6.2 Examples of functions of a real variable ‣ 6 Summary notes ‣ Block III Functions ‣ Introduction to Mathematics at University

6.2.1 Polynomials

Definition 6.8 (Polynomial, polynomial function).

Suppose $a_{0},a_{1},a_{2},\ldots,a_{n}$ are real numbers, with $a_{n}\neq 0$ . The expression

a_{0}+a_{1}x+a_{2}x^{2}+\cdots+a_{n}x^{n}

is called a polynomial with real coefficients. The numbers $a_{0},a_{1},a_{2},\ldots,a_{n}$ are called coefficients. The integer $n$ is called the degree of the polynomial. The special polynomial $0$ has degree $0$ .

We can use this algebraic expression to define a polynomial function $p:\mathbb{R}\to\mathbb{R}$ given by the formula

p(x)=a_{0}+a_{1}x+a_{2}x^{2}+\cdots+a_{n}x^{n}.

If a polynomial has degree $n$ , we call the coefficient in front of $x^{n}$ the leading coefficient of the polynomial.

Note that there is a subtle difference between a polynomial and a polynomial function. Polynomials (i.e. expressions of the form $a_{0}+a_{1}x+a_{2}x^{2}+\cdots+a_{n}x^{n}$ as above) can be treated as mathematical objects in their own right and are studied in the field of abstract algebra. Whereas, polynomial functions must also have a domain and codomain. In practice, mathematicians are often informal about the difference between a polynomial and polynomial function. This informality rarely causes problems, especially when the domain and codomain are $\mathbb{R}$ , but it is good to keep clear in your mind what sort of mathematical object you are working with.

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A polynomial of degree $0$ defines a constant function.
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A polynomial of degree $1$ defines an affine function. For example, $p:\mathbb{R}\to\mathbb{R}$ where $p(x)=1-5x$ defines an affine function.
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A polynomial of degree $2$ defines a quadratic function. For example, $q:\mathbb{R}\to\mathbb{R}$ with $q(x)=5-2x^{2}$ defines a quadratic function.
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Polynomials of degree $3$ , $4$ and $5$ are respectively known as cubics, quartics and quintics.

The graph of an affine function is a straight line in the plane. However, since these lines need not go through the origin, they do not define what mathematicians call “linear functions” in many other areas (e.g. “linear maps” in linear algebra.) Hence, we use the word “affine” to avoid future confusion.

6.2.2 Manipulating and combining polynomials

Polynomials can be added, subtracted, multiplied and manipulated in general using elementary algebra. For example, to multiply the polynomials $x^{2}+3$ and $2x-5$ you could expand $(x^{2}+3)(2x-5)$ and then collect together terms of the same degree, as follows:

(x^{2}+3)(2x-5)=x^{2}(2x-1)+3(2x-5)=2x^{3}-x^{2}+6x-15.

Exercise 6.9.

Write $4x^{2}-12x+7$ in the form $a(x-b)^{2}+c$ , for constants $a$ , $b$ and $c$ . [This manipulation is known as ‘completing the square’.]

Solution (please try for yourself before looking)

First observe that

a(x+b)^{2}+c=ax^{2}+2abx+(ab^{2}+c).

By comparing this with $4x^{2}-12x+7$ and equating corresponding coefficients, we see that $a=4$ . Then $2ab=-12$ which means $b=-3/2$ . Finally, $ab^{2}+c=7$ . So, after substituting our values for $a$ and $b$ , we obtain $c=-2$ . Therefore,

4x^{2}-12x+7=4\left(x-\frac{3}{2}\right)^{2}-2.

[Expand out the right hand side of this equation to check that it does indeed give you $4x^{2}-12x+7$ ].

In Block 1, we were introduced to the notion of a divisor within the set of real numbers. This notion of a divisor can also be applied to polynomials.

Definition 6.10 (Polynomial divisor).

Let $a_{0}+a_{1}x+a_{2}x^{2}+\cdots+a_{n}x^{n}$ and $b_{0}+b_{1}x+b_{2}x^{2}+\cdots+b_{m}x^{m}$ be polynomials, where $a_{0}+a_{1}x+a_{2}x^{2}+\cdots+a_{n}x^{n}\neq 0$ .

