Title: Default for 2.7 Some Common Functions
Course: HELM Workbooks 2023-2024

Polynomial functions

A very important type of function is the polynomial. Polynomial functions are made up of multiples of non-negative whole number powers of a variable, such as \(3x^2\), \(−7x^3\) and so on. You are already familiar with many such functions. Other examples include:

Polynomial functions

The class of the polynomial functions plays a very important role. Polynomial functions are made up of multiples of non-negative whole number powers of a variable, such as \(3x^2\), \(7x^3\) and so on. You are already familiar with many such functions. Other examples include:

- \(P_0(t)=6\),
- \(P_1(t) = 3t+ 9\),
- \(P_2(x) = 3x^2−x+ 2\),
- \(P_4(z) = 7z^4+z^2−1\),

where \(t\), \(x\) and \(z\) are independent variables.

If the powers of the independent variable are fractional or negative, then the function does not represent a polynomial; i.e. \(f(x)=x^{-1}\) and \(g(x)=x^{3/2}\) are not polynomials. The function \(P_0(t)=6\) is a polynomial instead, in fact we can regard it as \(P_0(t)=6 t^0\).

By convention, a polynomial is written with the powers either increasing or decreasing. For example the polynomial

\(3x+ 9x^2−x^3+ 2\)

would be written as

\(−x^3+ 9x^2+ 3x+ 2\)         or         \(2 + 3x+ 9x^2−x^3\).

In general, we have the following definition:

Key Point

A polynomial expression has the form:

\[a_n x^n+a_{n−1} x^{n−1}+a_{n−2} x^{n−2}+...+a_2 x^2+a_1 x+a_0\]

where \(n\) is a non-negative integer, a_n, a_{n−1}, ..., a_1, a_0 are constants and \(x\) is a variable.

A polynomial function \(P(x)\) has the form

\[P_n(x)=a_n x^n+a_{n−1} x^{n−1}+a_{n−2} x^{n−2}+...+a_2 x^2+a_1 x+a_0.\]

The degree of a polynomial or polynomial function is the value of the highest power. Referring to the examples listed above, the polynomial \(P_2\) has degree \(2\), because the term with the highest power is \(3x^2\), \(P_4\) has degree \(4\), \(P_1\) has degree \(1\) and \(P_0\) has degree \(0\). Polynomials with low degrees have special names given in the following table. We assume that \(a \neq 0\) in each case.

 | degree | name
\(a\) | 0 | constant
\(ax+b\) | 1 | linear
\(ax^2+bx+c\) | 2 | quadratic
\(ax^3+bx^2+cx+d\) | 3 | cubic
\(ax^4+bx^3+cx^2+dx+e\) | 4 | quartic

Q: Exercise

Typical graphs of some polynomial functions are shown in the following figure. The first graph shows the lines \(Q_1(x)=-x+4\) and \(P_1(x)=2x+3\), the second one the functions \(P_2(x)=-x^2+3\) and \(Q_2(x)=-x^2+2x\), in the last one we have the plots of \(P_3(x)=x^3\) and \(Q_3(x)=-x^3+7x-6\).

[Image: The graphs of the functions listed in the previous paragraph]

State for each polynomial graph if it is odd or even, and if it has a periodicity.

1. \(P_1\): [...][...],
\(Q_1\): [...][...]
2. \(P_2\): [...][...] ,
\(Q_2\): [...][...]
3. \(P_3\): [...][...] ,
\(Q_3\): [...][...]
Also offered: Periodic
A: Neither odd nor even; Non periodic; Neither odd nor even; Non periodic; Even; Non periodic; Neither odd nor even; Non periodic; Odd; Non periodic; Neither odd nor even; Non periodic

Exercise

Select the correct description associated with each of the following functions.

