Properties Of Continuous Functions

Continuous functions have important properties. For example, if a function is continuous on a closed interval, it attains a maximum value and a minimum value on that interval. This property is very useful when dealing with optimization problems. Continuous functions have the intermediate value property; that is, whenever they take on two values, they also take all values in between. One immediate application of the intermediate value property is an approximate method of finding roots called the bisection method. Also, we will learn later that continuous functions are integrable.

Quick Reference

Theorem	Hypotheses	Conclusion
Bolzano's Theorem	$f$ continuous on $[a, b]$ , $f (a)$ and $f (b)$ have different signs	$\exists c \in [a, b]$ such that $f (c) = 0$ .
Intermediate Value Theorem (IVT)	$f$ continuous on closed interval $[a, b]$ , $k$ between $f (a)$ and $f (b)$	$\exists c \in [a, b]$ such that $f (c) = k$ .
Extreme Value Theorem (EVT)	$f$ continuous on closed interval $[a, b]$	$f$ attains both an absolute maximum and an absolute minimum in $[a, b]$ .

The Intermediate Value Theorem

The following theorem states an important property of continuous functions.

If $f$ is continuous on $[a, b]$ and if $f (a)$ and $f (b)$ have different signs, then there exists a point $c$ in $[a, b]$ such that $f (c) = 0$ .

In other words, if $f$ is a continuous function on $[a, b]$ and if $f (a) > 0$ and $f (b) < 0$ (or conversely if $f (a) < 0$ and $f (b) > 0$ ), then $f$ takes on 0 at least once in that interval. Equivalently, we can say that $f$ has a root (or zero) in that interval.

Geometrically this theorem is intuitive because it merely tells us that the curve of a continuous function, which begins below the $x$ -axis and ends above it, must intersect the $x$ -axis at some point in between (see the following figure).

If f is continuous and f(a)f(b) < 0 then the graph of f cuts the x-axis somewhere between a and b — If $f$ is continuous and $f (a) f (b) < 0$ then the graph of $f$ cuts the $x$ -axis somewhere between $a$ and $b$ .

Bolzano's Theorem guarantees the existence of one zero (or root), but the equation $f (x) = 0$ may have more than one solution (see the following figure).

If f is continuous and f(a)f(b) < 0, Bolzano's theorem assures us that there is at least one solution for the equation f(x) = 0 between a and b, but there may be more than one solution as we see in this figure — If $f$ is continuous and $f (a) f (b) < 0$ , Bolzano's theorem assures us that there is at least one solution for the equation $f (x) = 0$ between $a$ and $b$ , but there may be more than one solution as we see in this figure.

It is possible that $f (a) f (b) > 0$ (that is, the signs of $f (a)$ and $f (b)$ are the same) but $f (x) = 0$ has a solution in $[a, b]$ (see the following figure).

Even if f(a)f(b) loading= — Even if $f (a) f (b) > 0$ , the equation $f (x) = 0$ may have a solution.

Note that if $f$ is discontinuous even at one point in $[a, b]$ , the theorem may not hold anymore. For example, consider the function $f (x) = 1 / (x - 2)$ . Here $f (1) f (3) < 0$ and $f$ is continuous everywhere except at $x = 2$ . Here Bolzano's Theorem does not hold and the graph of $f$ does not intersect the $x$ -axis between 1 and 3 (see the following figure).

Graph of f(x) = 1/(x - 2). Here Bolzano's theorem does not apply because f is discontinuous at x = 2 — Graph of $f (x) = 1 / (x - 2)$ . Here Bolzano's theorem does not apply because $f$ is discontinuous at $x = 2$ .

As we can see from the following figure, if $f (a) f (b) < 0$ , the continuity of $f$ on the open interval $(a, b)$ is not enough to assure us that $f (x) = 0$ has a solution between $a$ and $b$ . To apply Bolzano's Theorem, the left-continuity at $a$ and the right-continuity at $b$ are also required.

