Differential and Integral Calculus: Continuity

3. Continuity

Table of Content

Limit of a function
Limits and continuity
Properties of continuous functions

In this section we define a limit of a function $f\colon S\to \mathbb{R}$ at a point $x_0$ . It is assumed that the reader is already familiar with limit of a sequence, the real line and the general concept of a function of one real variable.

Limit of a function

For a subset of real numbers, denoted by $S$ , assume that $x_0$ is such point that there is a sequence of points $(x_k)\in S$ such that $x_k\to x_0$ as $k\to \infty$ . Here the set $S$ is often the set of all real numbers, but sometimes an interval (open or closed).

Example 1.

Note that it is not necessary for $x_0$ to be in $S$ . For example, the sequence $x_k = 1/k\to 0$ as $k\to \infty$ in $S=]0,2[$ , and $x_k\in S$ for all $k=1,2,\ldots$ but $0$ is not in $S$ .

Limit of a function

We consider a function $f$ defined in the set $S$ . Then we define the limit of the function $f\colon S\to \mathbb{R}$ at $x_0$ as follows.

Definition 1: Limit of a function

Suppose that $S\subset \mathbb{R}$ and $f\colon S\to \mathbb{R}$ is a function. Then we say that $f$ has a limit $y_{0}$ at $x_{0}$ , and write $\lim_{x \to x_{0}}f(x)=y_{0},$ if, $f(x_{k})\to y_{0}$ as $k\to \infty$ for every sequence $(x_{k})$ in $S\setminus\{x_0\}$ , such that $x_{k}\to x_{0}$ as $k\to \infty$ .

Example 2.

The function $f\colon \mathbb{R} \to \mathbb{R}$ defined by $f(x)=x^2$ has a limit $0$ at the point $x=0$ .

Function $y=x^2$ .

Example 3.

The function $g\colon\mathbb{R}\to \mathbb{R}$ defined by $g(x)= \left\{\begin{array}{rl}0 & \text{ for }x$ does not have a limit at the point $x=0$ . To formally prove this, take sequences $(x_k)$ , $(y_k)$ defined by $x_k=1/k$ and $y_k=-1/k$ for $k=1,2,\ldots$ . Then the both sequences are in $S=\mathbb{R}$ , but $f(x_k)=1$ and $f(y_k)=0$ for any $k$ .

Example 4.

The function $f(x)=x \sin(1/x)$ , $x>0$ does have the limit $0$ at $0$ .

Example 5.

The function $g(x)= \sin(1/x)$ , $x>0$ does not have a limit at $0$ .

One-sided limits

An important property of limits is that they are always unique. That is, if $\lim_{x\to x_0} f(x)=a$ and $\lim_{x\to x_0} f(x)=b$ , then $a=b$ . Although a function may have only one limit at a given point, it is sometimes useful to study the behavior of the function when $x_k$ approaches the point $x_0$ from the left or the right side. These limits are called the left and the right limit of the function $f$ at $x_0$ , respectively.

Limit rules

The following limit rules are immediately obtained from the definition and basic algebra of real numbers.

Theorem 2: Limit rules

Let $c\in \mathbb{R}, \lim_{x\to x_{0}} f(x)=a$ and $\lim_{x\to x_{0}} g(x)=b.$ Then

Example 8.

Finding limits by calculating $f(x_0)$ :

a) $\lim_{x\to 2}(5x-3)=10-3=7.$

b) $\lim_{x\to -2}\frac{3x+2}{x+5} = \frac{-6+2}{-2+5}=-\frac{4}{3}.$

c) $\lim_{x\to 2} \frac{x^2-4}{x-2} = \lim_{x\to 2} \frac{(x+2)(x-2)}{x-2} = \lim_{x\to 2}(x+2) = 4.$

Limits and continuity

In this section, we define continuity of the function. The intutive idea behind continuity is that the graph of a continuous function is a connected curve. However, this is not sufficient as a mathematical definition for several reasons. For example, by using this definition, one cannot easily decide if $\tan(x)$ is a continuous function or not.

For continuity of a function $f$ at a given point $x_0$ , it is required that:

$f(x_0)$ is defined,
$\lim_{x \to x_0} f(x)$ exists (and is finite),
$\lim_{x \to x_0} f(x) = f(x_0)$ .

