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Functions and single-variable calculus A course resource for Math 218

Ellen Goldstein, Aaron Greicius

Section 1.4 Algebraic functions

Our current repertoire of functions is rather limited, consisting as it does only of constant, linear, and quadratic functions. In this section we will significantly expand this to include power, polynomial, and rational functions. These function types can then be further combined using arithmetic operations to form an even larger family of functions: algebraic functions.
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Subsection Learning Goals

  • Learn the definition of power, polynomial and rational functions.
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  • Review the algebraic rules for fractions and exponents.
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  • Learn to distinguish between power, polynomial, and rational functions.
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Subsection Power functions

We will begin with power functions. In its most general form, a power function is a function of the form
\begin{equation} f(x)=cx^r\text{,}\tag{1.31} \end{equation}
where \(c\) and \(r\) are real constants. The important defining detail here is that in the formula (1.31) the input variable \(x\) is raised to some fixed power: that is, in a power function, the input variable \(x\) appears as the base in the power expression, and not the exponent. Later we will introduce exponential functions, which are also defined via a power expression, but where the input variable appears as the exponent, as opposed to the base.
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As familiar as the function formula (1.31) may seem, it takes significant meaning to give meaning to this for an aritrary power \(r\text{.}\) For now we will settle for a a definition for rational powers \(r\text{:}\) i.e., for real numbers \(r\) that can we written as \(r=m/n\text{,}\) where \(m\) and \(n\) are integers. In other words, Definition 1.4.2 will give meaning to functions of the form \(f(x)=\sqrt{2}x^{-11/7}\text{,}\) but not to functions of the form \(f(x)=x^{\sqrt{2}}\text{,}\) or \(f(x)=3x^\pi\text{,}\) since \(\sqrt{2}\) and \(\pi\) (famously) are not rational numbers. We will eventually extend our definition of power functions to include such irrational powers, but to do so we will need the help of calculus.
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Definition 1.4.1. Rational numbers.

A real number \(r\in \R\) is rational if it can be written as a quotient of two integers: i.e., we have
\begin{equation} r=\frac{m}{n}\tag{1.32} \end{equation}
for some integers \(m,n\in \Z\text{.}\)
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We denote by \(\Q\) the set of all rational numbers: i.e.,
\begin{equation} \Q=\{r\in \R\mid r \text{ is rational}\}=\{\tfrac{m}{n}\mid m,n\in \Z \text{ and } n\ne 0\}\text{.}\tag{1.33} \end{equation}
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Definition 1.4.2. Power functions: rational power.

A power function (with rational power) is any function of the form \(f(x)=cx^q\text{,}\) where \(c\ne 0\) and \(q\in \Q\text{,}\) and where the expression \(x^q\) is defined below in a case-wise manner.
  • Zero power.
    We define \(x^0=1\) for all \(x\text{.}\)
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  • Positive integer.
    If \(n\) is a positive integer, then
    \begin{equation*} x^n=\underset{n \text{ times}}{\underbrace{x\cdot x\cdots x}}\text{.} \end{equation*}
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  • Odd \(n\)-th root.
    If \(n\) is a positive odd integer, then
    \begin{equation*} x^{1/n}=\sqrt[n]{x}\text{,} \end{equation*}
    where \(\sqrt[n]{x}\) is the unique real \(n\)-th root of \(x\text{.}\)
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  • Even \(n\)-th root.
    If \(n\) is a positive even integer, and \(x\geq 0\) then
    \begin{equation*} x^{1/n}=\sqrt[n]{x}\text{,} \end{equation*}
    where \(\sqrt[n]{x}\) is the unique positive \(n\)-th root of \(x\text{.}\)
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  • Positive rational.
    If \(m\) and \(n\) are relatively prime positive integers
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    Integers are relatively prime if they share no integer factors besides \(\pm 1\text{.}\) This technical condition is here to ensure the definition is well defined.
    , then
    \begin{equation*} x^{m/n}=(x^{1/n})^m=(x^m)^{1/n}\text{.} \end{equation*}
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  • Negative rational.
    If \(m\) and \(n\) are relatively prime positive integers, then
    \begin{equation*} x^{-m/n}=\left(x^{m/n}\right)^{-1}=\frac{1}{x^{m/n}}\text{.} \end{equation*}
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Looking carefully at the various cases that define \(x^q\) for a rational power \(q\text{,}\) we see that some restrictions on what \(x\) can be have been snuck in. Namely, when \(n\) is a positive even integer, we see that \(x^{1/n}=\sqrt[n]{x}\) is only defined for nonnegative values of \(x\text{.}\) Additionally, for negative rational powers \(q=-m/n\text{,}\) since by definition
\begin{equation*} x^{-m/n}=\frac{1}{x^{m/n}}\text{,} \end{equation*}
we see that this expression is defined if and only if \(x^{m/n}\) is defined and nonzero. As a consequence of these restrictions, we need to take care when computing the implied domain of a power function \(f(x)=cx^q\text{.}\)
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Example 1.4.3. Power functions: implied domain.