Here although f is continuous on the open interval (a,b), because it is not left-continuous at a and right-continuous at b, Bolzano's theorem does not apply — Here although $f$ is continuous on the open interval $(a, b)$ , because it is not left-continuous at $a$ and right-continuous at $b$ , Bolzano's theorem does not apply.

A slight generalization of Bolzano's theorem is called the Intermediate Value Theorem:

Let $f$ be a continuous function on the closed interval $[a, b]$ . If $k$ is a number between $f (a)$ and $f (b)$ , then there exists a point $c$ in $[a, b]$ such that $f (c) = k$ .

In other words, $f$ takes on any given value between $f (a)$ and $f (b)$ . The graph of $f$ between $(a, f (a))$ and $(b, f (b))$ is unbroken and any horizontal line between $y = f (a)$ and $y = f (b)$ intersects the graph of $f$ at least once (see the following figure).

If f is continuous on the closed interval [a,b], for any number k between f(a) and f(b) there is at least one number c between a and b such that f(c) = k — If $f$ is continuous on the closed interval $[a, b]$ , for any number $k$ between $f (a)$ and $f (b)$ there is at least one number $c$ between $a$ and $b$ such that $f (c) = k$ .

Let's consider the function $g$ defined by:

g (x) = f (x) - k .

Because $g$ is a continuous function on $[a, b]$ and has different signs at the two ends of the interval, it follows from Bolzano's theorem that there is a point $c$ in $[a, b]$ such that $g (c) = 0$ or $f (c) = k$ .

As an example of the application of the Intermediate Value Theorem, consider a moving vehicle. If the speedometer shows 100 kilometers per hour, then for any speed $v$ between 0 and 100 km/hr, there must be a time when the speed of the car was exactly $v$ . If you are 5 feet 8 inches tall, there must be a time when you were exactly 5 feet 2.5 inches.

Use Bolzano's Theorem to show that the function $f (x) = x^{3} - 2 x - 1$ has a zero in the interval $[0, 2]$ .

Solution

$f$ is a polynomial, so it is continuous everywhere including on the interval $[0, 2]$ . On the other hand:

f (0) = 0^{3} - 2 (0) - 1 = - 1, f (2) = 2^{3} - 2 (2) - 1 = 3.

Because $f (0)$ and $f (2)$ have opposite signs, the conditions of Bolzano's Theorem are satisfied and we conclude that $f$ has a zero in $[0, 2]$ as shown in the following figure.

Graph of y = x^3 - 2x - 1 — Graph of $y = x^{3} - 2 x - 1$ .

Show that if $f$ is continuous on $[0, 1]$ and if $0 \leq f (x) \leq 1$ for every $x$ in $[0, 1]$ , then there is at least a point $c$ ( $0 \leq c \leq 1$ ) such that $f (c) = c$ .

Solution

If $f (0) = 0$ or $f (1) = 1$ then $c = 0$ or $c = 1$ . If $f (0) \neq 0$ and $f (1) \neq 1$ , we introduce a new function $g$ :

g (x) = f (x) - x .

We have:

g (0) = f (0) - 0 = f (0) > 0

because $f (x) \geq 0$ and $f (0) \neq 0$ , and:

g (1) = f (1) - 1 < 0

because $f (x) \leq 1$ and $f (1) \neq 1$ . Since $g$ is a continuous function on the closed interval $[0, 1]$ and $g (0)$ and $g (1)$ have opposite signs, it follows from Bolzano's Theorem that there is a point $c$ between 0 and 1 such that $g (c) = f (c) - c = 0$ or $f (c) = c$ .

The geometric interpretation is simple. If $f (0) \neq 0$ and $f (1) \neq 1$ , then the graph of $f$ has to cut the line $y = x$ at some point between 0 and 1 (see the following figure).

If f is continuous on [0,1] and 0 <= f(x) <= 1, then the equation f(x) = x has at least one solution (0 <= x <= 1) — If $f$ is continuous on $[0, 1]$ and $0 \leq f (x) \leq 1$ , then the equation $f (x) = x$ has at least one solution ( $0 \leq x \leq 1$ ).

