #f(b)#.

Explanation:

In order to remember or understand it better, please know that the math vocabulary uses a lot of images. For instance, you can perfectly imagine an increasing function! It's the same here, with intermediate you can imagine something between 2 other things if you know what I mean. Don't hesitate to ask any questions if it's not clear!

Jan 5, 2016

Answer:

You could say that it basically says the Real numbers have no gaps.

Explanation:

The intermediate value theorem states that if #f(x)# is a Real valued function that is continuous on an interval #[a, b]# and #y# is a value between #f(a)# and #f(b)# then there is some #x in [a,b]# such that #f(x) = y#.

In particular Bolzano's theorem says that if #f(x)# is a Real valued function which is continuous on the interval #[a, b]# and #f(a)# and #f(b)# are of different signs, then there is some #x in [a,b]# such that #f(x) = 0#.

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Consider the function #f(x) = x^2-2# and the interval #[0, 2]#.

This is a Real valued function which is continuous on the interval (in fact continuous everywhere).

We find that #f(0) = -2# and #f(2) = 2#, so by the intermediate value theorem (or the more specific Bolzano's Theorem), there is some value of #x in [0, 2]# such that #f(x) = 0#.

This value of #x# is #sqrt(2)#.

So if we were considering #f(x)# as a rational valued function of rational numbers then the intermediate value theorem would not hold, since #sqrt(2)# is not rational, so is not in the rational interval #[0, 2] nn QQ#. To put it another way, the rational numbers #QQ# have a gap at #sqrt(2)#.

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The big thing is that the intermediate value theorem holds for any continuous Real valued function. That is there are no gaps in the Real numbers.