Determine the implied domain of the given power function.
  1. \(\displaystyle f(x)=20x^{-5}\)
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  2. \(\displaystyle f(x)=7x^{1/12}\)
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  3. \(\displaystyle f(x)=\pi\, x^{4/5}\)
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  4. \(\displaystyle f(x)=x^{-3/2}\)
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Solution.
  1. By definition, we have
    \begin{equation*} f(x)=20x^{-5}=20\cdot \frac{1}{x^5}=\frac{20}{x^5}\text{.} \end{equation*}
    Since \(x^5\) is defined for all \(x\) and equal to 0 exactly when \(x=0\text{,}\) we see that the implied domain of \(f\) is
    \begin{equation*} D=\{x\in \R\mid x\ne 0\}=(-\infty, 0)\cup (0,\infty)\text{.} \end{equation*}
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  2. Since \(12\) is an even positive integer, the expression \(x^{1/12}=\sqrt[12]{x}\) is defined only on \([0,\infty)\text{.}\) It follows that the implied domain of \(f\) is
    \begin{equation*} D=[0,\infty)\text{.} \end{equation*}
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  3. By definition, we have
    \begin{equation*} f(x)=\pi\, x^{4/5}=\pi(x^4)^{1/5}\text{.} \end{equation*}
    Since \(x^4\) is defined for all \(x\text{,}\) and \(t^{1/5}\) is defined for all \(t\text{,}\) it follows that \(x^{4/5}\) is defined for all \(x\text{.}\) We conclude that the implied domain of \(f\) is \(D=\R\text{.}\)
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  4. By definition, we have
    \begin{align*} f(x) \amp =x^{-3/2}\\ \amp = \frac{1}{x^{3/2}}\\ \amp = \frac{1}{(x^3)^{1/2}}\\ \amp =\frac{1}{\sqrt{x^3}}\text{.} \end{align*}
    Since \(\sqrt{t}\) is defined if and only if \(t\geq 0\text{,}\) and since \(x^3\geq 0\) if and only if \(x\geq 0\text{,}\) we see that \(x^{3/2}=\sqrt{x^3}\) is defined if and only if \(x\geq 0\text{.}\) For the expression \(f(x)\) to be defined, we need further that \(x^{3/2}=\sqrt{x^3}\) is nonzero. This adds the additional restraint that \(x\ne 0\text{.}\) We conclude that the implied domain of \(f\) is
    \begin{equation*} D=\{x\in \R\mid x\geq 0 \text{ and } x\ne 0\}=(0,\infty)\text{.} \end{equation*}
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To evaluate a power function at an input, we must make our way through the various cases in the definition of a power expression \(x^q\text{.}\) Depending on the complexity of \(q\text{,}\) we may at some point have to compute a reciprocal and/or an \(n\)-th root. Our ability to write down a nice \(n\)-th root of some given value determines how “nicely” our computed value is expressed in the end.
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Example 1.4.4. Evaluating power functions.

Evaluate the given power function \(f\) at the given input \(x\text{.}\)
  1. \(f(x)=17x^{2/3}\text{,}\) \(x=-8\)
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  2. \(f(x)=x^{-1/2}\text{,}\) \(x=27\)
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Solution.
  1. We compute
    \begin{align*} f(-8) \amp =17(-8)^{2/3}\\ \amp =17\sqrt[3]{(-8)^2}\\ \amp =17\sqrt[3]{64}\\ \amp =17\sqrt[3]{4^3}\\ \amp =17\cdot 4\\ \amp = 68\text{.} \end{align*}
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  2. We compute
    \begin{align*} f(27) \amp =(27)^{-1/2}\\ \amp = \frac{1}{27^{1/2}}\\ \amp = \frac{1}{\sqrt{3^3}}\\ \amp =\frac{1}{\sqrt{3}\sqrt{3^2}}\\ \amp =\frac{1}{3\sqrt{3}} \text{.} \end{align*}
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As we combine power functions with other functions, we will want to be able to simplify expressions wherever possible. The following power identities will come in handy in this respect.
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Remark 1.4.6. Iterated powers.

The restriction \(a\geq 0\) is indeed a necessary one for the identity \((a^b)^c=a^{bc}\text{.}\) For example consider the choices
\begin{align*} a \amp = -1 \amp b\amp=2 \amp c=\frac{1}{2}\text{.} \end{align*}
Using Definition 1.4.2, we have
\begin{align*} (a^b)^c \amp = ((-1)^2)^{1/2}\\ \amp = \sqrt{1}\\ \amp = 1\\ a^{bc} \amp =(-1)^{2\cdot \tfrac{1}{2}}\\ \amp = (-1)^1\\ \amp =-1 \text{.} \end{align*}
Thus \((a^b)^c\ne a^{bc}\) in this case. This counterexample illustrates again the subtleties of our definition of a power expression \(x^q\text{:}\) specifically, that when \(q=1/n\) for \(n\) a positive even integer, \(x^{1/n}=\sqrt[n]{x}\) is defined as the positive square root of \(x\) (assuming \(x\geq 0\)).
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Put another way, the failure of the identity \((a^b)^c=a^{bc}\) to hold in the example is a consequence of the fact that the equality
\begin{equation*} \sqrt{x^2}=x \end{equation*}
does not hold for all \(x\in \R\text{;}\) it is only true for \(x\in [0,\infty)\text{.}\) After we officially introduce the absolute value, we will prove an improved version of this identity that does hold for all real numbers: namely,
\begin{equation*} \sqrt{x^2}=\abs{x}\text{.} \end{equation*}
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Example 1.4.7.