The Extreme Value Theorem

When analyzing the behavior of a function $f$ over a domain $E$ (such as an interval), we are often interested in identifying its extreme outputs, which means the highest output value (the absolute maximum) and the lowest output value (the absolute minimum) attained by $f$ on that domain.

Intuitively:

The absolute maximum corresponds to the highest peak on the graph of $f$ over $E$ . Its value $f (p)$ is greater than or equal to the function's value at any other point $x$ in $E$ .
The absolute minimum corresponds to the lowest point (or valley) on the graph of $f$ over $E$ . Its value $f (q)$ is less than or equal to the function's value at any other point $x$ in $E$ .

Depending on the function and the domain, these extreme values can occur at interior turning points (peaks or valleys) or right at boundary endpoints, as illustrated in the following diagrams.

With this visual intuition in mind, we now state the formal definition.

Let $f$ be a function defined on a set $E$ . We say $f$ has an absolute maximum on (or in) $E$ , if there is at least one point $p$ in $E$ such that:

f (x) \leq f (p) for every x in E .

In this case, we say $p$ is the point of absolute maximum and $f (p)$ is the absolute maximum value (or simply maximum) of $f$ on $E$ .

Similarly we say $f$ has an absolute minimum on $E$ , if there exists a point $q$ in $E$ such that:

f (q) \leq f (x) for every x in E .

In this case, we say $q$ is the point of absolute minimum and $f (q)$ is the minimum value (or simply minimum) of $f$ on $E$ .

The term absolute extremum refers to either absolute maximum or absolute minimum.

Consider the function $f$ defined by:

f (x) = \frac{3}{2} \sqrt{1 - x^{2}} .

The graph of $f$ is sketched in the following figure.

Graph of f(x) = 3/2 sqrt(1-x^2) — Graph of $f (x) = \frac{3}{2} \sqrt{1 - x^{2}}$ . The maximum value of $f$ is $f (0) = 1.5$ and the minimum value of $f$ is $f (1) = f (- 1) = 0$ .

The absolute maximum of $f$ occurs at $x = 0$ (or $x = 0$ is the point of maximum).

The absolute minimum of $f$ occurs at $x = \pm 1$ (or the points of minimum are $x = 1$ and $x = - 1$ ).

The maximum value of $f$ is $f (0) = 1.5$ and its absolute minimum is $f (1) = f (- 1) = 0$ .

If $f$ is continuous on a closed interval $[a, b]$ , then $f$ attains both an absolute maximum and an absolute minimum in $[a, b]$ .

The above theorem states that if $f$ is continuous on $[a, b]$ , then there are numbers $p$ and $q$ in $[a, b]$ such that for all $x$ in $[a, b]$ : $f (q) \leq f (x) \leq f (p) .$
The Extreme Value Theorem is illustrated in the following figure.
It follows from the above theorem that if $f$ is a continuous function on $[a, b]$ then $f$ is bounded on $[a, b]$ . Let $M = f (p)$ and $m = f (q)$ , then: $m \leq f (x) \leq M for all x in [a, b] .$

If f is continuous on a closed interval [a,b], then f attains both its absolute maximum M and absolute minimum m in [a,b] — If $f$ is continuous on a closed interval $[a, b]$ , then $f$ attains both its absolute maximum $M$ and absolute minimum $m$ in $[a, b]$ . That is, there are numbers $p$ and $q$ in $[a, b]$ such that $f (p) = M$ and $f (q) = m$ .

Although the above theorem is intuitively plausible, a proof of this theorem is not within the scope of an elementary course.

The extreme value theorem states two conditions together are sufficient to ensure that a function f has both a minimum and a maximum value on an interval:
1. $I$ is a closed interval.
2. $f$ is continuous at every point of $I$ (for the endpoints, we need left-continuity or right-continuity).
If any of these two conditions fails, the theorem may not hold anymore.
If $f$ is continuous on $(a, b)$ , then whether or not it may have absolute extrema.

Frequently Asked Questions

Does Bolzano's Theorem find the root, or just guarantee one exists?