1. \([value]\)
2. \([value]\)
3. \(\displaystyle[value]\)
4. \([value]\)
5. \(\displaystyle[value]\)
Feedback: A polynomial is a function of the form \(P_n=a_n x^n+a_{n-1} x^{n-1}+...+a_2 x^2+a_1 x+a_0\); its degree is given by the highest of the degrees of the polynomial's monomials with non-zero coefficients. For the given functions: 1. \([value]:\) [value], in fact, \(4\) represents the highest power of \(x\). 2. \([value]:\) [value]; in this case, we have the constant function, so we can interpret it as \(h(x)=[value]x^0\) and we can see explicitly that the degree is zero. 3. \(\displaystyle[value]:\) [value].This function does not represent a polynomial because of the inverse powers of x, \(\dfrac{[value]}{[value]}\) and \(\dfrac{[value]}{[value]}\). 4. \([value]:\) [value]. The highest of the degrees of the monomials is \(3\), the dominant monomial is \([value][value]^3\). 5. \(\displaystyle[value]:\) [value]. Same case as \([value]\).

Rational functions

A rational function is formed by dividing one polynomial by another. Examples include:

\(R_1(x)=\dfrac{x+6}{x^2+1}\),      \(R_2(t)=\dfrac{t^3-1}{2t+3}\),      \(R_3(z)=\dfrac{2z^2+z-1}{z^2+z-2}\)  .

For convenience we have labelled these rational functions \(R_1\), \(R_2\) and \(R_3\).

Key Point

A rational function has the form:

\[R(X)=\dfrac{P(x)}{Q(x)}\]

where \(P\) and \(Q\) are polynomial functions and \(Q)\) is nonzero.

\(P\) is called the numerator and \(Q\) is called the denominator.

The graphs of rational functions can take a variety of different forms and can be difficult to plot by hand when you first encounter them. We will see later how to analyse these functions but in the short term the use of a graphics calculator or computer software can help. If you have access to a plotting package or calculator it would be useful to obtain graphs of these functions for yourself. In what follows we explore some of the features of the graphs.

Example

Given the rational function \(R_1(x)=\dfrac{x+2}{x^2+1}\) and its graph in the picture, answer the following questions.

[Image]

1. For what values of \(x\), if any, is the denominator zero?
2. For what values of \(x\), if any, is the denominator negative?
3. For what values of \(x\) is the function negative?
4. What is the value of the function when \(x\) is zero?
5. What happens to the function as \(x\) gets larger and larger?

Solution

1. \(x^2+ 1\) is never zero.
2. \(x^2+ 1\) is never negative, it is always positive.
3. Only when the numerator \(x+ 2\) is negative which is when \(x\) is less than \(−2\).
4. \(2\), because that is when the numerator \(x+ 2 = 0\).
5. \(R_1\) approaches zero because the \(x^2\) term in the denominator becomes very large. (This is seen by substituting larger and larger values e.g. \(10,100,1000...\)).

Note that for large \(x\) values the graph gets closer and closer to the \(x\)-axis. We say that the \(x\)-axis is a horizontal asymptote of this graph.

Answering questions such as (a) to (c) above will help you to sketch graphs of rational functions.

Exercise

Consider the rational function function \(R_2(t)=\dfrac{t^3-1}{2t+3}\), and its graph. The solid curve represents the rational function, while the dotted line is an asymptote.

[Image: 2.7.2.2 R(t)]

1. What is the function value when \(t= 1\)?
2. What is the value of the denominator when \(t=−\dfrac{3}{2}\)?
3. What do you think happens to the graph of the function when \(t=−\dfrac{3}{2}\)?
Feedback: 1. If we substitute \(t=1\) in \([value]\), we have that \(t^3-1=1-1=0\), then \([value]\). 2. If we substitute \(t=−\dfrac{3}{2}\) in the denominator of our rational function, we have \(2t+3=2*(-\dfrac{3}{2})+3=-3+3=0\). 3. Note that we must exclude the value \(t=−\dfrac{3}{2}\) from the domain of this function because division by zero is not defined. At this point, as you can see, the graph shoots off towards very large positive values (we say it tends to positive infinity) if the point is approached from the left, and towards very large negative values (we say it tends to negative infinity) if the point is approached from the right. The dotted line in the graph of \(R_2(x)\) has equation \(t=−\dfrac{3}{2}\). It is approached by the curve as \(t\) approaches \(−\dfrac{3}{2}\) and is known as a vertical asymptote.

Exercise

Consider the graph and the algebraic form of the rational function \(R_3(z)= \dfrac{2z^2+z-1}{(z-1)(z+2)}\).