In other words:

Definition 2: Continuity

A function $f\colon S\to \mathbb{R}$ is continuous at a point $x_{0}\in S$ , if $\lim_{x\to x_{0}}f(x)=f(x_{0}).$ A function $f\colon S\to \mathbb{R}$ is continuous, if it is continuous at every point $x_{0}\in S$ .

Example 1.

Let $c\in \mathbb{R}$ . Functions $f,g,h$ defined by $f(x)=c$ , $g(x)=x$ , $h(x)=|x|$ are continuous at every point $x\in \mathbb{R}$ .

Why? If $x_{k}\to x_{0}$ , then $f(x_{k})=c$ and $\lim_{k\to \infty}f(x_k)= c=f(x_{0})$ . For $g$ , we have $g(x_{k})=x_{k}$ and hence, $\lim_{k\to\infty} g(x_k)=x_{0}=g(x_{0})$ . Similarly, $h(x_{k})=|x_{k}|$ and $\lim_{k\to\infty}h(x_k)= |x_{0}|=h(x_{0})$ .

Continuous functions $y=c$ , $y=x$ and $y=|x|$ .

Continuous functions $y=c$ , $y=x$ and $y=|x|$ .

Example 2.

Let $x_{0}\in \mathbb{R}$ . We define a function $f\colon\mathbb{R}\to \mathbb{R}$ by $f(x)= \left\{\begin{array}{rl}2 & \text{ for }x \lt x_{0}, \\ 3 & \text{ for }x\geq x_{0}.\end{array}\right.$ Then $\lim_{x \to x_{0}^{-}}f(x)=2,\text{ and } \lim_{x \to x_{0}^{+}}f(x)=3.$ Therefore $f$ is not continuous at the point $x_{0}$ .

Some basic properties of continuous functions of one real variable are given next. From the limit rules (Theorem 2) we obtain:

Theorem 3.

The sum, the product and the difference of continuous functions are continuous. Then, in particular, polynomials are continuous functions. If $f$ and $g$ are polynomials and $g(x_{0})\neq 0$ , then $f/g$ is continuous at a point $x_{0}$ .

A composition of continuous functions is continuous if it is defined:

Theorem 4.

Let $f\colon \mathbb{R}\to\mathbb{R}$ and $g\colon \mathbb{R}\to \mathbb{R}$ . Suppose that $f$ is continuous at a point $x_{0}$ and $g$ is continuous at $f(x_{0})$ . Then $g\circ f\colon \mathbb{R}\to \mathbb{R}$ is continuous at a point $x_{0}$ .

Proof.

Let $(x_k)$ be a sequence such that $x_{k}\to x_{0}$ .

Now $f(x_{k})\to f(x_{0})$ , because $f$ is continuous at the point $x_{0}$ .

Because $g$ is continuous at $f(x_{0})$ , we get $\lim_{k\to\infty}(g\circ f)(x_{k})=\lim_{k\to\infty} g(f(x_{k})) = g(f(x_{0}))=(g\circ f)(x_0).$

Hence $g\circ f$ is continous at $x_0$ .

$\square$

Note. If $f$ is continuous, then $|f|$ is continuous.

Why?

Write $g(x):=|x|$ . Then $(g\circ f)(x)=|f(x)|$ .

Note. If $f$ and $g$ are continuous, then $\max (f,g)$ and $\min (f,g)$ are continuous. (Here $\max (f,g)(x):=\max \{f(x),g(x)\}$ .)

Why?

Write $\begin{cases}(a+b)+|a-b|=2\max(a,b), \\ (a+b)-|a-b|=2\min(a,b). \end{cases}$

\text{Function }f(x)= \left\{\begin{array}{rl}2 & \text{ for }x\lt x_{0}, \\
3 & \text{ for }x\geq x_{0}. \end{array}\right. — $\text{Function }f(x)= \left\{\begin{array}{rl}2 & \text{ for }x\lt x_{0}, \\ 3 & \text{ for }x\geq x_{0}. \end{array}\right.$

'}], { strokeColor : colors[0], fontSize : 16, visible : true }); l[1] = board.create('text', [-4, 3, function() { return '

$y=x$ '}], { strokeColor : colors[1], fontSize : 16, visible : false }); l[2] = board.create('text', [-4, 3, function() { return '