Bolzano's Theorem is an existence theorem—it guarantees that at least one root $c \in (a, b)$ exists such that $f (c) = 0$ , but it does not provide an explicit algebraic formula for finding $c$ .

However, Bolzano's Theorem forms the mathematical foundation for numerical root-finding algorithms, such as the Bisection Method. By evaluating $f (x)$ at the midpoint of $[a, b]$ and selecting the half-interval where the function values have opposite signs, the interval containing the root can be halved repeatedly to approximate $c$ to any desired precision.

Can a discontinuous function satisfy the conclusion of the Intermediate Value Theorem?

Yes! The Intermediate Value Theorem (IVT) establishes that continuity on a closed interval $[a, b]$ is a sufficient condition for a function to take on every intermediate value between $f (a)$ and $f (b)$ . However, continuity is not a necessary condition.

A discontinuous function can still satisfy the conclusion of the IVT. For instance, the function:

f(x) = \begin{cases} \sin\left(\frac{1}{x}\right) & \text{if } x \neq 0, \\ 0 & \text{if } x = 0 \end{cases}

has an essential discontinuity at $x = 0$ , yet on any interval containing $0$ , it attains every value between $- 1$ and $1$ . Functions that satisfy the intermediate value property (whether continuous or not) are said to possess the Darboux property.

Why must the interval be closed for the Extreme Value Theorem?

The Extreme Value Theorem requires the domain to be a closed and bounded interval $[a, b]$ . If the interval is open or half-open (such as $(a, b)$ or $[a, b)$ ), a continuous function may approach a limit at an open endpoint without ever attaining that value within the interval.

For example:

Consider $f (x) = x$ on the open interval $(0, 1)$ . As $x \to 0^{+}$ , $f (x) \to 0$ , and as $x \to 1^{-}$ , $f (x) \to 1$ . Because $0 \notin (0, 1)$ and $1 \notin (0, 1)$ , $f$ never reaches $0$ or $1$ , so it has no absolute minimum and no absolute maximum on $(0, 1)$ .
Consider $f (x) = \frac{1}{x}$ on $(0, 1]$ . While $f$ is continuous on $(0, 1]$ , as $x \to 0^{+}$ , $f (x) \to + \infty$ , making $f$ unbounded and without an absolute maximum.

What happens if the function is continuous on a closed interval but the interval is infinite?

If the domain is an infinite closed interval—such as $[a, + \infty)$ , $(- \infty, b]$ , or $(- \infty, + \infty)$ —the Extreme Value Theorem is not guaranteed to hold, even if $f$ is continuous at every point. This is because the domain is not bounded.

For example:

The function $f (x) = x$ is continuous on the closed infinite interval $[0, + \infty)$ . It attains an absolute minimum of $0$ at $x = 0$ , but as $x \to + \infty$ , $f (x) \to + \infty$ , so it has no absolute maximum.
The function $f (x) = \arctan x$ is continuous on $(- \infty, + \infty)$ . Its range is $(- \frac{π}{2}, \frac{π}{2})$ . As $x \to \pm \infty$ , $f (x)$ approaches $\pm \frac{π}{2}$ , but it never attains either value. Thus, $f$ has neither an absolute maximum nor an absolute minimum on $ℝ$ .

Exercises

Use Bolzano's Theorem to prove that the polynomial equation $x^{4} - 3 x - 5 = 0$ has at least one real root in the interval $[1, 2]$ .

Answer

Let $f (x) = x^{4} - 3 x - 5$ . $f (1) = - 7 < 0$ and $f (2) = 5 > 0$ . Since $f$ is a polynomial, it is continuous on $[1, 2]$ , so by Bolzano's Theorem, there exists $c \in [1, 2]$ such that $f (c) = 0$ .

Solution

Define the function $f (x) = x^{4} - 3 x - 5$ .

Step 1: Check continuity

Since $f (x)$ is a polynomial function, it is continuous on all of $ℝ$ , and in particular, continuous on the closed interval $[1, 2]$ .