[Image]

1. What happens to the function when \(z=-2\) and \(z=1\)?
2. Which values of \(z\) should be excluded from the domain of this function?
3. Substitute some values for \(z\) (e.g. \(10, 100...\)). What happens to \(R_3\) as \(z\) becomes larger and larger (namely, as \(z\) goes to infinity)?
4. We can observe that the function \(R_3(z)\) has an horizontal asymptote. The equation of this asymptote is: \(y=\)
5. What is the name given to the lines \(z=1\) and \(z=−2\)?
Feedback: 1. When \(z=-2\) and \(z=1\) the value of the function blows up to infinity because the denominator goes to zero (if we substitute \(z=-2\), for example, in \((z-1)(z+2)=(-2-1)(-2+2)=-3*0=0\)) . To be more precise, when we approach \(z=-2\) from the left and \(z=1\) from the right, the function approaches +∞. If instead we approach \(z=-2\) and \(z=1\) from the right and from the left respectively, the value of the function goes to \(- \infty\). 2. The points \(z=-2\) and \(z=1\) are not in the domain since they represent singularities for the function (the denominator of the function is zero in these cases so the value of the function explodes). 3. If you look at the graph you can notice that for positive and negative large values of \(z\) (\(z \to \pm \infty\)), the value of the function tends at a constant value. 4. The constant value of the previous item is \(y=2\). This line is called to be an horizontal asymptot for the function. 5. The lines \(z=1\) and \(z=−2\) are called vertical asymptotes.

The previous exercises are intended to give you some guidance so that you will be able to sketch rational functions yourself. Each function must be looked at individually but some general guidelines are given in the following KeyPoint.

Key Point

Sketching rational functions

- Find the value of the function when the independent variable is zero. This is generally easy to evaluate and gives you a point on the graph.
- Find values of the independent variable which make the denominator zero. These values must be excluded from the domain of the function and give rise to vertical asymptotes.
- Find the values of the independent variable which make the dependent variable zero. This gives you points where the graph cuts the horizontal axis (if at all).
- Study the behaviour of the function when the independent variable is large and positive and when it is large and negative.
- Are there any vertical or horizontal asymptotes? (Oblique asymptotes may also occur but these are beyond the scope of this section).

Exercise

It is particularly important for engineers to find values of the independent variable for which the denominator is zero. These values are known as the poles of the rational function.

Find the poles of the following rational functions (in case the function has more than one pole, write the solution in the form {pole1, pole2,...}, including the curly braces):

1. \(\displaystyle[value]\) has pole \(t=\) .
2. \(\displaystyle[value]\) has poles \(s=\) .
3. \(\displaystyle[value]\) has poles \(x=\) .
4. \(\displaystyle[value]\) has poles \(x=\) .
Feedback: The poles of a rational function are the points for which the denominator becomes zero and, hence, the function diverges. The poles of the rational functions given are: 1. \(\displaystyle[value]\), in fact for this value of \(t\) we have that the denominator is zero: \((t-[value])=([value]-[value])=0\) 2. \(\displaystyle[value]\), in fact if we substitute in the denominator the value \(s=[value]\) or \(s=-[value]\), the denominator is zero. 3. \(\displaystyle[value]\), if we substitute in the denominator \(x=-[value]\) or \(x=-([value])\), the denominator becomes zero and then these two points represent the poles of our function. 4. \(\displaystyle[value]\), same reasons as for the previous cases.

The modulus function

The modulus of a number is the size of that number with no regard paid to its sign. For example, the modulus of \(−7\) is \(7\). The modulus of \(+7\) is also \(7\). We can write this concisely using the modulus sign \(| |\). So we can write \(|−7|= 7\) and \(|+ 7|= 7\). The modulus function is defined as follows:

Key Point

Modulus function

The modulus function is defined as:

\(f(x)=|x|=x\),    for \(x\geq 0\),

\(f(x)=|x|=-x\),   for \(x<0\).

The graph of such a function is shown in the following picture.