Simplify each expression to one involving only positive exponents. You may assume all the identities of Theorem 1.4.5 apply in each case.
  1. \(\displaystyle 2k^4\cdot 3k^5\)
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  2. \(\displaystyle 4a^3b^{-2}\cdot 3a^{-4}b^3\)
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  3. \(\displaystyle (7x^3y)^{-2}\)
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  4. \(\displaystyle \dfrac{2h^3j^{-3}k^4}{3jk^{1/2}}\)
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Solution.
  1. We compute
    \begin{align*} 2k^4\cdot 3k^5 \amp =2\cdot 3\, k^4\cdot k^5 \\ \amp = 6k^9 \amp (\text{common base})\text{.} \end{align*}
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  2. We compute
    \begin{align*} 4a^3b^{-2}\cdot 3a^{-4}b^3 \amp =4\cdot 3\, a^3\cdot a^{-4}b^{-2}b^3 \\ \amp =12a^{-1}b^1 \amp (\text{common base}) \\ \amp =\frac{12b}{a} \amp \end{align*}
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  3. We compute
    \begin{align*} (7x^3y)^{-2} \amp =7^{-2}(x^3)^{-2}y^{-2} \amp (\text{common exp.}) \\ \amp =7^{-2}x^{-6}y^{-2} \amp (\text{iter. powers})\\ \amp = \frac{1}{7^2x^6y^2} \amp (\text{reciprocal})\\ \amp =\frac{1}{49x^6y^2}\text{.} \end{align*}
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  4. We compute
    \begin{align*} \frac{2\,h^3\,j^{-3}\,k^4}{3\,j\,k^{1/2}} \amp =\frac{2}{3}\, h^3j^{-3-1}\,k^{4-1/2} \\ \amp =\frac{2}{3}h^3j^{-4}k^{7/2}\\ \amp = \frac{2h^3k^{7/2}}{3j^4}\text{.} \end{align*}
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The identities in Theorem 1.4.5 are a convenient way to recall some identities involving radicals. As an example, below we translate certain instances of the identities from Theorem 1.4.5 into radical notation using the definition \(a^{1/n}=\sqrt[n]{a}\text{:}\)
\begin{align*} (ab)^{1/n} \amp =a^{1/n}\,b^{1/n} \amp \sqrt[n]{ab} \amp =\sqrt[n]{a}\cdot\sqrt[n]{b}\\ \left( \frac{a}{b}\right)^{1/n}\amp = \frac{a^{1/n}}{b^{1/n}} \amp \sqrt[n]{\frac{a}{b}} \amp =\frac{\sqrt[n]{a}}{\sqrt[n]{b}}\\ a^{m/n} \amp = (a^m)^{1/n}=(a^{1/n})^m \amp a^{m/n}\amp=\sqrt[n]{a^m} =(\sqrt[n]{a})^m\text{.} \end{align*}
We make this observation official in Corollary 1.4.8.
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Notice that we say a function \(f\) is a power function if it can be written in the form \(f(x) = Ax^k\) for some constants \(A\) and \(k\text{;}\) it may not start out written that way. Sometimes it can take a great deal of algebra to determine whether or not a function is a power function "in disguise."
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Example 1.4.9.

For example, is the function
\begin{equation*} f(x) = \frac{(2x)^3+8x^{17/5}}{x^{2/5}+1} \end{equation*}
a power function? If so, what are the numbers \(A\) and \(k\text{?}\)
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Solution.
We can expand, factor, and then make a lucky cancellation to write this expression in the form \(Ax^k\text{:}\)
\begin{equation*} f(x) = \frac{(2x)^3+8x^{17/5}}{x^{2/5}+1} = \frac{8x^3+8x^{17/5}}{x^{2/5}+1} = \frac{8x^3(1+x^{17/5-3})}{x^{2/5}+1} = \frac{8x^3(1+x^{2/5})}{x^{2/5}+1} = 8x^3 \end{equation*}
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Checkpoint 1.4.10.

The expression \(\displaystyle{\ \sqrt{x^5 y^2} \sqrt[3]{x^5 y^3} \sqrt[5]{x^4} \ }\) equals \(\displaystyle{\ x^ry^s \ }\)
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where \(\ r\text{,}\) the exponent of \(\ x\text{,}\) is:
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and \(\ s\text{,}\) the exponent of \(\ y\text{,}\) is:
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Answer 1.
\({\frac{5}{2}}+{\frac{5}{3}}+{\frac{4}{5}}\)
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Answer 2.
\(1+1\)
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Solution.
\(\displaystyle{\begin{gathered} \sqrt{x^5 y^2} \sqrt[3]{x^5 y^3}\sqrt[5]{x^4} = (x^5 y^2)^{1/2} (x^5 y^3)^{1/3} (x^4)^{1/5} \\ = x^{5/2} y^{2/2} x^{5/3} y^{3/3} x^{4/5} = x^{(5/2+5/3+4/5)} y^{(2/2+3/3)} \\ = x^{149/30} y^{2/1}. \end{gathered} }\)
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Therefore \(r = {{\frac{149}{30}}}\) and \(s={2}\text{.}\)
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Subsection Polynomial functions

