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) = x2, 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.
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].
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