Calculus I, Chapter 3: The Average Rate of Change and Limits
Previous Section: 3.10 Infinite Limits
The Intermediate Value Theorem (IVT) is a theorem which states that a continuous function on an interval cannot skip values. This should seem obvious. If the function has no breaks, holes, asymptotes or otherwise, the function should not skip any values.
That is to say, if I were to drive 500 miles. There is a point at which I am 500 miles away, and a point at which I am 0 miles away. Therefore, there is also a point that I am 203 miles away, or any other number between 0 and 500.
A Close Look at the IVT
A visual could look something like the following:
We are able to claim that any y-values between e and f must occur for the function between the x-values of a and b. For example, if we choose an arbitrary y-value, y=d, we can say that there is some point c between a and b so that f(c)=d.
In a really simplified way, we are saying that y-values inside of the box correspond to x-values that are also inside of the box, as long as the function doesn’t skip any values, because it is continuous.
Below is the formal definition for the Intermediate Value Theorem written two ways. First, using a logical shorthand, and secondly written out in words:
(f(x) continuous on [a, b] ) ∧ ( f(a) ≠ f(b) )) →(∀M ∈(f(a), f(b)) ∃c ∈(a, b) s.t. f(c) = M
If f(x) is continuous on a closed interval [a,b] and f(a) ≠ f(b), then for every value M between f(a) and f(b), there exists at least one value c between a and b such that f(c) = M.
This theorem is useful for showing the existence of certain values. For example, showing the existence of an irrational number or an obscure value for a difficult equation. The Intermediate Value Theorem is also especially useful for finding roots of unwieldy equations. With modern day technology, this is easy to repeat in order to get more precise intervals for estimation.
These could be structured in variety of ways. Let’s talk about some informal and formal methods of demonstrating the Intermediate Value Theorem.
Informally
The Intermediate Value Theorem demonstrates an idea that is easy to understand and is fairly obvious. In light of that, many problems can be solved pretty easily. In cases where a function is clearly continuous and the interval of interest clearly contains the point we can simply state the conclusion.
Example 1: Does the function f(x)=x^2+2 contain the value 300?
Solution:
x^2+2 is a continuous, quadratic function with a minimum of 2, so it tends to infinity as x increases. Therefore, the value 300 is an output of the function.
Example 2:
Formally
Many areas of advanced mathematics contain formal proofs in order to demonstrate a theorem or show that a formula is true. The Intermediate Value Theorem can serve as a starting point for us to use formal proof in an obvious setting. Let us practice some more complicated examples using a more formal approach.
Example 3: Show that for all integers n, sin(nx) = cosx for some x ∈[0,π]
Solution:
Let f(x) = sin(nx) – cos(x)
f(x) is continuous on the interval since both sine and cosine are continuous functions.
Let x ∈[0,π]
f(x) ∈ [-1, 1] and 0 ∈ [-1, 1]
By the IVT ∃c∈(0,π) such that f(c) =0.
Thus, sin(nx)-cos(x) =0
Finally, sin(nx) = cos(x)
Take a minute to notice the structure of this proof:
Before the proof begins we identify a strategy. The problem as is doesn’t have a specific value we are searching for, only that the two expressions are equal. By subtracting one expression from both sides we have that the difference between them is 0. So, our goal is to show that sin(nx)-cos(x) =0.
We start by identifying the function to be used.
We take a moment to state that it is continuous and why.
State the interval for the input values.
We identify the range of the function and state that the value of interest is contained in that range (the function goes from -1 to 1, and 0 is in between -1 and 1).
This is enough to satisfy the requirements of the Intermediate Value Theorem, so there is a value in the input range that produces the output we want.
We have shown that our goal is true.
Thus, the problem we set out to demonstrate is also true.
Having a specific structure when approaching proofs that are similar is very helpful in giving yourself the tools to succeed. Following this structure will work on almost any Intermediate Value Theorem problem and provides a scaffolding or skeleton to work with.
Example 4: Show that ∛2 exists and estimate it
Let f(x) = x^3
f(x) is continuous on (-∞,∞)
Let x ∈ [1,3]
Then f(x) ∈ [1,29]
2 ∈ [1,29] Thus, by the IVT ∃c∈(1,3) such that f(c) =2.
So c3 = 2, and c = ∛2 and 1< ∛2 <3
In this example we are able to estimate the cube root of 2, but this is not particularly helpful thanks to modern day technology. It should be noted that problems like these, for estimating a value like this, are not particularly useful but are helpful in terms of understanding and applying the theorem. Recall that our goal here, as with any topic really, is to learn and practice a theorem, but a real application would look vastly different.