4.9 Properties of continuous functions

In this section, we learn

If a function is continuous on a closed interval and has opposite signs at the endpoints, it must be zero somewhere in between.
If a function $f$ is continuous on a closed interval $[a, b]$ , it takes on any value between $f (a)$ and $f (b)$ at some point in the interval $(a, b)$ .
If a function is continuous on a closed interval, it has both a maximum and a minimum on that interval.

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The Intermediate Value Theorem

The following theorem states an important property of continuous functions.

Theorem (Bolzano’s Theorem): 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 Figure 1).

Figure 1: 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 Figure 2).

Figure 2 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 Figure 3)

Figure 3 Even if $f (a) f (b) > 0$ , the equation of $f (x) = 0$ 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 Figure 4).

Figure 4 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 Figure 5, 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.

Figure 5 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:

Theorem (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.

Show that the Intermediate Value Theorem is a direct result of Bolzano's theorem

Let consider the function $g$ defined by $g (x) = f (x) - k .$ Because $g$ is a continuous function on $[a, b]$ and have 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 $f (c) = k$ .

A few applications in everyday life:

As an example of the application of the Intermediate Value Theorem, consider a moving vehicle. If the speedometer shows 100 kilometer 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.

Example 1

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 Figure. 6

Figure 6 Graph of $y = x^{3} - 2 x - 1$

Example 2

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 figures).

Figure 7: 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

Definition 1: 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 all $x$ 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.

For example, consider the function $f$ defined by $f (x) = \sqrt{1 - x^{2}} .$ The absolute maximum of $f$ occurs at $x = 0$ and its absolute minimum occurs at $x = \pm 1$ . The maximum value of $f$ is $f (0) = 1$ and its absolute minimum is $f (1) = f (- 1) = 0$ . The graph of $f$ is sketched in the following figure.

Figure 7 Graph of $f (x) = \sqrt{1 - x^{2}}$ . The maximum value of $f$ is $f (0) = 1$ and the minimum value of $f$ is $f (1) = f (- 1) = 0$ .

Theorem (Extreme Value Theorem) 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 Figure 8.

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 $m = f (q)$ , so for all $x$ in $[a, b]$ $m \leq f (x) \leq M$


(a)	(b)	(c)

Figure 8: 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 extremum
The above theorem guarantees the existence of extreme-values, but it does not tell us anything about how to find them. Later on, we will develop some tools to help us find the extreme values of functions.

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The Intermediate Value Theorem

Show that the Intermediate Value Theorem is a direct result of Bolzano's theorem

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The Extreme Value Theorem