Step 2: Evaluate values at endpoints

Evaluating at $x = 1$ :

\begin{aligned} f(1) &= 1^4 - 3(1) - 5 \\ &= 1 - 3 - 5 \\ &= -7 < 0. \end{aligned}

Evaluating at $x = 2$ :

\begin{aligned} f(2) &= 2^4 - 3(2) - 5 \\ &= 16 - 6 - 5 \\ &= 5 > 0. \end{aligned}
Step 3: Apply Bolzano's Theorem
Since

f (1) < 0

and

f (2) > 0

, the function values have opposite signs at the endpoints. By Bolzano's Theorem, there exists at least one number

c \in (1, 2)

such that

f (c) = 0

, meaning

c^{4} - 3 c - 5 = 0

Show that the equation $\cos x = x$ has at least one solution in the interval $[0, \frac{π}{2}]$ .

Answer

Let $g (x) = \cos x - x$ . $g (0) = 1 > 0$ and $g (\frac{π}{2}) = - \frac{π}{2} < 0$ . By Bolzano's Theorem, there is a solution $c \in [0, \frac{π}{2}]$ .

Solution

Rewrite the equation $\cos x = x$ as $g (x) = 0$ , where:

g (x) = \cos x - x .

Step 1: Check continuity
Both

\cos x

and

x

are continuous functions on

ℝ

, so their difference

g (x)

is continuous on the closed interval

[0, \frac{π}{2}]

Step 2: Evaluate values at endpoints

At $x = 0$ :

\begin{aligned} g(0) &= \cos(0) - 0 \\ &= 1 - 0 \\ &= 1 > 0. \end{aligned}

At $x = \frac{π}{2}$ :

\begin{aligned} g\left(\frac{\pi}{2}\right) &= \cos\left(\frac{\pi}{2}\right) - \frac{\pi}{2} \\ &= 0 - \frac{\pi}{2} \\ &= -\frac{\pi}{2} < 0. \end{aligned}
Step 3: Apply Bolzano's Theorem
Since

g (0) > 0

and

g (\frac{π}{2}) < 0

g (x)

changes sign on

[0, \frac{π}{2}]

. By Bolzano's Theorem, there exists a point

c \in (0, \frac{π}{2})

such that

g (c) = 0

, which implies

\cos c = c

Find the absolute maximum and absolute minimum values of $f (x) = x^{3} - 3 x + 2$ on the closed interval $[0, 2]$ .

Answer

Absolute maximum value: $4$ (at $x = 2$ )

Absolute minimum value: $0$ (at $x = 1$ )

Solution

Since $f (x) = x^{3} - 3 x + 2$ is a polynomial, it is continuous on the closed interval $[0, 2]$ . By the Extreme Value Theorem, $f$ attains an absolute maximum and absolute minimum on $[0, 2]$ .

Step 1: Evaluate endpoints

At $x = 0$ :

f (0) = 0^{3} - 3 (0) + 2 = 2.

At $x = 2$ :

f (2) = 2^{3} - 3 (2) + 2 = 8 - 6 + 2 = 4.

Step 2: Evaluate interior point $x = 1$
Factoring

f (x)

f (1) = 1^{3} - 3 (1) + 2 = 1 - 3 + 2 = 0.

Step 3: Compare values
- The absolute maximum value on

[0, 2]

4

, which occurs at

x = 2

.
- The absolute minimum value on

[0, 2]

0

, which occurs at

x = 1

Consider the function $f (x) = \frac{1}{x - 1}$ on the interval $[0, 2]$ .

(a) Show that $f (0)$ and $f (2)$ have opposite signs.

(b) Does $f (x) = 0$ have any solution in $[0, 2]$ ?

(c) Explain why this result does not contradict Bolzano's Theorem.

Answer

(a) $f (0) = - 1 < 0$ and $f (2) = 1 > 0$

(b) No, $f (x) = \frac{1}{x - 1} \neq 0$ for all $x \in [0, 2]$ .

(c) $f$ is discontinuous at $x = 1 \in [0, 2]$ , so the continuity hypothesis of Bolzano's Theorem fails.

