# 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 on 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.

## 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.

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 $f(c)=k$.

• 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.

Use Bolzano's Theorem to show that the function $f(x)=x^{3}-2x-1$ has a zero in the interval $[0,2]$. $f$ is a polynomial, so it is continuous everywhere including on the interval $[0,2]$. On the other hand, $f(0)=0^{3}-2(0)-1=-1,\quad f(2)=2^{3}-2(2)-1=3.$ 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 $y=x^{3}-2x-1$

Show that if $f$ is continuous on $[0,1]$ and if $0\leq f(x)\leq1$ for every $x$ in $[0,1]$, then there is at least a point $p$ ($0\leq p\leq1$) such that $f(c)=c$. If $f(0)=0$ or $f(1)=1$ then $c=0$ or $c=1$. If $f(0)\neq0$ and $f(1)\neq1$, we introduce a new function $g$ $g(x)=f(x)-x.$ We have $g(0)=f(0)-0=f(0)>0$ because $f(x)\geq0$ and $f(0)\neq0$ and $g(1)=f(1)-1<0$ because $f(x)\leq1$ and $f(1)\neq1$. 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 $f(0)\neq0$ and $f(1)\neq1$, then the graph of $f$ has to cut the line $y=x$ at some point between 0 and 1 (see the following figures).

## The Extreme Value Theorem

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 $x=\pm1$. 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 $[a,b]$ $f(q)\leq f(x)\leq f(p).$
• 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 $[a,b]$ $m\leq f(x)\leq M$

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 the entire points of $I$ (for the end points, 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.