$y=|x|$ '}], { strokeColor : colors[2], fontSize : 16, visible : false }); g[0] = board.create('functiongraph', [f0, -6, 6], { visible : true, strokeWidth : 1.5, strokeColor : colors[0], highlight : false }); g[1] = board.create('functiongraph', [f1, -6, 6], { visible : false, strokeWidth : 1.5, strokeColor : colors[1], highlight : false }); g[2] = board.create('functiongraph', [f2, -6, 6], { visible : false, strokeWidth : 1.5, strokeColor : colors[2], highlight : false }); var currentGraph = g[0]; var currentLabel = l[0]; select.on('drag', function() { currentGraph.setAttribute({ visible : false }); currentLabel.setAttribute({ visible : false }); currentGraph = g[select.Value()]; currentLabel = l[select.Value()]; currentGraph.setAttribute({ visible : true }); currentLabel.setAttribute({ visible : true, useMathJax : true }); select.setAttribute({ fillColor : colors[select.Value()] }); board.update(); }); board.fullUpdate(); })(); /* Example 2. */ (function() { var board = JXG.JSXGraph.initBoard('jxgbox16', { boundingbox : [-3.5, 4.5, 3.5, -1.25], showcopyright : false, shownavigation : false}); var xaxis = board.create('axis', [[0, 0], [1, 0]]); xaxis.removeAllTicks(); var yaxis = board.create('axis', [[0, 0], [0, 1]], { drawZero : true, ticks : { majorHeight : 5, minorTicks : 0, ticksDistance : 1.0 } }); yaxis.defaultTicks.ticksFunction = function() { return 1; }; var xtick = board.create('segment', [[1, .05],[1, -.1]], { strokeWidth : 1, strokeColor : 'black', strokeOpacity : .4, highlight : false }); var f = function(x) { return 2; } var g = function(x) { return 3; } board.create('functiongraph', [f, -3.5, 1], { strokeColor : 'black', strokeWidth : 2, highlight : false }); board.create('functiongraph', [g, 1, 3.5], { strokeColor : 'black', strokeWidth : 2, highlight : false }); board.create('point', [1, f(1)], { name : '', fillColor : 'white', strokeColor : 'black', strokeWidth : .5, size : 2, fixed : true, showInfobox : false }); board.create('point', [1, g(1)], { name : '', fillColor : 'black', strokeColor : 'black', strokeWidth : .5, size : 2, fixed : true, showInfobox : false }); var x0 = board.create('text', [1, 0, function() { return '

$x_{0}$ '; }], { useMathJax : true }); board.fullUpdate(); })();

Delta-epsilon definition

The so-called $(\varepsilon,\delta)$ -definition for continuity is given next. The basic idea behind this test is that, for a function $f$ continuous at $x_0$ , the values of $f(x)$ should get closer to $f(x_0)$ as $x$ gets closer to $x_0$ .

This is the standard definition of continuity in mathematics, because it also works for more general classes of functions than ones on this course, but it not used in high-school mathematics. This important definition will be studied in-depth in Analysis 1 / Mathematics 1.

$(\varepsilon,\delta)$ -test:

Theorem 5: $(\varepsilon,\delta)$ -definition

Let $f: S\to \mathbb{R}$ . Then the following conditions are equivalent:

$\lim_{x\to x_0} f(x)= y_0$ ,
For all $\varepsilon> 0$ there exists $\delta >0$ such that if $0 < |x-x_0| < \delta$ , then $|f(x) - y_0|$ for all $x\in S$ .

Proof.

The proof is taken in two steps:

(a) Suppose first that (2) is not true. We show that (1) is not true either.

Because (2) was assumed not to be true, there exists $\varepsilon >0$ such that for any $\delta>0$ , and for some $x_\delta\in S$ we have $0$ but $|f(x_\delta)-y_0| \geq \varepsilon$ .

It follows that $\lim_{k\to\infty} f(x_k)\neq y_0$ for any sequence $(x_k)\in S\setminus\{x_0\}$ such that $\lim_{k\to \infty} x_k=x_0$ . Therefore (1) is not true.

(b) Suppose then that (2) is true. Fix $\varepsilon >0$ .

Now there exists $\delta = \delta_\varepsilon>0$ such that, if $|x-x_0|$ , then $|f(x)-y_0| < \varepsilon$ for any $x\in S$ .

