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### The Intermediate Value Theorem

```
Date: 09/14/98 at 05:42:14
From: BOOK
Subject: Intermediate Value Theorem

I don't understand about the "Intermediate Value Theorem" at all. Can
you please explain it to me? Also for the function:

f(x) = 1/(x-3) + 2,    x not equal to 3

verify that f(0) < f(5), find the value of f(x) such that there's no x
value in the interval where x is greater or equal than 0, but less than
or equal to 5 such that the function takes this value. For the last
question, can you explain why this case appears to be a counterexample
to the Intermediate Value Theorem. I really need your help.

Thank you very much for your kindness.
```

```
Date: 09/14/98 at 08:41:58
From: Doctor Anthony
Subject: Re: Intermediate Value Theorem

We will examine the Intermediate Value Theorem using the equation:

f(x) = 1/(x-3) + 2

First check f(0) and f(5):

f(0) = 1 and 2/3
f(5) = 2.5

So f(0) < f(5) as required.

Now we need to find a value of f(x), say c, such that there's no x
value in the interval [0,5] where the function takes this value. In
other words, there is no value x in [0,5] such that f(x) = c.

The curve has an asymptote at y = 2, and the graph does not cross the
line y = 2. This is because there is no value of x such that
1/(x-3) = 0. Thus there is no value of x (except +/- infinity) where
f(x) = 2.

The Intermediate Value Theorem would suggest that if f(0) = 1.666 and
f(5) = 2.5, then there should be a value of x between x = 0 and x = 5
for which f(x) equals 2, since 2 is a value between f(0) and f(5). But
we have seen that this is not so. The reason is that f(x) is not
continuous in the interval 0 < x < 5. Because one of the conditions
before you apply the Intermediate Value Theorem is that the function
is continuous in the interval concerned, the Intermediate Value
Theorem does not apply to this case, and so, this example is not
really a counterexample.

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 09/14/98 at 12:43:27
From: Doctor Teeple
Subject: Re: Intermediate Value Theorem

Hi,

I'd like to expand a bit on interpreting the Intermediate Value
Theorem.

According to the CRC Concise Encyclopedia of Mathematics, the
Intermediate Value Theorem states that if a function f is continuous
on a closed interval [a,b] and c is between f(a) and f(b) inclusive,
then there exists at least one x in [a,b] such that f(x) = c. So
another condition to check is that f(a) < f(b). Often, before trying
to prove a theorem, it is a good idea to try examples to convince
yourself that the theorem is true. For example, examine the following
graph:

We start by picking two points on the x-axis, a and b, and noting f(a)
and f(b). Then pick a point of the curve in the closed interval. We
could pick either of the endpoints or a point in between. We label
this point c. Since we know that [a,b], [f(a),f(b)] are closed and
that f is continuous, we know, by the Intermediate Value Theorem, that
there exists a point x in [a,b] such that f(x) = c. If you examine the
graph, you can find x quite easily.

Try playing around with these conditions to see what happens if one or
more of them fail. For example, what if we took an open interval
(a,b)? Could you find a graph where the conclusion of the Intermediate
Value Theorem fails?

In your example, f(x) = 1/(x-3) + 2. Here is the graph:

We can note right away that we are relaxing the condition that f(x) is
continuous. So we know that the Intermediate Value Theorem will not
apply because all of the conditions have not been met. Let's keep
going to see how the theorem may fail.

With your example, we start by picking our two points as 0 and 5,
noting that f(0) = 1.6667 and f(5) = 2.5.

From the graph, we can see that there is an asymptote at y = 2. Note
that 2 is in [1.6667, 2.5], and we know that there is no x value in
[0, 5] such that f(x) = 2.

Thus the Intermediate Value Theorem fails with this example when we
relax the condition of continuity.

- Doctor Teeple, The Math Forum
Check out our web site! http://mathforum.org/dr.math/
```
Associated Topics:
High School Analysis

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