Before giving the formal definition of a polynomial, let’s take a look at some examples of what we will call polynomial functions:
\begin{align*} f(x) \amp =1-5x\\ f(x) \amp =x^3\\ f(x) \amp =3\\ f(x) \amp =2x^4+1\\ f(x) \amp =x^7+x^6+x^5+x^4+x^3+x^2+x+1\\ f(x) \amp = 4.23x^{517}-\sqrt{2}x^{24}+x^{6}-\pi\text{.} \end{align*}
We next make a list of some common features of these functions.
  • In contrast to linear and quadratic functions, these polynomials are seen to be sums of an unlimited (but finite) number of terms.
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  • Each term is either constant or a number times a power of \(x\text{.}\)
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  • The power of \(x\) is always a positive whole number, though the constant in front may be any real number whatsoever.
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Lastly, we synthesize these observations in a formal definition.
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Definition 1.4.11. Polynomials.

A polynomial is a function \(f\) that can be written in the form
\begin{equation} f(x) = a_nx^n+a_{n-1}x^{n-1}+\cdots+a_2x^2+a_1x+a_0\tag{1.41} \end{equation}
for some integer \(n\geq 0\) and real numbers \(a_0, a_1, \dots a_n\text{.}\) For each \(0\leq k\leq n\text{,}\) \(a_kx^k\) is called the \(k\)-th term of \(f\) and \(a_k\) is called its \(k\)-th coefficient; \(a_0\) is called the constant term of \(f\text{.}\)
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Furthermore, if \(a_n\ne 0\text{,}\) then \(a_nx^n\) is called the leading term of \(f\text{,}\) \(a_n\) its leading coefficient, and \(n\) its degree, denoted \(\deg f\text{.}\)
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Remark 1.4.12. Constant, linear, quadratic functions.

Observe that constant, linear, and quadratic functions are each members of the larger family of polynomials. In more detail, they are polynomials for which the \(n\) appearing in Definition 1.4.11 is equal to \(0\text{,}\) \(1\text{,}\) or \(2\text{,}\) as illustrated in Table 1.4.13.
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Table 1.4.13. Small degree polynomials
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\(n\) Type Formula
0 Constant \(f(x)=a_0\)
1 Linear \(f(x)=a_1x+a_0\)
2 Quadratic \(f(x)=a_2x^2+a_1x+a_0\text{,}\) \(a_2\ne 0\)
In this sense Definition 1.4.11 is a generalization of these earlier definitions. As such our parametric manner of understanding linear and quadratic (and constant) functions applies more generally to polynomials. In this more general setting, the coefficients \(a_0,a_1,\dots, a_n\) appearing in the general formula
\begin{equation*} f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots +a_1+a_0 \end{equation*}
function as parameters: by varying our choice of the coefficients \(a_k\text{,}\) we get an infinite family of polynomials.
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Remark 1.4.14. Zero function and zero polynomial.

A function \(f\colon D\rightarrow \R\) is called a zero function if \(f(x)=0\) for all \(x\in D\text{.}\) Observe that a zero function is also a polynomial according to Definition 1.4.11. Indeed we have
\begin{equation*} f(x)=a_0 \end{equation*}
where \(a_0=0\text{.}\) Because of this, zero functions are also called zero polynomials.
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Since the formula for the zero polynomial has no nonzero coefficients, the notion of degree given in Definition 1.4.11 does not apply: i.e., technically for us the zero polynomial does not have a degree. There is not a fast and fixed convention on this point in the mathematics literature: some texts declare the degree of the zero polynomial to be undefined; still others declare the degree of the zero polynomial to be \(-\infty\text{!}\)
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Remark 1.4.15. Larger degree polynomials.

Using the new language of degree, we see that quadratic functions are just polynomials of degree 2. We have similar names for polynomials \(f\) of higher degree.
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Table 1.4.16. Polynomials of higher degree
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Formula Degree Type
\(f(x)=a_2x^2+a_1x+a_0, a_2\ne 0\) \(\deg f=2\) Quadratic
\(f(x)=a_3x^3+a_2x^2+a_1x+a_0, a_3\ne 0\) \(\deg f=3\) Cubic
\(f(x)=a_4x^4+a_3x^3+\cdots +a_0, a_4\ne 0\) \(\deg f=4\) Quartic
\(f(x)=a_5x^5+a_4x^4+\cdots +a_0, a_5\ne 0\) \(\deg f=5\) Quintic
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Example 1.4.17. Polynomials.