Choose a sequence $(x_k)$ such that $\lim_{k\to\infty} x_k=x_0$ .

Then there exists $N_\delta \in \mathbb{N}$ such that for $k\geq N_\delta$ , we have $|x_k-x_0|$ .

For $k\geq N_\delta$ , it follows that $|f(x_k)-y_0| < \varepsilon$ , and hence $\lim_{k\to\infty}f(x_k)=y_0$ . Therefore (1) is true.

$\square$

Example 3.

From Theorem 3 we already know that the function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = 4x$ is continuous. We can also use the $(\varepsilon,\delta)$ -definition to prove this.

Proof. Let $x_0 \in \mathbb{R}$ and $\varepsilon > 0$ . Now $|f(x) - f(x_0)| = |4x - 4x_0| = 4|x - x_0| < \varepsilon,$ when $|x - x_0| < \delta \text{, where } \delta = \frac{\varepsilon}{4}.$

So for all $\varepsilon > 0$ there exists $\delta > 0$ such that if $|x - x_0| < \delta$ , then $|f(x) - f(x_0)| < \varepsilon$ for all $x \in \mathbb{R}$ . Thus by Theorem 5 $\lim_{x \to x_0} f(x) = f(x_0)$ for all $x_0 \in \mathbb{R}$ and by definition this means that the function $f: \mathbb{R} \to \mathbb{R}$ is continuous.
$\square$

Interactivity. $(\varepsilon, \delta)$ in example 3.

'} ], { strokeColor : 'black', fontSize : 12, fixed : true }) var s = board.create('slider', [[-3, 1], [-1.25, 1], [0.05, 2, 2]], { name : '

$\\varepsilon', unitLabel : '$ ', fillColor : '#bd4444', snapWidth : 0.01, label : { useMathJax : true, strokeColor : '#bd4444', offset : [0, 25] }}); var x1 = board.create('glider', [1/2, 0, xaxis], { name: '

$x_{0}$ ', strokeColor : 'black', strokeWidth : .5, fillColor : '#446abd', showinfobox : false, label : { useMathJax : true, strokeColor : '#446abd', offset : [5, 25] } }); /*board.suspendUpdate();*/ var y1 = board.create('point', [0, function(){return f(x1.X());}], { name : '

$f(x_{0})$ ', size : 2, strokeColor : 'black', strokeWidth : .5, fillColor : '#bd4444', showinfobox : false, highlight : false, label : { useMathJax : true, strokeColor : '#bd4444', offset : [5, 25] }}); var y2 = board.create('point', [0, function(){return f(x1.X())-s.Value();}], { visible : false }); var y3 = board.create('point', [0, function(){return f(x1.X())+s.Value();}], { visible : false }); var z1 = board.create('point', [function(){return y1.Y()/4;}, function(){return y1.Y();}], { visible : false }); var z2 = board.create('point', [function(){return y2.Y()/4;}, function(){return y2.Y();}], { visible : false }); var z3 = board.create('point', [function(){return y3.Y()/4;}, function(){return y3.Y();}], { visible : false }); var v1 = board.create('segment', [z1, y1], { strokeColor : '#bd4444', strokeWidth : 1, highlight : false }); var v2 = board.create('line', [z2, y2], { strokeColor : '#bd4444', dash : 2, strokeWidth : 1, highlight : false }); var v3 = board.create('line', [z3, y3], { strokeColor: '#bd4444', dash : 2, strokeWidth : 1, highlight : false }); var epsilon = board.create('polygon', [function() { return [-.5, y2.Y()]; }, function() { return [2.5, y2.Y()]; }, function() { return [2.5, y3.Y()]; }, function() { return [-.5, y3.Y()]; }], { fillColor : '#bd4444', fillOpacity : .3, highlight : false, vertices : { visible : false }, borders : { visible : false }}); var h1 = board.create('segment', [function() { return x1; }, function() { return z1; }], { strokeColor : '#446abd', strokeWidth : 1, highlight : false }); var h2 = board.create('segment', [function() { return [z2.X(), 0]; }, function() { return [z2.X(), 8]}], { strokeColor : '#446abd', dash : 2, strokeWidth : 1, highlight : false }); var h3 = board.create('segment', [function() { return [z3.X(), 0]; }, function() { return [z3.X(), 8]}], { strokeColor : '#446abd', dash : 2, strokeWidth : 1, highlight : false }); var delta = board.create('polygon', [h2.point1, h2.point2, h3.point2, h3.point1], { fillColor : '#446abd', fillOpacity : .3, highlight : false, vertices : { visible : false }, borders : { visible : false }}); var txt = board.create('text', [-2.5, .7, function() { return '

$\\delta = \\epsilon/4 = ' + (s.Value()/4).toFixed(3) + '$ '; }], { strokeColor : '#446abd', useMathJax : true, fixed : true }); /*board.unsuspendUpdate();*/ board.fullUpdate(); })();

Example 4.

