Important Theorems on Continuity
Continuity is a fundamental property of functions, ensuring smooth and predictable behavior without breaks or jumps. However, beyond just defining continuity, certain powerful theorems provide deeper insights into how continuous functions behave in different settings. These theorems help answer important questions, such as whether a function reaches its highest and lowest values, whether it takes all intermediate values, and whether it remains bounded.
In this chapter, we explore key theorems related to continuity, including:
- Extreme Value Theorem (EVT): Guarantees that a continuous function on a closed interval attains both a maximum and a minimum.
- Boundedness Theorem: Ensures that a continuous function on a closed interval remains within a fixed range of values.
- Intermediate Value Theorem (IVT): Ensures that a continuous function takes every value between its starting and ending points.
- Bolzano’s Theorem: A special case of IVT that guarantees the existence of a root if the function changes sign.
These theorems have important applications in finding roots of equations, optimization problems, and understanding function behavior. They also highlight why the distinction between closed and open intervals is crucial—certain properties hold only on closed intervals and fail on open ones.
We will not formally prove these theorems here because their proofs require deeper mathematical ideas, such as the completeness property of real numbers, the nested interval property (NIP), and even a more precise definition of continuity itself. These concepts belong to a more advanced branch of mathematics called Real Analysis, which you will encounter in higher mathematics courses. However, we will provide strong reasoning and intuitive explanations to help develop a solid understanding of why these theorems must be true.
Extreme Value Theorem
If a function \( f(x) \) is continuous on a closed interval \( [a, b] \), then it attains both a maximum and a minimum value on \( [a, b] \). That is, there exist points \( c, d \in [a, b] \) such that
This guarantees that the function must reach its highest and lowest values somewhere in the interval.
A function being continuous on a closed interval means that it has no breaks, jumps, or infinite behavior on \( [a, b] \). The key observation is that in a closed interval, the function cannot escape to infinity or have missing values at the endpoints, ensuring that it remains bounded. Since the function is continuous, it cannot avoid reaching its maximum or minimum value somewhere in the interval.
If a function is defined on an open interval \( (a, b) \), it may not attain its maximum or minimum values because the function may become unbounded near the endpoints. Unlike a closed interval, where both endpoints are included, an open interval allows the function to behave unpredictably near \( x = a \) or \( x = b \), potentially approaching infinity or oscillating wildly.
For example, consider the function
At any point in \( (0, \pi/2) \), this function is well-defined and continuous. However, as \( x \to 0^+ \), we have \( \sin x \to 0 \), causing \( 1/\sin^2 x \) to grow arbitrarily large. Similarly, as \( x \to (\pi/2)^- \), \( \cos x \to 0 \), causing \( 1/\cos^2 x \) to diverge to infinity. This means that
Thus, \( f(x) \) does not attain a maximum value on \( (0, \pi/2) \), even though it is continuous. This demonstrates why the Extreme Value Theorem fails for open intervals—a continuous function on an open interval may lack maximum or minimum values due to unbounded behavior near the endpoints.
The Extreme Value Theorem applies only to closed intervals, where continuity ensures that a function is both bounded and attains its extreme values. In contrast, on an open interval, the function may become arbitrarily large or small near the endpoints, preventing it from reaching a maximum or minimum.
Boundedness Theorem
If a function \( f(x) \) is continuous on a closed interval \( [a, b] \), then \( f(x) \) is bounded on \( [a, b] \). This means that there exist real numbers \( m \) and \( M \) such that
This ensures that the function does not diverge to infinity anywhere in the interval.
This theorem is an immediate consequence of the Extreme Value Theorem (EVT). The EVT states that a continuous function on a closed interval attains both a maximum and a minimum value. If \( f(x) \) reaches its highest value at some point \( x = c \) and its lowest value at some point \( x = d \), then for all \( x \in [a, b] \), the function must satisfy
Thus, choosing \( m = f(d) \) and \( M = f(c) \), we see that \( f(x) \) is bounded.
