Mat210 Section 1.4 - The Intermediate Value Theorem
In this section, we will make use of continuity when we show that certain types of functions have solutions, also called "zeros." A handy theorem called the Intermediate Value Theorem (IVT) gives us an idea of when and where we can expect to find solutions to functions. Let's take a look at the theorem:
Let f be a function continuous in the interval [a,b] and assume that f(a) and f(b) have different signs.
Then there is at least one c in the open interval (a,b) such that f(c) = 0.
Wow! This may seem a bit confusing at first, so let's go through one step at a time.
First, f must be continuous in the given interval, so remember that means from section 1.3. This must be true for all values of x in [a,b], which is a closed and bounded interval. In other words, closed and bounded means the endpoints of the interval are included (that's what those brackets mean!) and all points in between are also included.
Next, we need the outputs at the endpoints (a and b) to be different signs. And why do you suppose that is needed? Let's take a look at a graph so we know why these two conditions are so important.
We can see from the graph that one endpoint is above the x-axis and the other is below the x-axis. And since the graph is continuous, which means there are no holes, jumps or breaks, then the graph must pass through the x-axis. That is precisely what the conclusion of the IVT states. If we have these two conditions, then there is a c in the interval (a,b) that will have an output of 0, or y = f(c) = 0. Recall that we call this a root, or zero, since at that particular x value, the graph of the function will pass through the x-axis. It's possible for the function to have more than one root in (a,b), so we will say at least one root.
Cutting to the chase here...
From a common sense perspective, suppose we have a list of numbers, and suppose some of the numbers in your list are negative ( - ) , and some are positive ( + ). And you have a whole lot of numbers, a huge number of values. One of those numbers is bound to be zero, right? So here is the result: if a function is continuous, it takes on every possible y value from start to finish across an interval. If a y value starts out negative, and ends up positive, or the reverse, somewhere in that interval the y value must be zero. And that is what the intermediate value theorem (IVT) is all about.
Use the IVT to determine if f has at least one root on (1,4).
In other words, we need to show that there is at least one solution to the equation f(x) = 0.
(Note: the given interval is an open interval since the root cannot occur on the boundary.)
First we need to show that f is continuous on [1,4].
There are two dicontinuities.There is a hole at x = 0, which means f(0) does not exist. And there is an asymptote at x = -2, which means the limit does not exist at x = -2. But these two discontinuities are not in the given interval [1,4]. So f is continuous on [1,4].
Next we need to check (calculate) the endpoint y values of our closed and bounded interval.
y = f(1) = -1/3
y = f(4) = 1/12
These two outputs do have different signs. So now we can conclude that the IVT will apply since the above conditions hold true. So that means f has at least one root on the interval [1,4]. Graphically this means that there is at least one place where the graph crosses the x axis in the interval. And so it does:
Can you use the IVT to show that f does NOT have a root on (-1,1)? No, all you can show is that the IVT cannot be applied.
a picture says a thousand words. Graph this function. Then look at
the behavior of the graph around x = 0. To illustrate the problem,
this time we show a screen grab of the graph from a TI -84
calculator, and the syntax for getting the graph to look right. While
you do not have to graph this on a graphing calculator, by hand
always works, It is handy.
Sample syntax for entering the function on a TI-84
A good viewing window
The graph of
First, let's check to see if f is a continuous function. Consider the following limits:
Since the left and right hand limits are not equal, that means the function is not continuous at x = 0. We can see this clearly on the graph as well. Since the graph is not continuous in the interval (-1,1), the IVT does not apply.
Botton line: remember that if a function is not continuous on the interval being examined, the IVT does not apply. It's not necessarily the case that there is not a root on (-1,1). We just can't prove it using the IVT.
Now you try one! Show that the function f has at least one root on (1,3).
Last Update: July 15, 2008 la copyright 2007 (c) Sharon Walker, Leslie Arce and the Department of Mathematics and Statistics at ASU - all rights reserved