Let $x_{0}\in \mathbb{R}$ . We define a function $f\colon\mathbb{R}\to \mathbb{R}$ by $f(x)= \left\{\begin{array}{rl}2 & \text{ for }x \lt x_{0}, \\ 3 & \text{ for }x \geq x_{0}.\end{array}\right.$ In Example 2 we saw that this function is not continuous at the point $x_0$ . To prove this using the $(\varepsilon,\delta)$ -test, we need to find some $\varepsilon > 0$ and some $x_\delta \in \mathbb{R}$ such that for all $\delta > 0$ , $|x_\delta - x_0| < \delta$ , but $|f(x_\delta) - f(x_0)| > \varepsilon$ .

Proof. Let $\delta > 0$ and $\varepsilon = 1/2$ . By choosing $x_\delta = x_0 - \delta /2$ , we have $0 < |x_\delta-x_0| = |x_0 - \frac{\delta}{2} + x_0| = \frac{\delta}{2} < \delta,$ and $|f(x_\delta) - f(x_0)| = |2 - 3| = 1 > \varepsilon.$ Therefore by Theorem 5 $f$ is not continuous at the point $x_{0}$ .
$\square$

Interactivity. $(\varepsilon, \delta)$ in example 4.

', fillColor : '#446abd', label : { useMathJax : true, strokeColor : '#446abd' }}); var x1 = board.create('glider', [0, 0, xaxis], { name:'

$x_{0}$ ', strokeWidth : .3, strokeColor : 'black', fillColor : '#446abd', showinfobox : false, highlight : false, fixed : true, label : { useMathJax : true, offset : [5, 25], strokeColor : '#446abd' }}); board.suspendUpdate(); var x2 = board.create('point', [ function(){return x1.X()-s.Value();}, 0], { visible : false }); var x3 = board.create('point', [function(){return x1.X()+s.Value();},0], { visible : false }); var y1 = board.create('point', [-2, function() { return f(x1.X()); }], { size : 2, name: '

$f(x_{0})$ ', strokeWidth : .3, strokeColor : 'black', fillColor : '#446abd', showinfobox : false, highlight : false, label : { useMathJax : true, offset : [5, 25], strokeColor : '#446abd' } }); var yd = board.create('point', [-2, function() { return f(x2.X()); }], { size : 2, name: '