Determine if the given function \(f\colon \R\rightarrow\R\) is a polynomial. If yes, determine the constant term of \(f\text{,}\) and if applicable, the leading term, leading coefficient, and degree.
  1. \(\displaystyle f(x)=-\sqrt{2}\)
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  2. \(\displaystyle f(x)=0\)
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  3. \(\displaystyle f(x)=x(3x+1)(x-2)^2\)
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Solution.
  1. As discussed in Remark 1.4.12, all constant functions are polynomials (corresponding to the choice \(n=0\)). In this case the constant term of \(f\) is \(\sqrt{2}\text{.}\) Since \(\sqrt{2}\ne 0\text{,}\) this is also the leading term and leading coefficient of \(f\text{,}\) and we have \(\deg f=0\text{.}\)
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  2. As discussed in Remark 1.4.14, the zero function (or zero polynomial) is indeed a polynomial. However, since no nonzero coefficients appear in its formula, the notions of leading term, leading coefficient, and degree do not apply.
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  3. The given function is indeed a polynomial, though we need to do some algebra to recognize that it has a formula of the form (1.41):
    \begin{align*} f(x) \amp = x(3x+1)(x-2)^2\\ \amp = (3x^2+x)(x^2-4x+4) \\ \amp = 3 x^4 - 11 x^3 + 8 x^2 + 4 x+0\text{.} \end{align*}
    Once in this form, we see that the constant term of \(f\) is \(0\text{,}\) its leading term is \(3x^4\text{,}\) its leading coefficient is \(3\text{,}\) and its degree is \(4\text{.}\)
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Checkpoint 1.4.18.

For the polynomial \(g(x) = x^5(1-2x)(1-x^3)\text{,}\) determine the leading coefficient, leading term, degree, constant term, and coefficients \(a_1\) and \(a_5\text{.}\)
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Subsection Rational functions

Just as rational numbers are defined as quotients of integers, a rational function is defined as a quotient of polynomials.
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Definition 1.4.19. Rational function.

A function \(f\) is a rational function if it can be expressed as a quotient of two polynomials: i.e., \(f\) is rational if there are polynomials \(p(x)=\anpoly\) and \(q(x)=\bmpoly\) satisfying
\begin{equation*} f(x)=\frac{p(x)}{q(x)}=\frac{\anpoly}{\bmpoly} \end{equation*}
for all \(x\) in the domain of \(f\text{.}\)
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Remark 1.4.20. Implied domain of rational function.

Suppose \(f(x)=p(x)/q(x)\) is a rational function, where
\begin{align*} p(x) \amp = \anpoly \amp q(x)\amp= \bmpoly\text{.} \end{align*}
Since \(p(x)\) and \(q(x)\) are defined for all \(x\in \R\text{,}\) we see that the expression defining \(f(x)\) is defined as long as \(q(x)\ne 0\text{.}\) Thus the implied domain of \(f\) is
\begin{equation*} D=\{x\in \R\mid q(x)\ne 0\}\text{.} \end{equation*}
To compute \(D\text{,}\) we need to be able to find the zeros of the denominator polynomial \(q(x)\text{.}\) It turns out that this is not always an easy task, especially if the degree of \(q\) is large. We will provide techniques for doing this in the next section.
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Just like our previous definitions for functions, the key realization is that a rational function can be written in such a form. It does not have to begin that way.
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Example 1.4.21. Rational function: like denominators.

Verify that \(f(x) = x-1+\frac{1}{2x+1}\) is a rational function by writing it as \(f(x)=p(x)/q(x)\text{,}\) where \(p\) and \(q\) are polynomials.
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Solution.
To express \(f\) as a quotient of polynomials, we will convert the given formula of \(f\) into a single quotient using the technique of like denominators.
\begin{align*} x-1+\frac{1}{2x+1} \amp = (x-1)\cdot\frac{2x+1}{2x+1}+\frac{1}{2x+1}\\ \amp = \frac{(x-1)(2x+1)}{2x+1}+\frac{1}{2x+1} \\ \amp = \frac{(x-1)(2x+1)+1}{2x+1}\\ \amp = \frac{(2x^2-x-2x-1)+1}{2x+1}\\ \amp = \frac{2x^2-3x}{2x+1}\text{.} \end{align*}
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Example 1.4.22. Rational function: clear denominators.

Verify that
\begin{equation*} f(x) = \frac{\frac{3}{x+2}-\frac{1}{x}}{x-1} \end{equation*}
is a rational function by writing it as \(f(x)=p(x)/q(x)\text{,}\) where \(p\) and \(q\) are polynomials.
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Solution.
We use the algebraic technique of clearing denominators:
\begin{align*} f(x) \amp = \frac{\frac{3}{x+2}-\frac{1}{x}}{x-1} \amp \\ \amp = \frac{x(x+2)}{x(x+2)}\cdot \frac{\frac{3}{x+2}-\frac{1}{x}}{x-1} \\ \amp = \frac{x(x+2)\cdot \frac{3}{x+2}-x(x+2)\cdot \frac{1}{x}}{x(x+2)(x-1)}\\ \amp = \frac{3x-(x+2)}{x(x+2)(x-1)} \\ \amp = \frac{2x-2}{x^3+x^2-x}\text{.} \end{align*}
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Remark 1.4.23. Fraction arithmetic.

