Calculus >

## What is the Intermediate Value Theorem?

The intermediate value states that:

If f is continuous over [a,b], and y is a number between f(a) and f(b), then there is a number, m, in the interval [a,b] such that f(c) = y.

In other words, if you have a continuous function and have a particular “y” value, there must be a particular “x” value to match it. With intermediate value theorems, you aren’t looking for a certain solution (a number), you are just proving that a number exists (or doesn’t exist). You do this by:

- Checking for continuity.
- Evaluating the function at some point.

**Example #1**: For the function f(x) = x^{2}, show that there is a number “m” between 2 and 3 such that f(m) = 7.

Step 1: Draw a graph to help you visualize the problem.

This is an optional step, but it’s good to wrap your head around exactly what you are looking for.

Step 2: Check for continuity. *If the function isn’t continuous, you can’t use the intermediate value theorem.* This function is a polynomial, so we can use the theorem.

Step 3: Evaluate the function at the lower and upper values given. For this example, you’re given x=2 and x=3, so:

f(2) = 4

f(3) = 9

7 is between 4 and 9, so there **must **be some number m between 2 and 3 such that f(c) = 7.

*That’s it!*

## Using the Intermediate Value Theorem to Prove Roots Exist

A second application of the intermediate value theorem is to prove that a root exists.

**Sample problem #2**: Show that the function f(x) = ln(x) – 1 has a solution between 2 and 3.

Step 1: Solve the function for the lower and upper values given:

ln(2) – 1 = -0.31

ln(3) – 1 = 0.1

You have both a **negative y value** and a **positive y value**. Therefore, the graph crosses the x axis at some point. A quick look at the graph and you can see this is true:

Step 2: Check that the graph is continuous. The function ln(x) is defined for all values of x>0, so it is continuous on the interval [2,3].

*That’s it!*

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