$f(x_{\\delta})$ ', strokeWidth : .3, strokeColor : 'black', fillColor : '#bd4444', showinfobox : false, highlight : false, label : { useMathJax : true, offset : [5, 25], strokeColor : '#bd4444' } }); /* doesn't really do anything atm... */ var dist = function() { if (Math.abs(f(x1)-f(x2.X())) != 0 || Math.abs(f(x1)-f(x3.X())) != 0) { return .5; } else { return 0; } }; var y2 = board.create('point', [0, function() { return y1.Y()-dist(); }], { visible : false }); var y3 = board.create('point', [0, function() { return y1.Y()+dist(); }], { visible : false }); var endpoint1 = board.create('point', [0, 2], { name : '', fixed : true, size : 2, fillColor : 'white', strokeWidth : .5, strokeColor : 'black', showinfobox : false }); var endpoint2 = board.create('point', [0, 3], { name : '', fixed : true, size : 2, fillColor : 'black', strokeWidth : .5, strokeColor : 'black', showinfobox : false }); var v1 = board.create('segment', [x1, function() { return [x1.X(), y1.Y()]; }], { strokeColor : '#446abd', strokeWidth : 1, highlight : false }); var v2 = board.create('line', [x2, function() { return [x2.X(), x2.Y()+1]; }], { strokeColor : '#446abd', dash : 2, strokeWidth : 1, highlight : false }); var v3 = board.create('line', [x3, function() { return [x3.X(), x3.Y()+1]; }], { strokeColor : '#446abd', dash : 2, strokeWidth : 1, highlight : false }); var delta = board.create('polygon', [function() { return [x2.X(), -5]; }, function() { return [x2.X(), 6]; }, function() { return [x3.X(), 6]; }, function() { return [x3.X(), -5]; }], { highlight : false, fixed : true, vertices : { visible : false }, borders : { visible : false }, fillColor : '#446abd', fillOpacity : .2 }); var h1 = board.create('segment', [y1, function() { return [x1.X(), y1.Y()]; }], { strokeColor : '#446abd', strokeWidth : 1, highlight : false }); var h2 = board.create('line', [function() { return y2; }, function() { return [y2.X()+1, y2.Y()]; }], { strokeColor : '#bd4444', dash : 2, strokeWidth : 1, highlight : false }); var h3 = board.create('line', [function() { return y3; }, function() { return [y3.X()+1, y3.Y()]; }], { strokeColor : '#bd4444', dash : 2, strokeWidth : 1, highlight : false }); var epsilon = board.create('polygon', [[-5, 3.5],[10, 3.5],[10, 2.5], [-5, 2.5]], { highlight : false, fixed : true, vertices : { visible : false }, borders : { visible : false }, fillColor : '#bd4444', fillOpacity : .2 }); var txt = board.create('text', [4.2, -1.5, function() { return '

$\\epsilon = 1/2$ '/*Math.max(Math.abs(y2.Y() - y1.Y()), Math.abs(y1.Y() - y3.Y())).toFixed(2)*/; }], { strokeColor: '#bd4444', useMathJax : true, fixed : true }); board.unsuspendUpdate(); board.fullUpdate(); })();

Properties of continuous functions

This section contains some fundamental properties of continuous functions. We start with the Intermediate Value Theorem for continuous functions, also known as Bolzano's Theorem. This theorem states that a function that is continuous on a given (closed) real interval, attains all values between its values at endpoints of the interval. Intuitively, this follows from the fact that the graph of a function defined on a real interval is a continuous curve.

Theorem 6: Intermediate Value Theorem

If $f\colon [a,b]\to \mathbb{R}$ is continuous and $f(a) \lt s \lt f(b)$ , then there is at least one $c\in ]a,b[$ such that $f(c)=s$ .

Proof.

Let $S:=\{x\in [a,b]: f(x)\lt s\}$

We note that $S$ is not an empty set because $a\in S$ , and $S\subset [a,b]$ .

Hence there must exist $c:=\sup(S)\in [a,b]$ .

If $(x_k)$ is a sequence of points in $S$ and $\lim_{k\to \infty}x_k=c$ , then $f(c)= \lim_{k\to \infty}f(x_{k})$ , because $f$ is continuous.

By the definition of $c$ , we have $f(c) \leq s$ .

For $k>1/(b-c)$ , we have $1/k < b-c$ and, hence $c+1/k\in [a,b]\setminus S$ because $a\leq c < b$ .

Again, because $f$ is continuous, $f(c) = \lim_{k\to \infty} f(c+1/k)$ .

As $c < c+1/k$ for all $k>0$ , it follows from the definition of $c$ that $s \leq f(c)$ .

Thus $s\leq f(c)\leq s$ , or $f(c) = s$ . The theorem is proved.

$\square$

Interactivity. Theorem 6.

The Intermediate Value Theorem.

Example 1.

Let function $f:\mathbb{R} \to \mathbb{R}$ , where $f(x) = x^5 - 3x - 1.$ Show that there is at least one $c \in \mathbb{R}$ such that $f(c) = 0$ .

Solution. As a polynomial function, $f$ is continuous. And because $f(1) = 1^5 - 3 \cdot 1 - 1 = -3 < 0$ and $f(-1) = (-1)^5 - 3 \cdot (-1) - 1 = 1 > 0,$ by the Intermediate Value Theorem there is at least one $c \in ]-1, 1[$ such that $f(c) = 0$ .

Example 2.

Let $f(x)=x^3-x=x(x^2-1)=x(x-1)(x-1)$ .

