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Section 4.2 Constructing Functions

Overview

We have learned previously that functions provide a map between two variables of a system. In modeling, the functions are almost always defined by formulas, with the dependent variable being equal to an expression involving only the independent variable. As we analyze these functions with calculus, the rules of computation will depend on how a function algebraically put together.

This section focuses on how expressions and functions are constructed. We start by reviewing elementary functions that represent basic operations on the independent variable. Expressions formed by these elementary operations are combined with the basic arithmetic operations and function composition. Common constructions often take the same form, so we consider parametrized families of functions, particularly including power functions and exponential functions.

Subsection 4.2.1 Elementary Functions

Every expression defining a function can be interpreted as a combination of various operations. Operations that act on a single expression are functions. Operations that combine multiple expressions include the binary arithmetic operations, particularly addition and multiplication. In order to characterize expressions, we first review the elementary operations that can be considered as elementary functions.

Subsubsection 4.2.1.1 Constant and Identity Functions

The simplest operations are the constant functions and the identity function. As an operation, the constant function ignores the variable and always gives the same value for the output. The identity function, on the other hand, has no net change with the variable and returns an output that matches the input.

Definition 4.2.1

A constant function is a function that has the same output value for every input value, \(f(x)=c\) for some constant \(c\text{.}\)

Figure 4.2.2 The constant function \(f(x)=3\) as a map \(x \overset{f}\mapsto 3\text{.}\)
Figure 4.2.3 The graph of the constant function \(y=f(x)=3\) in the \((x,y)\) plane.
Definition 4.2.4

The identity function is a function where the output value is the same as the input value, \(f(x)=x\text{.}\)

Figure 4.2.5 The identity function \(f(x)=x\) as a map \(x \overset{f}\mapsto x\text{.}\)
Figure 4.2.6 The graph of the identity function \(y=f(x)=x\) in the \((x,y)\) plane.

We earlier learned that inverse functions \(f\) and \(f^{-1}\) have composition

\begin{equation*} f \circ f^{-1}(x) = x, \quad f^{-1}\circ f(x) = x; \end{equation*}

this is equivalent to saying that \(f \circ f^{-1}\) and \(f^{-1}\circ f\) are equivalent to an identity function.

Subsubsection 4.2.1.2 Elementary Arithmetic Operations

The four basic arithmetic operations of addition, subtraction, multiplication, and division can be used as functions. Because these binary operations require two operands (the expressions being acted on), the simplest functions involve operating on the variable and a particular constant.

For example, \(x \mapsto x+4\) is an elementary operation that adds the constant 4 to the independent variable. Similarly, \(x \mapsto 4x\) is an elementary operation that multiplies the input by 4. Because subtraction is really addition with an additive inverse (the negation) of a number, an operation like \(x \mapsto x-4\) is equivalent to \(x \mapsto x+-4\text{.}\) Likewise, division is really multiplication with a multiplicative inverse (the reciprocal) of a number, so an operation like \(x \mapsto x \div 4\) is equivalent to \(x \mapsto \frac{1}{4} x\text{.}\)

We therefore start with two families of constant arithmetic operations: the constant sum and the constant multiple.

Definition 4.2.7

For every real number (constant) \(c\text{,}\) we can define the constant sum operation

\begin{equation*} x \mapsto x+c \end{equation*}

and the constant multiple operation

\begin{equation*} x \mapsto cx\text{.} \end{equation*}

A constant sum represents a mapping that maintains a constant offset between the input and output. For example, the function \(x \mapsto x-3\) has an output that is always 3 units to the left of the input. This mapping is illustrated in the following interactive figure.

Figure 4.2.8 The constant sum \(f(x)=x-3\) as a map \(x \overset{f}\mapsto x-3\text{.}\)

A constant multiple represents a mapping that maintains a constant scaling or ratio between the input and output. For example, the function \(x \mapsto 2x\) has an output that is always twice the value of the input. This mapping is illustrated in the following interactive figure.

Figure 4.2.9 The constant multiple \(f(x)=2x\) as a map \(x \overset{f}\mapsto 2x\text{.}\)

There are two more arithmetic operations possible with constants. Taking a constant and subtracting the variable, as in \(x \mapsto 4-x\text{,}\) is not equivalent to a constant sum because we are not adding something to \(x\text{.}\) Similarly, dividing a constant by a variable, as in \(x \mapsto 4 \div x\text{,}\) is not equivalent to a constant multiple. We can, however, obtain these functions and others by introducing the reciprocal operation as an elementary arithmetic operation and then using composition of these elementary operations.

Definition 4.2.10

The reciprocal or inverse operation is the function

\begin{equation*} x \mapsto \div x = \frac{1}{x}\text{,} \end{equation*}

defined for all \(x \ne 0\text{.}\)

Example 4.2.11

Express the functions \(x \mapsto 4-x\) and \(x \mapsto 4 \div x\) as compositions of elementary arithmetic operations.