We say the polynomial $a_{0}+a_{1}x+a_{2}x^{2}+\cdots+a_{n}x^{n}$ is a divisor of $b_{0}+b_{1}x+b_{2}x^{2}+\cdots+b_{m}x^{m}$ if

\frac{b_{0}+b_{1}x+b_{2}x^{2}+\cdots+b_{m}x^{m}}{a_{0}+a_{1}x+a_{2}x^{2}+% \cdots+a_{n}x^{n}}

can be written as a polynomial.

Exercise 6.11.

Given that $x+7$ is a divisor of $x^{3}-37x+84$ , factorise $x^{3}-37x+84$ completely as a product of polynomials of degree 1.

Solution (please try for yourself before looking)

First we divide $x^{3}-37x+84$ by $x+7$ , and then factorise to get

\frac{x^{3}-37x+84}{x+7}=x^{2}-7x+12=(x-4)(x-3).

Hence $x^{3}-37x+84=(x+7)(x-4)(x-3)$ .

Polynomial expressions have many written forms. for example,

	Expanded	$\displaystyle ax^{2}+bx+c$
		$\displaystyle a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+\cdots+a_{n}x^{n}$
	Factored	$\displaystyle a(x-b)(x-c)$
		$\displaystyle a(x-r_{1})(x-r_{2})(x-r_{3})\cdots(x-r_{n})$
	Completed square	$\displaystyle a(x-b)^{2}+c$
	$\displaystyle\mbox{About the point }x=a\quad$	$\displaystyle a_{0}+a_{1}(x-a)+a_{2}(x-a)^{2}$
		$\displaystyle a_{0}+a_{1}(x-a)+a_{2}(x-a)^{2}+\cdots+a_{n}(x-a)^{n}$
		$\displaystyle a_{0}+a_{1}x+a_{2}x(x-1)+a_{3}x(x-1)(x-2)+\cdots$
		$\displaystyle+a_{n}x(x-1)(x-2)\cdots(x-(n-1))$
	Nested form	$\displaystyle a_{0}+x(a_{1}+xa_{2})$
		$\displaystyle a_{0}+x(a_{1}+x(a_{2}+x(a_{3}+\cdots x(a_{n-1}+a_{n}x)\cdots)))$

A significant goal of elementary algebra is to transform an expression from one form into an equivalent expression in another form. We can show all of the above forms are equivalent to the expanded form by multiplying out the brackets. More work is needed to re-write an expanded polynomial into one of the other forms.

6.2.3 Rational functions

Definition 6.12 (Rational function).

A rational expression is defined as the ratio of two polynomials, e.g.

\frac{x^{2}-1}{x^{3}+7}.

A rational function is a function defined by a formula that is the ratio of two polynomials. That is $f$ is a rational function if

f(x)=\frac{p(x)}{q(x)},

where $p$ and $q$ are polynomial functions, and $q(x)\neq 0$ .

Note that the implied domain of a rational function $f(x)=p(x)/q(x)$ is all real numbers except for those where the denominator $q(x)$ is zero (in symbols, $\mathbb{R}\setminus\{x\in\mathbb{R}:q(x)=0\}$ ). The codomain of a rational function is usually $\mathbb{R}$ .

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Zeros of $f$ are the values $x$ where $f(x)=0$ . Equivalently, these are the values $x$ where $p(x)=0$ and $q(x)\neq 0$ .
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Poles (or singularities) of $f$ are the values $x$ where denominator $q(x)=0$ .
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If the same factor appears in both the numerator and the denominator, it can cancel out algebraically. This algebraic cancellation might remove values of $x$ for which $q(x)=0$ , in which case the graph has a removable discontinuity (a hole) at that point.

What happens to the rational function $f$ when its input $x$ is large depends on the degrees of $p$ and $q$ . Using $\deg p$ and $\deg q$ to denote the degree of $p$ and $q$ respectively, we have the following:

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If $\deg p<\deg q$ , then $f(x)$ approaches 0 as $x$ approaches $\infty$ or $-\infty$ .
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If $\deg p=\deg q$ , then $f(x)$ approaches the ratio of the leading coefficients as $x$ approaches $\infty$ or $-\infty$ .
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If $\deg p>\deg q$ and the leading coefficient of $p$ is positive then $f(x)$ grows to $\infty$ as $x$ approaches $\infty$ . Otherwise, if the leading coefficient of $p$ is negative, then $f(x)$ descends towards $-\infty$ as $x$ approaches $\infty$ . In both cases, the graph of $f$ looks like a polynomial when $x$ is large (this can be made precise by long division).