If a function is continuous on an open interval \( (a, b) \), it may not be bounded because it could tend to infinity near the endpoints. For example, consider
This function is continuous in \( (a, b) \) but is unbounded as \( x \to a^+ \) or \( x \to b^- \), since the denominator approaches zero, making \( f(x) \) tend to infinity. This example shows why the Boundedness Theorem requires a closed interval, just like the Extreme Value Theorem.
Intermediate Value Theorem (IVT)
If a function \( f(x) \) is continuous on a closed interval \( [a, b] \) and if \( f(a) \neq f(b) \), then for any value \( k \) between \( f(a) \) and \( f(b) \), there exists at least one \( c \in (a, b) \) such that
This means that a continuous function must take every intermediate value between \( f(a) \) and \( f(b) \) as \( x \) moves from \( a \) to \( b \).
A function being continuous means that it has no jumps or gaps. If \( f(a) < k < f(b) \) (or \( f(a) > k > f(b) \)), the function must pass through \( k \) at some point in the interval. Since there are no sudden jumps, the function cannot skip over \( k \)—it must equal \( k \) at least once.
A useful way to think about IVT is to imagine walking on a hilly road from point \( x = a \) to \( x = b \). If you start at height \( f(a) \) and end at height \( f(b) \), you must have passed through every height between them. You cannot suddenly teleport over certain heights; every intermediate height must be reached somewhere.
Consider the function
We check its values at the endpoints:
Now, let \( k = 3 \). Since \( f(-1) = 0 \) and \( f(2) = 6 \), and \( 3 \) lies between \( 0 \) and \( 6 \), the Intermediate Value Theorem guarantees that there exists some \( c \in (-1,2) \) such that \( f(c) = 3 \), even though we do not yet know the exact value of \( c \).
The Intermediate Value Theorem does not apply if the function is not continuous. For example, consider
On \( [0,2] \), we have \( f(0) = 1 \) and \( f(2) = 4 \), but there is no \( x \) where \( f(x) = 2 \) or \( f(x) = 3 \), since the function jumps at \( x = 1 \). Thus, continuity is essential for IVT to hold.
Thus, we can conclude that the Intermediate Value Theorem ensures that a continuous function cannot skip values. It is particularly useful for proving that equations have solutions and is widely used in root-finding methods.
The Intermediate Value Theorem (IVT) states that if a function \( f(x) \) is continuous on a closed interval \( [a, b] \) and \( f(a) \neq f(b) \), then for any value \( k \) that lies between \( f(a) \) and \( f(b) \), there exists at least one \( c \in (a, b) \) such that
However, IVT does not guarantee anything if \( k \) does not lie between \( f(a) \) and \( f(b) \). If \( k \) is greater than both \( f(a) \) and \( f(b) \) or smaller than both, the theorem does not apply, and we cannot conclude whether \( f(x) \) attains the value \( k \) in \( (a, b) \) or not.
For example, consider \( f(x) = x^3 \) on \( [-1,1] \). Here,
If we take \( k = 0.5 \), IVT guarantees that there exists some \( c \in (-1,1) \) such that \( f(c) = 0.5 \). However, if we ask whether \( f(x) \) attains \( k = 2 \) within \( (-1,1) \), IVT does not apply, even though \( f(x) \) is continuous, because 2 does not lie between \( f(-1) = -1 \) and \( f(1) = 1 \). In this case, the function simply does not reach 2 in the given interval.
Thus, the Intermediate Value Theorem does not claim that a function attains any arbitrary value—it only applies when the value lies between \( f(a) \) and \( f(b) \).
Example
Let \( f(x) \) be a continuous function on \( [a, b] \). Show that the equation
has at least one solution in \( [a, b] \).
Solution:
Since \( f(x) \) is continuous on \( [a, b] \), the Intermediate Value Theorem (IVT) applies. The given equation asks for some \( c \in [a, b] \) such that
Since arithmetic mean of two numbers always lies in the moddle of two numbers, clearly,
Since \( f(x) \) is continuous, by the Intermediate Value Theorem, \( f(x) \) must take every value between \( f(a) \) and \( f(b) \). In particular, it must take the value
at least once in \( [a, b] \). Thus, there exists some \( c \in [a, b] \) such that
This proves the existence of at least one solution.