[Image]

Exercise

1. Complete the table with the values of the function \(f(x) =|x−2|\) for values of \(x\) between \(−3\) and \(5\).

\(x\) | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5
\(f(x)\) | 5 |  | 3 | 2 |  | 0 |  |  |

2. Which is the graph of \(f(x)\)? Select the correct graph.
Feedback: 1. It is enough to substitute the value of \(x\) given and then compute the absolute value. Consider, i.e., \(x=-3\), then we have \(f(x)=|x-2|=|-3-2|=|-5|=5\). Now consider \(x=-2\), then \(f(x)=|-2-2|=|-4|=4\). You can proceed in this way for all the other values of \(x\). 2. You can sketch the graph of the function using the points found in the table. You can draw those points and then join them by lines.

Exercise

Select the graph associated with each of the following function.

1. \(f(x)=3|x|\)  
2. \(f(x)=|x+1|\)  
3. \(f(x)=7|x-3|\)  
Feedback: Sketching the graph of the absolute value of a function is not complicated. In these cases, we have absolute values of lines. It is enough to identify some points and to join them, as we see in the previous exercise. Remember that the absolute value is always positive, so it is in the upper plane.

The unit step function (Heaviside step function)

The unit step function is defined as follows:

Key Point

A polynomial expression has the form:

\(u(t)=1\),    for \(t \geq 0\),

\(u(t)=0\),   for \(t<0\).

This is also known as the Heaviside step function in which case is is denote by \(H(x)\). Note that different conventions may be used at the jump point but we will use the above.

Study this definition carefully. You will see that it is defined in two parts, with one expression to be used when \(t\) is greater than or equal to 0, and another expression to be used when \(t\) is less than 0. The graph of this function is shown in the following figure. Note that the part of \(u(t)\) for which \(t<0\) lies on the \(t\)-axis but, for clarity, is shown as a distinct dashed line.

[Image]

There is a jump, or discontinuity in the graph when \(t= 0\). That is why we need to define the function in two parts; one part when \(t\) is negative, and one part when \(t\) is non-negative. The point with coordinates \((0,1)\) is part of the function defined on \(t≥0\).

The position of the discontinuity may be shifted to the left or right. The graph of \(u(t−d)\) is shown in the following picture:

[Image]

In the previous two figures, the function takes the value 0 or 1. We can adjust the value 1 by multiplying the function by any other number we choose. The graph of \(2u(t−3)\) is:

[Image]

Exercise

Write down in the boxes the expression of the step function corresponding to the graph below (call the step function \(u\) and use the independent variable \(t\)).

1. [Image]
2. [Image]
3. [Image]
4. [Image]
5. [Image]
6. [Image]
7. [Image]
Feedback: A step function is a function whose graph looks like a series of steps because it consists of a series of horizontal line segments with jumps in-between. For this reason, it is also sometimes called a staircase function. A step function has a constant value on given intervals, but the constant is different for each interval. Given the step function \(u(t)\), this is equal to \(1\) for \(t \geq 0\), \(0\) otherwise. This is very clear in the first graph. Consider now the graph of the step function in point c); in this case, the function is equal to \(1\) for \(t \geq 1\), hence the step function can be written as \(u(t-1)\) (we can easily see that the argument of \(u\) is zero for \(t=1\)). The same arguments apply to the other graphs.

Exercise

State if the following functions are polynomials or not.

1. \([value]\)
2. \(\displaystyle[value]\)
3. \(\displaystyle[value]\)
4. \(\displaystyle [value]\)
Feedback: A polynomial is always a function of the form \(P_n(x)=a_n x^n+a_{n-1} x^{n-1}...a_2 x^2+a_1 x+a_0\), so inverse powers of \(x\), semi integer power of \(x\) or, in general, other functions as trigonometric functions, logarithms etc. are not allowed in a polynomial. For the given functions: 1. \([value]\) is a polynomial. 2. \(\displaystyle[value]\) is a polynomial. 3. \([value]\) is not a polynomial since the term [value] is not of the form \(a_n x^n\). 4. \(\displaystyle[value]\) is a polynomial.

Exercise

In each case below, identify the term which prevents the expression from being a polynomial.

