What is the Intermediate Value Theorem?
The basic idea behind the intermediate value theorem (IVT) is: suppose you have a segment (between points a and b, inclusive) of a continuous function, and that function crosses a horizontal line. Given these facts, then the intersection of the two lines—point c— must exist. The theorem is used for two main purposes:
- To prove that point “c” exists,
- To prove the existence of roots (sometimes called zeros of a function).
This may seem like an exercise without purpose, but the theorem has many real world applications.
A More Formal Definition
The textbook definition of the intermediate value theorem states that:
If f is continuous over [a,b], and y0 is a real number between f(a) and f(b), then there is a number, c, in the interval [a,b] such that f(c) = y0.
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. The theorem is also stated—a little bit more simply—as that a continuous function takes on all values between f(a) and f(b); there are no gaps or missing values.
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].
A simple real world example of how the theory works: You measure the weight of your new puppy and she is 15 lbs. Ten days later she weights 18 lbs. That gives you two points on your closed interval:
f(a) = f(0) = 15 lbs
f(b) = f(10) = 18 lbs
Let’s say you wanted to pinpoint the moment when your puppy weighed c = 16.5 lbs. The IVT tells you that this point c must exist. It might take you a while to pinpoint that exact time, but you know your time isn’t going to be wasted. This simple example can be extended to any problem in science that involves pinpointing a specific time, weight, or other metric. For example, carbon dating an object typically takes several days to process, so it’s nice to know from the outset that a solution is possible. Invoking the IVT before you start any complicated technical process saves wasting time and resources on hunting down solutions that may or may not exist.
Some more real life examples:
- At any point in time, there are two points diametrically opposite each other on the Earth’s surface that have exactly the same temperature (Devlin, 2007).
- Climatologists use the IVT to make predictions about how rising carbon dioxide levels will affect our planet (CK12).
- Any table that is wobbly because one leg isn’t touching the ground can be stabilized by rotating it (thus, saving napkins, and therefore, trees) (Devlin, 2007).
CK12 (n.d.) Intermediate Value Theorem: Existence of Solutions. Retrieved January 15, 2018 from: https://www.ck12.org/calculus/intermediate-value-theorem-existence-of-solutions/rwa/Ups-and-Downs/
Devlin, K. (2007). How to stabilize a wobbly table. Retrieved January 15, 2018 from: https://www.maa.org/external_archive/devlin/devlin_02_07.html
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