Example
Let \( f(x) \) be a continuous function on \( [a, b] \). Show that the equation
has at least one solution in \( [a, b] \) for all \( \lambda \in [0,1] \).
Solution:
Since \( f(x) \) is continuous on \( [a, b] \), the Intermediate Value Theorem (IVT) applies. We must show that the given right-hand side value lies between \( f(a) \) and \( f(b) \), ensuring that \( f(x) \) takes this value at some \( x \in [a, b] \).
It can be easily verified that for any \( \lambda \in [0,1] \),
lies between \( f(a) \) and \( f(b) \). Specifically, when \( \lambda = 0 \), the expression equals \( f(a) \), and when \( \lambda = 1 \), it equals \( f(b) \). For any \( 0 < \lambda < 1 \), the expression represents a weighted average of \( f(a) \) and \( f(b) \), ensuring that it stays within the range
Since \( f(x) \) is continuous on \( [a, b] \), the Intermediate Value Theorem guarantees that \( f(x) \) attains every value between \( f(a) \) and \( f(b) \) at least once. In particular, there exists some \( c \in [a, b] \) such that
Thus, the equation has at least one solution for all \( \lambda \in [0,1] \).
Example
Let \( f(x) = x^4 + 1 \). Show that for all \( a, b \in \mathbb{R} \) with \( a < b \), the equation
has at least one solution in \( [a, b] \).
Solution:
Define \( f(x) = x^4 + 1 \). The given equation can be rewritten as
Since \( \sqrt{(a^4+1)(b^4+1)} = \sqrt{f(a) f(b)} \) is the geometric mean of \( f(a) \) and \( f(b) \), it follows from the AM-GM inequality that
This guarantees that the value \( \sqrt{f(a) f(b)} \) lies between \( f(a) \) and \( f(b) \).
Since \( f(x) = x^4 + 1 \) is continuous on \( [a, b] \), by the Intermediate Value Theorem (IVT), \( f(x) \) attains every value between \( f(a) \) and \( f(b) \) at least once in \( [a, b] \). In particular, there exists some \( c \in [a, b] \) such that
which proves the existence of a solution to the given equation in \( [a, b] \).
Bolzano’s Theorem
A direct consequence of the Intermediate Value Theorem (IVT) is Bolzano’s Theorem, which provides a useful criterion for the existence of roots of a continuous function.
If a function \( f(x) \) is continuous on a closed interval \( [a, b] \) and satisfies
then there exists at least one \( c \in (a, b) \) such that
By the Intermediate Value Theorem, if \( f(x) \) is continuous on \( [a, b] \), it must take every value between \( f(a) \) and \( f(b) \). Since \( f(a) \) and \( f(b) \) have opposite signs, one is positive and the other is negative, meaning that somewhere in between, \( f(x) \) must cross zero.
Thus, by IVT, there must exist some \( c \in (a, b) \) such that \( f(c) = 0 \), proving the theorem.
For example:
Consider \( f(x) = x^3 - x - 1 \) on \( [1,2] \). We compute
Since \( f(1) \) and \( f(2) \) have opposite signs, Bolzano’s Theorem guarantees that there exists some \( c \in (1,2) \) such that \( f(c) = 0 \).
Example
Show that the equation
has at least one real solution.
Solution:
Define the function
We want to show that \( f(x) = 0 \) has at least one solution. Evaluating \( f(x) \) at some convenient points:
Since \( e^2 \approx 7.389 \), we see that
Thus, \( f(0) < 0 \) and \( f(2) > 0 \). Since \( f(x) \) is a continuous function (as it is a combination of the exponential function and polynomials, which are continuous everywhere), by Bolzano’s Theorem, there must exist at least one \( c \in (0,2) \) such that
Thus, the given equation has at least one real solution in \( (0,2) \).