Solution

We start with the function \(x \mapsto 4-x\text{.}\) The negation of a number is always equal to that number times \(-1\)—a constant multiple,

\begin{equation*} f_1(x) = -1 x\text{.} \end{equation*}

Because \(4-x = -x + 4\text{,}\) we can see that we take the output of \(f_1\) and apply the elementary arithmetic operation of adding 4,

\begin{equation*} f_2(x) = x+4\text{.} \end{equation*}

The function is question is the composition,

\begin{equation*} f_2 \circ f_1(x) = f_2(-x) = -x + 4 = 4-x\text{.} \end{equation*}

For the second function, \(x \mapsto 4 \div x\text{,}\) we recall \(4 \div x = 4 \cdot \frac{1}{x}\text{.}\) Our first elementary operation is the reciprocal,

\begin{equation*} g_1(x) = \frac{1}{x}\text{.} \end{equation*}

We then multiply by the constant 4,

\begin{equation*} g_2(x) = 4x\text{.} \end{equation*}

Our function is the composition

\begin{equation*} g_2 \circ g_1(x) = g_2(\frac{1}{x}) = 4 \cdot \frac{1}{x} = \frac{4}{x}\text{.} \end{equation*}

We can construct fairly complex operations by composition of various elementary operations. Our ability to recognize compositions will be very important in calculus, so we take advantage of this opportunity to develop this skill.

Example 4.2.12

Express \(\displaystyle x \mapsto 3 + \frac{4}{2x-5}\) as a composition of elementary arithmetic operations.

Solution

We start by looking for the variable \(x\) in the formula. Using the order of operations, we state in words (at least in our minds) what happens to that value.

  • Start with \(x\text{.}\)
  • Multiply by 2.
  • Subtract 5, or, in other words, add \(-5\text{.}\)
  • Divide 4 by the result.
  • Add 3.

The step where we divide 4 by the result is not an elementary operation. We break it into two steps: (i) find the reciprocal and (ii) multiply by 4.

Now that each step is an elementary operation, we can state our composition. First, we define the relevant functions, one for each step.

\begin{gather*} f_1(x)=2x\\ f_2(x)=x+-5\\ f_3(x)=\frac{1}{x}\\ f_4(x)=4x\\ f_5(x)=x+3 \end{gather*}

Our function is the composition

\begin{equation*} f_5 \circ f_4 \circ f_3 \circ f_2 \circ f_1(x) = 4 \cdot \frac{2x+-5}+3\text{.} \end{equation*}

It is common for learners to be confused by the functions used in composition. For example, many learners think that \(f_2(x)\) should have been \(2x+-5\text{.}\) Remember that the function representing the operation describes what happens to its input and does not take into account what we might separately know will become the input. The composition of the first two operations would be \(f_2 \circ f_1(x) = 2x+-5\text{.}\) It might be helpful to use a different placeholder for the input. We could have written \(\square\) as a placeholder for the input of subsequent operations to remind us that the operation acts on the input and not on the \(x\) of the ultimate formula.

\begin{gather*} f_1(x)=2x\\ f_2(\square)=\square+-5\\ f_3(\square)=\frac{1}{\square}\\ f_4(\square)=4\square\\ f_5(\square)=\square+3 \end{gather*}

When a function can be represented as a composition of elementary arithmetic operations, the inverse function can easily be found by the reverse composition of inverse arithmetic operations.

  • The inverse of a constant sum operation is another sum with the negation of that constant.
  • The inverse of a constant multiple operation is another multiple with the inverse or reciprocal of that constant. (Multiplying by zero can not be inverted.)
  • The inverse of a reciprocal operation is to apply the reciprocal operation again.
Example 4.2.13

Find the inverse function for \(\displaystyle f(x)=\frac{4}{2x-5}+3\text{.}\)

Solution

In the previous example, we identified that \(f(x)\) was a composition of the following operations, starting with the input \(x\text{:}\)

  1. Multiply by 2.
  2. Add \(-5\text{.}\)
  3. Reciprocal.
  4. Multiply by 4.
  5. Add 3.

The inverse function must apply the inverse operations in the reverse order.

