Intermediate Value Theorem

The Intermediate Value Theorem says that if f(x) is continuous on the interval [a, b]  and f(a) < 0 and f(b) > 0 (or f(a) > 0 and f(b) < 0), then there exists a number c in the interval [a, b] such that f(c) = 0.

In plain english, the Intermediate Value Theorem is saying that if a function is continuous and it's positive somewhere and negative somewhere else, then the function has to equal zero at some point between those two values.

The important part of this theorem is to note that f(x) must be continuous. If f(x) isn't continuous, we can't apply the theorem.

The Intermediate Value Theorem is most often used to prove that there is a solution to a given equation. The Intermediate Value Theorem only proves that there is a solution - it doesn't give us any indication of what that solution may be.

For the textbook questions, you'll probably have to use trial and error to find values for which f(x) is positive and for which f(x) is negative. Usually on the midterm and final exams, they give you the appropriate interval - just use the endpoints of the interval they give you.

Example:

Show that there is at least one solution to

Solution:

First, move everything to one side and call that function f(x):

Let

First, notice that f(x) is the sum of functions that are all continuous. So, f(x) is also a continuous function.

Now, (through trial and error) try and find a value of x that gives f(x) > 0 and another value of x that gives f(x) < 0:

So, f(1) > 0

So, f(-2) < 0

Since f(x) is continuous on the interval (-2, 1) and since f(1) > 0 and f(-2) < 0, by the Intermediate Value Theorem, there is a number c in the interval (-2, 1) such that f(c) = 0. And so the equation must have a solution.

Example:

Show that has at least one solution

Solution:

First, move everything to one side and set that equal to f(x). This this case, we'll have:

Now, we use trial and error to find two values of x - one that will make f(x) positive and one that will make f(x) negative:

So, f(0) < 0

Now we need to find a value of of x that makes f(x) > 0:

I don't know what cos(1000) is. But I do know that it's bigger than or equal to -1 (cosine of any number is always between -1 and 1)

So, we've shown that f(1000) > 0

Since f is a continuous function, the Intermediate Value Theorem tells us that there is some value between x = 0 and x = 1000 where f(x) = 0. At this value, we have a solution to

Example:

Show that there is at least one solution to on the interval [0, 1]

Solution:

First, define f(x) by moving everything in the equation above to the left side:

Now, we're trying to show that there's a value between 0 and 1 where f(x) = 0

So, we plug in x = 1 and x = 0 and hope that f(x) is positive at one of these values and negative at the other

So, since f(x) is a continuous function, by the Squeeze Theorem, there is a value c in the interval [0, 1] such that f(c) = 0