The examples above hopefully reactivated your fraction arithmetic skills. Let’s summarize some of the definitions and techniques utilized.
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Table 1.4.24. Fraction arithmetic
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Fraction multiplication \(\frac{a}{b}\cdot \frac{c}{d}=\frac{ac}{bd}\)
Multiplying by \(1=c/c\) \(\frac{a}{b}=\frac{c}{c}\cdot \frac{a}{b}=\frac{ca}{cb}\)
Fraction addition with like denominators \(\frac{a}{c}+\frac{b}{c}=\frac{a+b}{c}\)
Fraction addition via like denominators \(\frac{a}{b}+\frac{c}{d}=\frac{ad}{bd}+\frac{bc}{bd}=\frac{ad+bc}{bd}\)
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Subsection Algebraic functions

A function is algebraic if it can be expressed in terms of constants and a variable \(x\) by using only arithmetic operations \((+, -, \times\text{,}\) and \(\div)\) and rational constant powers of the variables. For example,
\begin{equation*} f(x) = \frac{3x^2-5}{1-x} \text{ and } g(x) = (\sqrt{x}+\sqrt[3]{x})^{7/2} \end{equation*}
are algebraic functions.
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There are four basic types of algebraic functions, which we have already investigated: linear functions, polynomial functions, rational functions, and power functions. More complicated algebraic functions can be constructed by using sums (\(+\)), differences (\(-\)), products (×), and quotients (\(\div\)) of these four basic types. Table 1.4.25 gives a general form and examples for each type:
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Table 1.4.25. A summary of algebraic functions
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Type General form Examples
Linear \(f(x) = mx+b\) \(f(x) = 2x-1\text{,}\) \(f(x) = 1.4x\)
Polynomial \(f(x) =\anpoly\) \(f(x) = 3x^5-2x^3+x-6\)
Rational \(f(x) = \frac{\anpoly}{\bmpoly}\) \(f(x) = \dfrac{3x^2-1}{x^5+x^3+1}\)
Power \(f(x) = cx^q\text{,}\) \(q\in \Q\) \(f(x) = 3x^3, f(x) = 1.7x^{-1.5}\)
Other Arithmetic comb. of previous types \(f(x) = \frac{1+\sqrt{x}}{1+x}\text{,}\) \(f(x) = x^{2/3}+5\)
There is a certain amount of overlap among the types listed in the table; for example, \(f(x) = x^2\) is both a power function and a polynomial function, and \(f(x) = \frac{1}{x} = x^{-1}\) is both a power function and a rational function.
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Functions that are not algebraic are called transcendental functions. In later chapters of this book, we will investigate four basic types of transcendental functions: exponential, logarithmic, trigonometric, and inverse trigonometric functions. We will not go into their definitions or general forms right now. However, Table 1.4.26 provides some examples.
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Table 1.4.26. A summary of transcendental functions
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Type Examples
Exponential \(f(x) = 2^x, f(x) = 3e^{4x}, f(x) = 1.2(3.4)^x\)
Logarithmic \(f(x) = \log_{10}x, f(x) = \ln x, f(x) = \log_2 x\)
Trigonometric \(f(x) = \sin x, f(x) = \cos x, f(x) = \cot x\)
Inverse Trigonometric \(f(x) = \arcsin x, f(x) = \cos^{-1} x, f(x) = \arctan x\)
Other \(f(x) = x+\sin x, f(x) = \ln(\sqrt{x}+12), f(x) = 2^x\arctan x\)

Example 1.4.27. Identifying algebraic function types.

For each function listed, determine whether it is a linear, power, polynomial, rational, algebraic or transcendental function. Some functions may satisfy more than one classification.
  1. \(\displaystyle f(x) = x^{-2}\)
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  2. \(\displaystyle g(x) = 3x^2+2^x+1\)
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  3. \(\displaystyle h(x) = \frac{\sqrt{x}}{1+x}\)
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  4. \(\displaystyle k(x) = (x-1)^2-x^2\)
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Solution.
  1. \(f(x) = x^{-2}\) is a power function of the form \(Ax^k\) with real-number coefficient \(A=1\) and rational-number exponent \(k=-2\text{.}\) Since \(x^{-2}=\frac{1}{x^2}\text{,}\) \(f\) is also a rational function of the form \(\frac{p(x)}{q(x)}\) with the polynomials \(p(x) = 1\) and \(q(x) = x^2\text{.}\) Thus \(f\) is an algebraic function that is both a power function and a rational function.
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  2. The function \(2^x\) is not algebraic, because it involves a nonconstant exponent. Therefore the function \(g(x) = 3x^2+2^x+1\) also fails to be algebraic and thus is transcendental.
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  3. \(h(x) = \frac{\sqrt{x}}{1+x}\) is a quotient of algebraic functions, but not a quotient of polynomials, since \(\sqrt{x}\) is not a polynomial. Therefore \(h(x)\) is an algebraic function, but not a rational function.
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  4. This function can be simplified: \(k(x) = (x-1)^2-x^2 = x^2-2x+1-x^2 = -2x+1\text{.}\) After the last simplification it is easy to see that \(k(x)\) is also a polynomial function, with degree \(n=1\text{,}\) and coefficients \(a_1 = -2\) and \(a_0 = 1\text{.}\) Furthermore, since we can write \(-2x+1 = \frac{-2x+1}{1}\text{,}\) \(k(x)\) is a rational function as well, with \(p(x) = -2x+1\) and \(q(x) = 1\text{.}\) Therefore \(k(x)\) is an algebraic function that is linear, polynomial, and rational.
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Checkpoint 1.4.28.