Solution

(a)

Evaluating $f$ at the endpoints $x = 0$ and $x = 2$ :

f (0) = \frac{1}{0 - 1} = - 1 < 0.

f (2) = \frac{1}{2 - 1} = 1 > 0.

Thus, $f (0)$ and $f (2)$ have opposite signs.

(b)

Setting $f (x) = 0$ :

\frac{1}{x - 1} = 0,

which has no real solution because a fraction with a numerator of $1$ can never equal $0$ .

(c)

Bolzano's Theorem requires $f$ to be continuous at every point of the closed interval $[0, 2]$ . Here, $f (x)$ has an infinite discontinuity at $x = 1$ , which lies inside $[0, 2]$ . Because the hypothesis of continuity is violated, Bolzano's Theorem does not apply.

Given the piecewise function on the interval $[- 3, 3]$ :

f(x) = \begin{cases} x^2 + 1 & \text{if } -3 \le x < 0 \\ -(x^2 + 3) & \text{if } 0 \le x \le 3 \end{cases}

(a) Find the values of $f (- 3)$ and $f (3)$ .

(b) Is there any point $c \in [- 3, 3]$ at which $f (c) = 0$ ?

(c) Explain why this result does not contradict Bolzano's Theorem.

Answer

(a) $f (- 3) = 10 > 0$ and $f (3) = - 12 < 0$

(b) No, $f (x)$ is never equal to $0$ anywhere on $[- 3, 3]$ .

(c) $f$ has a jump discontinuity at $x = 0$ , so the continuity hypothesis of Bolzano's Theorem is not satisfied.

Solution

(a)

Evaluating at the endpoints:

For $x = - 3$ (using the top branch $- 3 \leq x < 0$ ):

\begin{aligned} f(-3) &= (-3)^2 + 1 \\ &= 9 + 1 \\ &= 10 > 0. \end{aligned}

For $x = 3$ (using the bottom branch $0 \leq x \leq 3$ ):

\begin{aligned} f(3) &= -(3^2 + 3) \\ &= -(9 + 3) \\ &= -12 < 0. \end{aligned}
(b)
Checking if

f (x) = 0

anywhere on

[- 3, 3]

:
- On

[- 3, 0)

f (x) = x^{2} + 1 \geq 1 > 0

, so

f (x)

is strictly positive.
- On

[0, 3]

f (x) = - (x^{2} + 3) \leq - 3 < 0

, so

f (x)

is strictly negative.
Therefore,

f (x)

never equals

0

at any point in the interval

[- 3, 3]

(c)

Although $f (- 3)$ and $f (3)$ have opposite signs, $f (x)$ has a jump discontinuity at $x = 0$ because:

lim_{x \to 0^{-}} f (x) = 1 while f (0) = lim_{x \to 0^{+}} f (x) = - 3.

Because $f$ is not continuous at $x = 0 \in [- 3, 3]$ , Bolzano's Theorem does not apply.

Does the function $f (x) = \frac{x^{3}}{3} - \cos (π x) + 2$ take on the value $3.5$ within the interval $[- 1, 2]$ ?

Answer

Yes. Since $f$ is continuous on $[- 1, 2]$ with $f (- 1) = 2 \frac{2}{3} \approx 2.67$ and $f (2) = 3 \frac{2}{3} \approx 3.67$ , and $2.67 < 3.5 < 3.67$ , by the Intermediate Value Theorem there exists at least one $c \in [- 1, 2]$ such that $f (c) = 3 \frac{1}{2}$ .

Solution

Step 1: Check continuity

The function $f (x) = \frac{x^{3}}{3} - \cos (π x) + 2$ is a sum of a polynomial and a cosine function, both of which are continuous everywhere on $ℝ$ . Therefore, $f$ is continuous on the closed interval $[- 1, 2]$ .