This describes the horizontal or oblique asymptotes.

Example 6.13.

Consider

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

The denominator vanishes at $x=2$ , giving a vertical asymptote there.

To better understand the behaviour of $f$ we perform division of $x^{2}-1$ by $x-2$ :

\begin{array}[]{r@{\;}c@{\;}rrrrrr}&&&&&x&+&2\\ \cline{2-8}\cr x-2&\Big{)}&&x^{2}&&&-&1\\ &&&x^{2}&-&2x&&\\ \cline{3-8}\cr&&&&&2x&-&1\\ &&&&&2x&-&4\\ \cline{5-8}\cr&&&&&&&3\\ \end{array}

Hence

\frac{x^{2}-1}{x-2}=x+2+\frac{3}{x-2}.

As $x$ approaches $\infty$ or $-\infty$ , the term $\frac{3}{x-2}$ gets very close to 0, and hence the graph behaves like the line $y=x+2$ for large positive $x$ and large negative $x$ .

Our discussion of the behaviour of a function $f$ as its input approaches $\infty$ or $-\infty$ is very informal here. You will learn about this more rigorously when you study “limits” in the course Introduction to Mathematical Analysis (IMA).

6.2.4 Equivalent Expressions vs Equivalent Functions

In algebra, two expressions are called equivalent if they can be transformed into one another using the rules of algebraic manipulation; for example,

\frac{x^{2}-1}{x-1}\quad\text{and}\quad x+1

are algebraically equivalent because

\frac{x^{2}-1}{x-1}=\frac{(x-1)(x+1)}{x-1}=x+1.

However, when creating functions from these expressions we need to be more careful: a function is defined by its domain, codomain and graph (defined by the expression) and two functions are equal only when their domains, codomains and graphs are the same. The expression $\frac{x^{2}-1}{x-1}$ defines a function whose domain excludes $x=1$ , while the expression $x+1$ defines a function on all real numbers. So although the formulas agree everywhere else, if their corresponding functions are defined on their implied domains, they are technically different functions, because these implied domains are not the same. In abstract algebra we often focus on equivalence of expressions, while in calculus (and later analysis) we distinguish functions by their graph (defined by their rule, expression or formula), domain and codomain.

The following example illustrates the dangers of algebraic cancelling without carefully considering the domains.

Example 6.14.

Consider the following two functions $f:\mathbb{R}\setminus\{6\}\to\mathbb{R}$ and $g:\mathbb{R}\setminus\{13\}\to\mathbb{R}$ where

f(x)=\frac{x+10}{x-6}-5,

and

g(x)=\frac{4\,x-40}{13-x}.

Let’s try to find the point(s) (if any) at which the graphs of these functions cross, that is, we want to find all $x$ for which $f(x)=g(x)$ .

Sketching the graphs of $f$ (red, solid line) and $g$ (blue, dashed line) we see that the graphs appear to cross once. So we expect one solution to $f(x)=g(x)$ .

However, if we try to solve the equation $f(x)=g(x)$ , we get:

	$\displaystyle\frac{x+10}{x-6}-5=\frac{4\,x-40}{13-x}\iff$	$\displaystyle\frac{x+10-5\,\left(x-6\right)}{x-6}=\frac{4\,x-40}{13-x}$
	$\displaystyle\iff$	$\displaystyle\frac{4\,x-40}{6-x}=\frac{4\,x-40}{13-x}$
	$\displaystyle\iff$	$\displaystyle 6-x=13-x$
	$\displaystyle\iff$	$\displaystyle 6=13.$

What went wrong? Can you try to correct it to find the solution to $f(x)=g(x)$ ?

Exercise 6.15.

Identify the error in the above example and correct it to find the solution to $f(x)=g(x)$ .

Solution (please try for yourself before looking)

It is important to note that, since we are looking for solutions to $f(x)=g(x)$ , we must only consider those $x$ that lie in BOTH the domain of $f$ and the domain of $g$ . This means our possible options for solutions are those elements of $\mathbb{R}\setminus\{6,13\}$ .

In the attempt above, we perform the step

\frac{4\,x-40}{6-x}=\frac{4\,x-40}{13-x}\iff 6-x=13-x

by dividing both sides of the equation by $4x-40=4(x-10)$ . But this is 0 when $x=10$ and $10\in\mathbb{R}\setminus\{6,13\}$ . So we must consider $x=10$ as a separate case.