What is an example of a single algebraic function that is linear, polynomial, rational, and a power function (all of these at once). You have seen several…
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Exercises Exercises

1.

Is the function \(\displaystyle g(x) = \frac{ \left( -x^{4} \right)^{5}}{4}\) a power function? If it is, write it in the form \(g(x) = kx^p\) and enter exact values for \(k\) and \(p\text{.}\) If it is not a power function, enter NONE in both blanks. Do not leave any blanks empty.
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\(k =\)
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\(p =\)
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Answer 1.
\(-{\frac{1}{4}}\)
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Answer 2.
\(20\)
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Solution.
SOLUTION\(g(x) = \frac{ \left( -x^{4} \right)^{5}}{4} = \frac{ (-1)^{5} (x^{4})^{5} }{4} = - \frac{1}{4} \ x^{20}\)\(k = - \frac{1}{4}\)\(p = 20\)
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2.

Is the function \(\displaystyle f(x) = \frac{2}{\sqrt{16 x}}\) a power function? If it is, write it in the form \(f(x) = kx^p\) and enter exact values for \(k\) and \(p\text{.}\) If it is not a power function, enter NONE in both blanks. Do not leave any blanks empty.
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\(k =\)
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\(p =\)
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Answer 1.
\({\frac{1}{2}}\)
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Answer 2.
\(-{\frac{1}{2}}\)
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Solution.
SOLUTION\(f(x) = \frac{2}{\sqrt{16 x}} = \frac{ 2 }{ \sqrt{16} \ \sqrt{x}} = \frac{2}{4} \frac{1}{\sqrt{x}} = \frac{2}{4} x^{-1/2}\)\(k = \frac{2}{4}\)\(p = - \frac{1}{2}\)
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3.

Is the function \(\displaystyle Q(z) = \left( \frac{1}{2 \sqrt{z}} \right)^{2}\) a power function? If it is, write it in the form \(Q(z) = kz^p\) and enter exact values for \(k\) and \(p\text{.}\) If it is not a power function, enter NONE in both blanks. Do not leave any blanks empty.
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\(k =\)
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\(p =\)
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Answer 1.
\({\frac{1}{4}}\)
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Answer 2.
\(-1\)
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Solution.
SOLUTION\(Q(z) = \left( \frac{1}{2 \sqrt{z}} \right)^{2} = \left( \frac{1}{2} z^{-1/2} \right)^{2} = \frac{1}{4} z^{-2/2}\)\(k = \frac{1}{4}\)\(p = - \frac{2}{2}\)
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4.

Is the function \(\displaystyle H(t) = \frac{ t^{3} }{6 \sqrt{t^{5}}}\) a power function? If it is, write it in the form \(H(t) = kt^p\) and enter exact values for \(k\) and \(p\text{.}\) If it is not a power function, enter NONE in both blanks. Do not leave any blanks empty.
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\(k =\)
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\(p =\)
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Answer 1.
\({\frac{1}{6}}\)
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Answer 2.
\({\frac{1}{2}}\)
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Solution.
SOLUTION\(H(t) = \frac{ t^{3} }{6 \sqrt{t^{5}}} = \frac{1}{6} \frac{t^{3}}{\sqrt{t^{5}}} = \frac{1}{6} t^{(3 - 5/2)} = \frac{1}{6} t^{\frac{1}{2}}\)\(k = \frac{1}{6}\)\(p = \frac{1}{2}\)
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5.

Evaluate the expression \(\displaystyle{\ 27^{-4/3} }\text{.}\)
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Your answer must be an exact integer or rational number, not a decimal or algebraic expression.
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\(\displaystyle{27^{-4/3}}\) =
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Answer.
\({\frac{1}{81}}\)
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Solution.
\(\displaystyle{27^{-4/3} = \frac{1}{27^{4/3}} = \frac{1}{\big(3^3\big)^{4/3}} = \frac{1}{3^{(3\times 4/3)}} = \frac{1}{3^4}= {{\frac{1}{81}}}}\)
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6.

Simplify and write the following as a product of rational powers of \(b\) and \(y\text{.}\)
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If \(\displaystyle \frac{ \left(b^3 y^4\right)^{1/15}} {b^{-1/3}y^{1/5}} = b^m y^n\) then \(m =\) and \(n =\)
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Answer 1.
\({\frac{8}{15}}\)
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Answer 2.
\({\frac{1}{15}}\)
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Solution.
\(\displaystyle{\begin{aligned} \amp \displaystyle \frac{ \left(b^3 y^4\right)^{1/15}} {b^{-1/3}y^{1/5}} = \frac{ b^{3/15} y^{4/15} }{b^{-1/3} y^{1/5}} \\ \amp = b^{(3/15+1/3)}y^{(4/15-1/5)}=b^{{{\frac{8}{15}}}}y^{{{\frac{1}{15}}}}\end{aligned}}\)
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7.