Step 2: Evaluate values at endpoints

At $x = - 1$ :

\begin{aligned} f(-1) &= \frac{(-1)^3}{3} - \cos(-\pi) + 2 \\ &= -\frac{1}{3} - (-1) + 2 \\ &= -\frac{1}{3} + 3 \\ &= \frac{8}{3} \approx 2.67. \end{aligned}

At $x = 2$ :

\begin{aligned} f(2) &= \frac{2^3}{3} - \cos(2\pi) + 2 \\ &= \frac{8}{3} - 1 + 2 \\ &= \frac{8}{3} + 1 \\ &= \frac{11}{3} \approx 3.67. \end{aligned}
Step 3: Apply the Intermediate Value Theorem
The target value

k = 3.5

lies strictly between

f (- 1) = \frac{8}{3} \approx 2.67

and

f (2) = \frac{11}{3} \approx 3.67

\frac{8}{3} < 3.5 < \frac{11}{3} .

By the Intermediate Value Theorem, there exists at least one point

c \in (- 1, 2)

such that

f (c) = 3 \frac{1}{2}

Show that on any closed interval $[a, b]$ of length greater than $1$ (that is, $b - a > 1$ ), the function $f (x) = x - ⌊ x ⌋$ attains its minimum value but never reaches its maximum value.

(Recall that the floor function $⌊ x ⌋$ is also known as the greatest integer function, often denoted by $⌊ x ⌋$ or $[[x]]$ , which outputs the greatest integer less than or equal to $x$ .)

Answer

The function $f (x) = x - ⌊ x ⌋$ attains its absolute minimum value of $0$ at any integer $n \in [a, b]$ , but its range on $[a, b]$ is $[0, 1)$ . Because $f (x) < 1$ for all $x$ , $f$ never reaches its supremum of $1$ , and thus has no maximum value on $[a, b]$ . This occurs because $f (x)$ is discontinuous at every integer.

Solution

Step 1: Analyze the range of $f (x) = x - ⌊ x ⌋$

In any interval $[n, n + 1)$ where $n$ is an integer, the floor function evaluates to $⌊ x ⌋ = n$ . Thus, on $[n, n + 1)$ :

f (x) = x - n .

As $x$ increases from $n$ to $n + 1$ , $f (x)$ increases strictly from $f (n) = n - n = 0$ up to $1$ , but never reaches $1$ because $x < n + 1$ . As $x \to (n + 1)^{-}$ :

lim_{x \to (n + 1)^{-}} f (x) = lim_{x \to (n + 1)^{-}} (x - n) = 1.

However, $f (x) = x - ⌊ x ⌋ < 1$ for every real number $x$ . Therefore, the range of $f (x)$ on any domain is contained in $[0, 1)$ .

Step 2: Show that $f (x)$ attains its minimum value on $[a, b]$

Since the interval $[a, b]$ has length $b - a > 1$ , it must contain at least one integer $n \in [a, b]$ . At this integer point $n$ :

f (n) = n - ⌊ n ⌋ = n - n = 0.

Since $f (x) \geq 0$ for all $x$ , $0$ is the minimum value of $f$ on $[a, b]$ , and this minimum value is actually attained at $x = n$ .

Step 3: Show that $f (x)$ never reaches its maximum value

Since $b - a > 1$ , the interval $[a, b]$ contains at least one integer $n$ . As $x$ approaches $n$ from the left (i.e. as $x \to n^{-}$ with $x \in [a, b]$ ), the value $f (x)$ gets arbitrarily close to $1$ . Thus, the least upper bound (supremum) of $f (x)$ on $[a, b]$ is $1$ .

However, because $f (x) = x - ⌊ x ⌋ < 1$ for all real numbers $x$ , there is no point $x \in [a, b]$ for which $f (x) = 1$ . Consequently, $f (x)$ never reaches its maximum value on $[a, b]$ .

Step 4: Discontinuity and the Extreme Value Theorem

The figure below shows the graph of $f (x) = x - ⌊ x ⌋$ (the sawtooth function).

The function $f (x)$ has jump discontinuities at every integer $x = n$ , where:

f (n) = 0 and lim_{x \to n^{-}} f (x) = 1.

Because $f$ is not continuous on $[a, b]$ , the Extreme Value Theorem does not apply, which explains why $f$ can fail to attain a maximum value on a closed, bounded interval $[a, b]$ .

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The Intermediate Value Theorem

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