  1. Add \(-3\text{.}\)
  2. Multiply by \(\frac{1}{4}\text{.}\)
  3. Reciprocal.
  4. Add 5.
  5. Multiply by \(\frac{1}{2}\text{.}\)

The composition of these operations gives us an inverse function,

\begin{equation*} f^{-1}(x) = \frac{1}{2}\Big(\frac{1}{\frac{1}{4}(x+-3)}+5\Big)\text{.} \end{equation*}

We can rewrite this expression in a simpler equivalent form. We can simplify the fraction within a fraction by recognizing \(\frac{1}{4}(x+-3) = \frac{x-3}{4}\) and then finding the reciprocal:

\begin{align*} f^{-1}(x) &= \frac{1}{2}\Big(\frac{1}{\big(\frac{x-3}{4}\big)}+5\Big)\\ &= \frac{1}{2}\Big(\frac{4}{x-3}+5\Big). \end{align*}

If we wanted to distribute the multiplication by \(\frac{1}{2}\text{,}\) we would obtain a slightly simpler expression,

\begin{equation*} f^{-1}(x) = \frac{2}{x-3} + \frac{5}{2}\text{.} \end{equation*}

Subsubsection 4.2.1.3 Exponentials, Powers, and Logarithms

While studying sequences, we learned about the operations of exponentials 2.4.4 and powers 2.4.5. In effect, these two operations add powers involving constants to our collection of elementary operations. We also include the inverse operations.

Definition 4.2.14

The elementary power function with power \(p\) is the function that raises the variable to a constant power,

\begin{equation*} \mathrm{pow}_p(x)=x^p\text{.} \end{equation*}
Definition 4.2.15

The elementary exponential function with base \(b\text{,}\) where \(b \gt 0\) and \(b \ne 1\) is the function that raises a constant base to the power of the variable,

\begin{equation*} \exp_b(x)=b^x\text{.} \end{equation*}

The inverse operations for a power is a root, which is itself a power:

\begin{equation*} \mathrm{pow}_p^{-1}(x) = \sqrt[p]{x} = x^{(1/p)}\text{.} \end{equation*}

Technically, the power operation is not one-to-one for \(p\) an even power, so the root is only a true inverse operation for \(p\) odd. When \(p\) is even, the root is an inverse operation only for non-negative values.

The inverse operation for an exponential is a logarithm,

\begin{equation*} \exp_b^{-1}(x) = \log_b(x) \end{equation*}

We consider logarithms as elementary functions.

Subsubsection 4.2.1.4 Trigonometric Functions

The trigonometric functions are used in relation to triangles as well as cyclic behavior. There are two fundamental trigonometric functions, the sine and cosine functions, from which the others are defined. We will study these functions in more depth soon, but for the purpose of summary include the following table here.

\begin{align*} \sin(x) & & \text{sine} \\ \cos(x) & & \text{cosine} \\ \tan(x) &= \frac{\sin(x)}{\cos(x)} & \text{tangent} \\ \sec(x) &= \frac{1}{\cos(x)} & \text{secant} \\ \cot(x) &= \frac{\cos(x)}{\sin(x)} & \text{cotangent} \\ \csc(x) &= \frac{1}{\sin(x)} & \text{cosecant} \end{align*}

The trigonometric functions are periodic, which implies that they must not be one-to-one. Inverse trigonometric functions are defined to solve equations for a limited interval and provide additional elementary functions for our use.

\begin{align*} \sin^{-1}(x) &= \arcsin(x) & \text{arcsine} \\ \cos^{-1}(x) &= \arccos(x) & \text{arccosine} \\ \tan^{-1}(x) &= \arctan(x) & \text{arctangent} \\ \sec^{-1}(x) &= \mathop{\mathrm{arcsec}}(x) & \text{arcsecant} \end{align*}

The arccotangent and arccosecant functions can be defined but are not used in practice.

Subsection 4.2.2 Algebraic Combinations

Functions defined by a formula are generally formed by combining these operations and functions into more complicated expressions. One of the most valuable skills in calculus is the ability to recognize how a formula is constructed. Many rules in calculus are named according to which operation forms the expression of interest. The basic operations of combination are the arithmetic operations of addition (a sum), subtraction (a difference), multiplication (a product), and division (a quotient) along with the operation of function composition. Although powers are usually considered algebraic operations, we will think of powers in terms of power and exponential functions.

Most formulas involve more than one operation. An expression is classified by the last operation that would be applied. The order of operations determines the priority with which operations are applied. In algebra, you may have learned the acronym PEMDAS, which stands for Parentheses, Exponents, Multiplication, Division, Addition, and Subtraction. Subtraction is really the addition of an inverse, so differences can be classified as sums. The same technically applies for division being multiplication, but this is less frequently used. We will change the meaning of E to stand for Every function, including powers and exponentials, as all functions have higher precedence than the arithmetic operations. Composition occurs whenever the operation is a function acting on the output of another operation involving a variable.

Example 4.2.16

Classify each function by the last operation that is applied, and then classify each component expression. Make note of when a binary operation involves a constant instead of two variable expressions.