1. \(\displaystyle[value]\):
2. \(\displaystyle[value]\):
3. \(\displaystyle[value]\):
4. \(\displaystyle[value]\):
Feedback: A polynomial is always a function of the form \(P_n(x)=a_n x^n+a_{n-1} x^{n-1}...a_2 x^2+a_1 x+a_0\), so inverse powers of \(x\), semi integer power of \(x\) or, in general, other functions as trigonometric functions, logarithms etc. are not allowed in a polynomial. For the given functions: 1. \(\displaystyle[value],\) the term \(\displaystyle[value]\) is not allowed because it is an inverse power of \(x\). 2. \(\displaystyle[value],\) the term \(\displaystyle[value]\) is not allowed because it is not an integer power of \(x\). 3. \(\displaystyle[value],\) the term \(\displaystyle[value]\) is not allowed because it is an inverse power of \(x\). 4. \(\displaystyle[value],\) the term \(\displaystyle[value]\) is not allowed because it is a trigonometric function.

Exercise

State the degree of the following polynomials

1. \([value]\)
2. \([value]\)
3. \(\displaystyle[value]\)
4. \([value]\)
Feedback: The degree of a polynomial is the highest of the degrees of the polynomial's monomials with non-zero coefficients; namely, the degree of the polynomial is found by looking at the highest power of the independent variable. Hence, in our case we have (we can follow the same guide we use for the previous question on the degree of a polynomial) 1. For this polynomial, the degree is \([value]\). 2. This polynomial is the constant function, then the degree is zero. 3. For this polynomial, the degree is \(\displaystyle[value]\). 4. This polynomial is linear in \(t\), so its degree is 1.

Exercise

Is \([value]\) a polynomial of degree \(0\) with independent variable \(z\)?
Feedback: The constant function can always be considered as a polynomial of degree zero because it is a function of the form \(P(z)=[value]z^0\). Note that since we have \(P(z)\) then we know that \(z\) is the independent variable.

Exercise

Consider the following graphs once again.

[Image: plots of Q1, P1, P2, Q2, P3 and Q3 (same functions given in question 1)]

State if the functions are one-to-one or many-to-one.

1. \(P_1\)
2. \(Q_1\)
3. \(P_2\)
4. \(Q_2\)
5. \(P_3\)
6. \(Q_3\)
Feedback: A function is one-to-one if no two elements in the domain of the function correspond to the same element in the range of the function. Graphically, you can see that a function is one-to-one if, drawing whichever horizontal line, this horizontal line intersects the graph of the function under consideration only once. In the case of straight lines, as \(P_1(x)\) and \(Q_1(x)\), we have always one-to-one functions because, whichever horizontal line we draw, none of them intersects the graph twice. In the case of a parabola, as for \(P_2(x)\) and \(Q_2(x)\), we can draw an infinite number of lines that intersect our graph twice. For \(P_3(x)\), any line we draw intersects the graph of the function only once. In the case of the cubic function \(Q_3(3)\), we can draw lines intersecting the graph of the function at least twice, in fact, we can draw lines intersecting the graph also three times.

Exercise

What is meant by rational function?
Feedback: A rational function is any function which can be defined by a rational fraction, i.e. an algebraic fraction such that both the numerator and the denominator are polynomials. A rational function is not identifiable with a polynomial and it is not the only function characterised by poles.

Exercise

State the degree of the numerator and the degree of the denominator of the rational function \(\displaystyle[value]\).

1. The numerator is a polynomial of degree
2. The denominator has degree
Feedback: The numerator and the denominator are simply polynomials, hence the degree of both is given by the highest of the degrees of the polynomial's monomials with non-zero coefficients. Apply the same principles we used for the other exercises on finding the degrees of the polynomials.

Exercise

What is the pole of a rational function?
Feedback: It is not necessary to repeat what a pole of a rational function is at this stage. However, we can clarify some concept appearing between the options. An extremum of a function is any point at which the value of a function is largest (a maximum) or smallest (a minimum). The line tangent to the curve at a point at infinity is called asymptote.

Exercise

Referring to the functions \(R_1(x)=\dfrac{x+6}{x^2+1}\), \(R_2(t)=\dfrac{t^3-1}{2t+3}\) and \(R_3(z)=\dfrac{2z^2+z-1}{z^2+z-2}\) state if they are one-to one or many-to-one.