By the Intermediate Value Theorem we have $f(x)$ for $x$ or $0 \lt x \lt 1$ . Similarly, $f(x)>0$ for $-1 \lt x \lt 0$ or $1 \lt x$ , because:

$f(x)=0$ if and only if $x=0$ or $x=\pm 1$ , and
$f(-2)0, f(1/2)$ and $f(2)>0$ .

'; }], { anchor : l, strokeColor : 'red', fontSize : 13, fixed : true }); var fleft = board.create('text', [.2, 4.4, function() { return '

$f(a)$ '}], { strokeColor : 'black', fontSize : 13, fixed : true }); var fright = board.create('text', [9.1, 7.6, function() { return '

$f(b)$ '}], { strokeColor : 'black', fontSize : 13, fixed : true }); var xleft = board.create('text', [1.1, -.1, function() { return '

$a$ ' }], { strokeColor : 'black', fontSize : 13, fixed : true }); var xright = board.create('text', [9.1, -.1, function() { return '

$b$ ' }], { strokeColor : 'black', fontSize : 13, fixed : true }); l.on('drag', function() { if(l.point1.Y() >= 7) { l.point1.moveTo([0, 7]); l.point2.moveTo([1, 7]); } else if(l.point1.Y() <= 4) { l.point1.moveTo([0, 4]); l.point2.moveTo([1, 4]); } }); var intersections = []; intersections[0] = board.create('intersection', [l, g, 0], { name : '

$f(c)=s$ ', showinfobox : false, label : { fontSize : 13, offset : [0, -15], strokeColor : 'blue', strokeWidth : .5} }); intersections[1] = board.create('intersection', [l, g, 1], { name : '

$f(c)=s$ ', showinfobox : false, label : { fontSize : 13, offset : [8, 22], strokeColor : 'blue', strokeWidth : .5} }); intersections[2] = board.create('intersection', [l, g, 2], { name : '

$f(c)=s$ ', showinfobox : false, label : { fontSize : 13, offset : [0, -15], strokeColor : 'blue', strokeWidth : .5} }); board.unsuspendUpdate(); })(); /* Example 1. */ (function() { var board = JXG.JSXGraph.initBoard('jxgbox17', { boundingbox : [-3.5, 2.5, 3.5, -3.5], showcopyright : false, shownavigation : false}); var xaxis = board.create('axis', [[0, 0], [1, 0]], { ticks : { majorHeight : 7, minorTicks : 0, drawZero : true } }); var yaxis = board.create('axis', [[0, 0], [0, 1]], { ticks : { majorHeight : 7, minorTicks : 0, drawZero : true } }); xaxis.defaultTicks.ticksFunction = function() { return 1; }; yaxis.defaultTicks.ticksFunction = function() { return 1; }; var f = function(x) { return (Math.pow(x,5)-3*x-1); } board.create('functiongraph', [f, -3.5, 3.5], { strokeColor : 'black', strokeWidth : 2, highlight : false }); board.fullUpdate(); })(); /* Example 2. */ (function() { var board = JXG.JSXGraph.initBoard('jxgbox18', { boundingbox : [-3.2, 2.9, 3.2, -2.9], showcopyright : false, shownavigation : false}); var xaxis = board.create('axis', [[0, 0], [1, 0]], { ticks : { majorHeight : 7, minorTicks : 0, drawZero : true } }); var yaxis = board.create('axis', [[0, 0], [0, 1]], { ticks : { majorHeight : 7, minorTicks : 0, drawZero : true } }); xaxis.defaultTicks.ticksFunction = function() { return 1; }; yaxis.defaultTicks.ticksFunction = function() { return 1; }; var f = function(x) { return (x*x*x-x); } board.create('functiongraph', [f, -3.2, -1], { strokeColor : 'blue', strokeWidth : 2, highlight : false }); board.create('functiongraph', [f, -1, 0], { strokeColor : 'red', strokeWidth : 2, highlight : false }); board.create('functiongraph', [f, 0, 1], { strokeColor : 'blue', strokeWidth : 2, highlight : false }); board.create('functiongraph', [f, 1, 3.2], { strokeColor : 'red', strokeWidth : 2, highlight : false }); board.fullUpdate(); })();

Next we prove that a continuous function defined on a closed real interval is necessarily bounded. For this result, it is important that the interval is closed. A counter example for an open interval is given after the next theorem.

Theorem 7.

Let $f\colon [a,b]\to \mathbb{R}$ be continuous. Then $f$ is bounded.