The expression \(\displaystyle{\ \sqrt[3]{\sqrt{{729} x^5}} \ }\) equals \(\displaystyle{\ nx^r }\)
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where \(\ n\text{,}\) the leading coefficient, is:
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and \(\ r\text{,}\) the exponent of \(\ x\text{,}\) is:
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Answer 1.
\(3\)
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Answer 2.
\({\frac{5}{6}}\)
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Solution.
\(\displaystyle{\begin{gathered}\sqrt[3]{\sqrt{{729} x^5}} = \sqrt[3]{\sqrt{{3}^6 x^5}} = \big( ({3}^6 x^5)^{1/2}\big)^{1/3}\\ = ({3}^{6/2}x^{5/2})^{1/3} = {3}^{6/6} x^{5/6} = {3} x^{5/6} \end{gathered}}\)
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8.

Simplify the expression
\begin{equation*} 3 \sqrt[3]{4} \left( 4 \sqrt[3]{16} + 4 \sqrt[3]{7} \right) \end{equation*}
and express your answer in the simplest radical form \(A+ B \sqrt[3]{C},\) where \(A,\) \(B,\) and \(C\) are integers.
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Answer: \(A =\) , \(B =\) , and \(C =\)
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Answer 1.
\(4\cdot 3\cdot 4\)
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Answer 2.
\(3\cdot 4\)
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Answer 3.
\(7\cdot 4\)
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9.

Part 1 of 2:
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Expand and simplify \(\displaystyle h(t) = t^2 - t(t-5)\text{.}\)
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\(h(t)\) =
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Answer.
\(5t\)
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10.

Are the functions below polynomials? If they are, enter their degree. If not, enter NONE.
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\(f(x) = 9^x + 4\) has degree
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\(g(x) = x^{9} + 4\) has degree
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Answer 1.
\(\text{NONE}\)
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Answer 2.
\(9\)
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Solution.
SOLUTION\(f(x)\)\(9^x\)\(g(x)\)\(x\)\(x^9\text{.}\)
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11.

Are the functions below polynomials? If they are, enter their degree. If not, enter NONE.
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\(f(x) = 2 x^{3.4} + 3\) has degree
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\(g(x) = 2 x^{3} + 3.4\) has degree
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\(h(x) = 3 x^{-2} + 3.4\) has degree
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Answer 1.
\(\text{NONE}\)
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Answer 2.
\(3\)
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Answer 3.
\(\text{NONE}\)
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Solution.
SOLUTION\(f(x)\)\(x\text{.}\)\(g(x)\)\(x\)\(x^3\text{.}\)\(h(x)\)\(x\text{.}\)
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12.

Is the following expression a monomial, binomial, or trinomial?
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\({-19x^{4}+14x^{8}-15x^{3}}\)
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What is the degree of the expression?
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Answer 1.
\(\text{trinomial}\)
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Answer 2.
\(8\)
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Solution.
This polynomial has three terms \(\displaystyle{{-19x^{4}}}\text{,}\) \(\displaystyle{{14x^{8}}}\text{,}\) \(\displaystyle{{-15x^{3}}}\text{.}\) Since there are three terms, and the variable \(x\) in each term has a different exponent, this is a trinomial.
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The variable \(x\) is raised to the powers 4, 8 and 3 in this polynomial. The degree of the polynomial is the greatest of these powers, degree = 8.
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People often expect the terms of a polynomial to be ordered so the exponents appear in descending order (largest to smallest). This convention often makes the polynomial easier to read and avoids confusion. On the right side of the next equation the terms of our polynomial are re-ordered so the exponents appear in descending order. \(\displaystyle{ {-19x^{4}+14x^{8}-15x^{3}}={14x^{8}-19x^{4}-15x^{3}} }\) You may agree that the version on the right side is a little easier to read.
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13.

Given the table below, find a cubic equation in standard form for \(g(x)\text{.}\)
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x 6 -8 3 9
g(x) -374 1180 -41 -1319
\(g(x) =\) .
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Hint.
\(y = Ax^3 +Bx^2+Cx+D\text{.}\)
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Answer.
\(-2x^{3}+2x^{2}+\left(-3\right)x+4\)
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14.

Is the function below a rational function? If it is, write it in reduced form as a ratio of polynomials \(\displaystyle \frac{p(x)}{q(x)}\text{.}\) If it is not, enter NONE in both blanks. Do not leave any blanks empty.
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\(\displaystyle \frac{x-1}{\sqrt{x}-3} =\)
Answer 1.
\(\text{NONE}\)
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Answer 2.
\(\text{NONE}\)
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Solution.
SOLUTION\(\sqrt{x} - 3\)
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15.

Is the function below a rational function? If it is, write it in reduced form as a ratio of polynomials \(\displaystyle \frac{p(x)}{q(x)}\text{.}\) If it is not, enter NONE in both blanks. Do not leave any blanks empty.
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\(\displaystyle \frac{ x^2}{x-3} - \frac{2}{x-6} =\)
Answer 1.
\(x^{3}-6x^{2}-2x+6\)
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Answer 2.
\(x^{2}-9x+18\)
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Solution.
SOLUTION
\begin{equation*} f(x) = \frac{x^2}{x-3}+\frac{2}{x-6} = \frac{(x^2)(x-6)}{(x-3)(x-6)}-\frac{2(x-3)}{(x-6)(x-3)} = \frac{x^3-6x^2 - 2x + 6}{x^2-9x+18}. \end{equation*}
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