Indeed, if $x=10$ ,

f(x)=\frac{10+10}{10-6}-5=\frac{20}{4}-5=0

and

g(x)=\frac{4(10)-40}{13-10}=0=f(x).

So $x=10$ is a solution to $f(x)=g(x)$ .

Are there any more solutions? Well now, if we consider the case where $x\neq 10$ , we are permitted to divide both sides of our equation by $4x-40\neq 0$ and we end up with the working given in the above example, resulting in the false statement $6=13$ . Thus, there are no solutions except for $x=10$ to the equation $f(x)=g(x)$ , which matches what we see in the graphs of these functions.

6.2.5 Symmetry properties

Definition 6.16 (Periodic function).

Let $D$ be a subset of $\mathbb{R}$ and $f:D\to\mathbb{R}$ be a function. Suppose there is a real number $p>0$ such that $f(x+p)=f(x)$ for all $x\in D$ . Then the function $f$ is said to be $p$ -periodic or periodic with period $p$ .

For example, the function $\sin:\mathbb{R}\to\mathbb{R}$ is $2\pi$ -periodic because $\sin(x+2\pi)=\sin(x)$ for all $x\in\mathbb{R}$ . Note that $\sin$ is also periodic with period $4\pi$ , $6\pi$ , $8\pi$ , and so on.

The graph of a periodic function has translational symmetry in the direction of the $x$ -axis. This means that the shifting the graph along the $x$ -axis by its period produces the same graph. [Check that you understand why from the definition of a periodic function.]

Exercise 6.17.

Is the constant function $f:\mathbb{R}\to\mathbb{R}$ defined by $f(x)=3$ periodic? If so, what period does it have?

Solution (please try for yourself before looking)

The function $f$ is $p$ -periodic for any real number $p>0$ . This is because $f(x)=3=f(x+p)$ for every $p>0$ . In fact this is true for every constant function.

Definition 6.18 (Even function).

Let $D$ be a subset of $\mathbb{R}$ . A function $f:D\to\mathbb{R}$ is said to be even if $f(-x)=f(x)$ for all $x\in D$ .

Example 6.19.

Some examples of even functions are:

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$\cos:\mathbb{R}\to\mathbb{R}$ because $\cos(-x)=\cos(x)$ for all $x\in\mathbb{R}$ .
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$g:\mathbb{R}\to\mathbb{R}$ where $g(x)=x^{2}$ because $g(-x)=(-x)^{2}=x^{2}=g(x)$ for all $x\in\mathbb{R}$ .

The graph of an even function has reflectional symmetry about the $y$ -axis. This means that the part of the graph that lies to the left of the $y$ -axis is the same as the part of the graph that lies to the right of the $y$ -axis, causing the $y$ -axis to act as a mirror. [Check that you understand why this is true from the definition of an even function.]

Definition 6.20 (Odd function).

Let $D$ be a subset of $\mathbb{R}$ . A function $f:D\to\mathbb{R}$ is said to be odd if $f(-x)=-f(x)$ for all $x\in D$ .

Example 6.21.

Some examples of odd functions are:

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$\sin:\mathbb{R}\to\mathbb{R}$ because $\sin(-x)=-\sin(x)$ for all $x\in\mathbb{R}$ ;
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$\tan:\mathbb{R}\setminus\{\frac{\pi}{2}+n\pi:n\in\mathbb{Z}\}\to\mathbb{R}$ because $\tan(-x)=-\tan(x)$ for all $x$ in its domain.
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$f:\mathbb{R}\to\mathbb{R}$ where $f(x)=x^{3}$ since $f(-x)=(-x)^{3}=-x^{3}=-f(x)$ for all $x\in\mathbb{R}$ .

The graph of an odd function has half-turn rotational symmetry about the origin. This means that we can take any straight line $L$ through the origin, take the part of the graph that lies on one side of $L$ and rotate it by $180^{\circ}$ around the origin to match the part of the graph that lies on the other side of $L$ .

6.2.6 Quadratic polynomials, completed square form and symmetry

Any quadratic polynomial can be written in completed-square form:

f(x)=a(x-b)^{2}+c.

Completed-square form makes the parabola’s symmetry immediate: the square $(x-b)^{2}$ depends only on the distance $x-b$ from the vertical line $x=b$ .