Proof.

The proof is based on a counter-assumption: We suppose that $f$ is unbounded.

We need to prove that our counter-assumption leads to a contradiction.

Because $f$ is unbounded, for all $k$ there exists $x_k\in [a,b]$ so that $|f(x_k)|>k$ .

We obtain a bounded sequence $(x_{k})$ , $k=1,\ldots,\infty$ .

The sequence $(x_k)$ has a convergent partial sequence $(x_{k_{j}})$ , $j=1,\ldots,\infty$ .

Let $\lim_{j\to \infty}x_{k_{j}}= c \in [a,b]$ .

Since the function $f$ is continuous, we have $|f(c)|=\lim_{j\to \infty}|f(x_{k_{j}})|>k_{j} \to \infty$ as $j\to \infty$ . This is a contradiction, because $f(c)\in \mathbb{R}$ .

$\square$

Note. If $f\colon ]a,b[\to \mathbb{R}$ is continuous, it can be unbounded.

Example 4.

Let $f\colon ]0,1]\to \mathbb{R}$ , where $f(x)=1/x$ . Now $\lim_{x\to 0+}f(x)=\infty.$

Theorem 8.

Let $f\colon [a,b]\to \mathbb{R}$ be continuous. Then there exist points $c,d\in [a,b]$ such that $f(c)\leq f(x)\leq f(d)$ for all $x\in [a,b]$ , i.e. $f(c)$ is minimum and $f(d)$ is maximum of $f$ on the interval $[a,b]$ .

Proof.

Because function $f$ is bounded, there exists $y=\sup_{x\in [a,b]} f(x)>y-\frac{1}{k}$ .

Now with every $k\in \mathbb{N}$ there exists $x_{k}\in [a,b]$ so that $f(x_k)>y-\frac{1}{k}$ .

We take from the sequence $(x_k)$ , $k=1,2,3,\ldots$ , a convergent partial sequence $(x_{k_{j}})$ where $j=1,2,3,\ldots$ .

Let $\lim_{j\to \infty}x_{k_{j}}=d\in [a,b]$ . Now $y\geq f(d)$ and, because $f$ is continuous, $f(d) = \lim_{j\to \infty} f(x_{k_{j}})\geq \lim_{j\to \infty}(y-\frac{1}{k_j})=y,$ so $f(d)=y$ .

The existence of the minimum is proved by a similar argument.

$\square$

Example 5.

Let $f:[-1,2] \to \mathbb{R}$ , where $f(x) = -x^3 - x + 3.$ The domain of the function is $[-1,2]$ . To determine the range of the function, we first notice that the function is decreasing. We will now show this.

Let $x_1 < x_2$ . Then $x_{1}^3 < x_{2}^3$ and $-x_{1}^3 > -x_{2}^3.$

Because $x_1 < x_2$ , $-x_1^3-x_1 > -x_2^3 -x_2$ and $-x_1^3-x_1 +3 > -x_2^3 -x_2 +3.$ Thus, if $x_1 < x_2$ then $f(x_1) > f(x_2)$ , which means that the function $f$ is decreasing.

We know that a decreasing function has its minimum value in the right endpoint of the interval. Thus, the minimum value of $f:[-1,2] \to \mathbb{R}$ is $f(2) = -2^3 - 2 + 3 = -7.$ Respectively, a decreasing function has it's maximum value in the left endpoint of the interval and so the maximum value of $f:[-1,2] \to \mathbb{R}$ is $f(-1) = -(-1)^3 - (-1) + 3 = 5.$

As a polynomial function, $f$ is continuous and it therefore has all the values between it's minimum and maximum values. Hence, the range of $f$ is $[-7, 5]$ .

Example 6.

Suppose that $f$ is a polynomial. Then $f$ is continuous on $\mathbb{R}$ and, by Theorem 7, $f$ is bounded on every closed interval $[a,b]$ , $a \lt b$ . Furthermore, by Theorem 3, $f$ must have minimum and maximum values on $[a,b]$ .

Note. Theorem 8 is connected to the Intermediate Value Theorem in the following way:

If $f\colon [a,b]\to \mathbb{R}$ be continuous, then there exist points $x_1,x_2\in [a,b]$ such that $f([a,b])=[f(x_1),f(x_2)]$ .