1. \(R_1(x)=\dfrac{x+6}{x^2+1}\)
2. \(R_2(t)=\dfrac{t^3-1}{2t+3}\)
3. \(R_3(z)=\dfrac{2z^2+z-1}{z^2+z-2}\)
Feedback: A function is said to be many-to-one if two or more elements in the domain of the function correspond to the same element in the range of the function. You can check if the function is many-to-one using the graphical method we explain in the previous exercises, or you can proceed analytically. Analytically: 1. \(R_1(x)=\dfrac{x+6}{x^2+1}\): consider the value \(R_1(x)=4\), there are two values of \(x\) for which this function assumes the value \(4\), in fact, substituting, we have \(4=\dfrac{x+6}{x^2+1}\), so we have to solve the second order equation \(4x^2-x-2=0\) which has two dinstict real solutions; hence, we can conclude that for two points of the domain corresponds the same value of the function. 2. \(R_2(t)=\dfrac{t^3-1}{2t+3}\): you can apply the same arguments we use for \(R_1(x)\). 3. \(R_3(z)=\dfrac{2z^2+z-1}{z^2+z-2}\): you can apply the same arguments as for \(R_1(x)\) (and \(R_2(x)\)).

Exercise

Without using a graphical calculator plot graphs of \(f(x)=\dfrac{1}{x}\) and \(f(x)=\dfrac{1}{x^2}\), select the correct options describing these functions, and state the position of any poles.

1. \(f(x)=\dfrac{1}{x}\): ,
2. \(f(x)=\dfrac{1}{x^2}\): ,
3. The pole for both graphs is given by \(x=\)
Feedback: Even functions and odd functions are functions which satisfy particular symmetry relations. An even function is such that \(f(x)=f(-x)\). An odd function is charcterised by \(-f(x)=f(-x)\). A continuous function is a function that does not have any abrupt changes in value, known as discontinuities; more rigorously a function \(f\) with variable \(x\) is said to be continuous at the point \(c\) on the real line, if the limit of \(f(x)\), as \(x\) approaches that point \(c\), is equal to the value \(f(c)\). 1. \(f(x)=\dfrac{1}{x}\) is an odd function, in fact, changing \(x\) with \(-x\), we have \(\dfrac{1}{-x}=-\dfrac{1}{x}\), namely \(-f(x)=f(-x)\). Moerover, this function is discontinous because for \(x=0\) it is not define (\(x=0\) is not in the domain). \(x=0\) represents a vertical asymptote for the function: when the function approaches the line \(x=0\) from the left, it goes to \(- \infty \), when the function approaches \(x=0\) from the right, it goes to \(+ \infty \). 2. \(f(x)=\dfrac{1}{x^2}\) is an even function, in fact we have \(f(-x)=\dfrac{1}{(-x)^2}=\dfrac{1}{x^2}=f(x)\). This function is discontinous because it is not define for \(x=0\). The line \(x=0\) is a vertical asymptote for the function and it approaches \(+ \infty\) both from the left and from the right. 3. The pole corresponds to the point for which the denominator is zero. In order to confirm our findings, here are the graphs of the two functions: [value] The graph of \(y = \dfrac{1}{x}\) [value] The graph of \(y = \dfrac{1}{x^2}\)

Exercise

Is the modulus function one-to-one or many-to-one?
Feedback: The modulus function is always many-to-one. In fact, you can draw an infinite number of horizontal lines intersecting the graph of the function twice. Analytically, it is enough to note that if \(f(x) = \lvert x \rvert\), then \(f(-1) = \lvert -1 \rvert = 1 = \lvert 1 \rvert = f(1)\).

Note from the answers to parts (b) and (c) that we must exclude the value \(t = -3/2\) from the domain of this function because division by zero is not defined. At this point, as you can see, the graph shoots off towards very large postive values (we say it tends to positive infinity) if the point is approached from the left, and towards very large negative values (we say it tends to negative infinity) if the point is approached from the right. The dotted line in the graph of \(R_2(x)\) has equation \(t = -3/2\). It is approached by the curve as \(t\) approaches \(-3/2\) and is known as a vertical asymptote.