Now suppose $h>0$ . At any point $x=b+h$ the value of $f$ is $f(b+h)=ah^{2}+c$ and at the mirror point $x=b-h$ the value of $f$ is

f(b-h)=a(-h)^{2}+c=ah^{2}+c.

So, because the $+h$ and $-h$ enter only through the same squared term, we get $f(b+h)=f(b-h)$ . This shows that points in the graph that are equidistant from $x=b$ have the same height (or $y$ coordinate). Thus the line $x=b$ is the axis of (reflectional) symmetry for our quadratic and $(b,c)$ is known as the unique vertex (highest or lowest point on the graph of $f$ , depending on the sign of $a$ ).

Example 6.22.

Take $f(x)=2x^{2}-8x+5$ and complete the square:

	$\displaystyle 2x^{2}-8x+5$	$\displaystyle=2(x^{2}-4x)+5$
		$\displaystyle=2\left((x-2)^{2}-4\right)+5$
		$\displaystyle=2(x-2)^{2}-8+5$
		$\displaystyle=2(x-2)^{2}-3.$

Therefore the unique vertex is $(2,-3)$ and the axis of symmetry is $x=2$ . Notice that $f(0)=5$ and $f(4)=5$ , so $(0,5)$ and $(4,5)$ are symmetric about the line $x=2$ .

6.2.7 The symmetry of cubic polynomials

Consider a general cubic polynomial with $a\neq 0$ :

p(x)=ax^{3}+bx^{2}+cx+d.

If we translate this horizontally by some $s\in\mathbb{R}$ we get

p(x+s)=a(x+s)^{3}+b(x+s)^{2}+c(x+s)+d

=\left\{\begin{array}[]{ccccccc}ax^{3}&+&3asx^{2}&+&3as^{2}x&+&as^{3}\\ &+&bx^{2}&+&2bsx&+&bs^{2}\\ &&&+&cx&+&cs\\ &&&&&+&d\\ \end{array}\right.

and thus

p(x+s)=ax^{3}+(3as+b)x^{2}+(3as^{2}+2bs+c)x+(as^{3}+bs^{2}+cs+d).

Then, since $p(x)=p((x-s)+s)$ , it follows that

p(x)=a(x-s)^{3}+(3as+b)(x-s)^{2}+(3as^{2}+2bs+c)(x-s)+(as^{3}+bs^{2}+cs+d).

Notice, in passing, that the complicated looking coefficients can be calculated using differential calculus

p(x)=\underbrace{(as^{3}+bs^{2}+cs+d)}_{=p(s)}+\underbrace{(3as^{2}+2bs+c)}_{=% \frac{p^{\prime}(s)}{1!}}(x-s)+\underbrace{(3as+b)}_{=\frac{p^{(2)}(s)}{2!}}(x% -s)^{2}+\underbrace{a}_{=\frac{p^{(3)}(s)}{3!}}(x-s)^{3}.

(6.1)

This holds for any value of $s$ .

Now choose $s$ so that $3as+b=0$ i.e. $s=-\frac{b}{3a}$ . Then the $x^{2}$ coefficient becomes zero and we have

p\left(x\right)=a(x-x^{*})^{3}+\left(c-{\frac{b^{2}}{3a}}\right)(x-x^{*})+% \left(d-{\frac{b\,c}{3\,a}}+{\frac{2\,b^{3}}{27\,a^{2}}}\right)

where $x^{*}=-\frac{b}{3a}$ .

This form has no $x^{2}$ term, and is called the reduced cubic. The point $x^{*}=-\frac{b}{3a}$ is called the point of inflexion¹¹ 1 A point of inflection is a point on a smooth plane curve at which the curvature changes sign. Such points are typically studied in a calculus class. of the cubic.

We performed a horizontal translation by $x^{*}=-\frac{b}{3a}$ . Now we perform a vertical translation by subtracting the constant term. If we translate vertically and horizontally in this particular way, then every cubic polynomial can be written in the form

q(x)=\alpha x^{3}+\beta x,

for some $\alpha,\beta\in\mathbb{R}$ . This defines an odd function $q:\mathbb{R}\to\mathbb{R}$ since

q(-x)=\alpha(-x)^{3}+\beta(-x)=-\alpha x^{3}-\beta x=-(\alpha x^{3}+\beta x)=-% q(x).

This shows that every cubic has half-turn rotational symmetry about the point of inflexion.