  1. \(f(x)=x^2-3x\sin(x)\)
  2. \(g(x)=(2x+1)(x-3)\)
  3. \(h(x)=(x^2+3)^4\)
  4. \(j(x)=\displaystyle \frac{2xy}{\sqrt{3x-1}}\)
  5. \(k(x)=5e^{2x}\)
Solution
  1. The function \(f(x)=x^2-3x\sin(x)\) is a difference of the expressions \(x^2\) and \(3x\sin(x)\text{.}\) The first component expression \(x^2\) is a power function (\(p=2\)) of \(x\text{;}\) the second component expression \(3x\sin(x)\) is a product of \(3x\) and \(\sin(x)\text{.}\) (We could also have used a sum of \(x^2\) and \(-3x\sin(x)\text{.}\))
  2. The function \(g(x)=(2x+1)(x-3)\) is a product of the expressions \(2x+1\) and \(x-3\text{.}\) The first expression \(2x+1\) is the constant sum of \(2x\) and \(1\) while the second expression is the constant sum of \(x\) and \(-3\text{.}\)
  3. The function \(h(x)=(x^2+3)^4\) has the power (\(p=4\)) as its last operation. Because we treat powers as functions, this is a composition. The inner expression is \(u=x^2+3\text{,}\) and \(h(x)=u^4\text{.}\) If we define outer function \(O(x)=x^4\) and inner function \(I(x)=x^2+3\text{,}\) then \(h(x) = O \circ I(x)\text{.}\) The outer expression is an elementary power. The inner expression is a sum of \(x^2\) and the constant \(3\text{.}\)
  4. The function \(j(x)=\displaystyle \frac{2xe^x}{\sqrt{3x-1}}\) is a quotient of expressions \(2xe^x\) and \(\sqrt{3x-1}\text{.}\) The first expression \(2xe^x\) is a product of \(2x\) and \(e^x\text{;}\) the second expression \(\sqrt{3x-1}\) is a square root (a function) of the expression \(3x-1\text{,}\) meaning this is a composition. The outer function would be \(O(x)=\sqrt{x}\) and the inner function would be \(I(x)=3x-1\text{.}\) We could also think of the square root as an elementary power function, \(O(x)=x^{\frac{1}{2}}\text{.}\)
  5. The expression \(5e^{2x}\) is a constant multiple of \(5\) with \(e^{2x}\text{.}\) the expression \(e^{2x}\) is a natural exponential function (base \(e\)) in composition, \(e^u\text{,}\) with the expression \(u=2x\text{.}\)

Although binary operations like addition and multiplication are defined in terms of two operands, we often see them in expressions involving more than two terms, such as \(a+b+c\) or \(3xy\text{.}\) By convention, the operations are performed left to right as \((a+b)+c\) or \((3x)y\text{.}\) Because addition and multiplication are commutative and associative, this order doesn't matter and we act as if it were one sum or one product. In calculus, however, all of the rules are based on the binary nature of the operations. When classifying the structure of a formula, we should identify exactly two operands.

Linear Combinations and Polynomials

One of the most common ways to combine expressions in mathematics is to create a sum of constant multiples of those expressions. Such a combination is called a linear combination. The calculus operations of limits, integrals, and derivatives all satisfy a linearity in that they preserve linear combinations. It is therefore useful to recognize them.

Definition 4.2.17

Given a finite set of expressions, \(u=(u_1, u_2, \ldots, u_n)\text{,}\) and the same number of constants, \(c = (c_1, c_2, \ldots, c_n)\text{,}\) the linear combination of the expressions \(u\) with coefficients \(c\) is the sum of constant multiples of the expressions

\begin{equation*} c_1 u_1 + c_2 u_2 + \cdots + c_n u_n. \end{equation*}

The non-negative integer powers of \(x\) are the powers \(x^0=1\text{,}\) \(x^1=x\text{,}\) \(x^2\text{,}\) \(x^3\text{,}\) etc., and form the basis for an important family of functions called polynomials.

Definition 4.2.18

Let \(n\) be a non-negative integer. A polynomial of degree \(n\) is a function that can be written in the form

\begin{equation*} f(x) = a_n x^n + \cdots + a_2 x^2 + a_1 x + a_0\text{,} \end{equation*}

where \(a_0, a_1, \ldots, a_n\) are constants called the coefficients. The term with the highest power \(a_n x^n\) is called the leading term and \(a_n\) is called the leading coefficient.

Polynomials are just linear combinations of non-negative integer powers of \(x\text{.}\) A constant multiple of just one such power is called a monomial. The sum of two constant multiples is called a binomial.

Example 4.2.19

The polynomial \(f(x)=3x^3+x^2-5x+8\) is a linear combination of the powers \((1, x, x^2, x^3)\text{.}\) The degree of the polynomial is \(n=3\text{,}\) and the coefficients are \((c_0, c_1, c_2, c_3) = (8, -5, 1, 3)\text{.}\) The leading coefficient is \(c_3=3\text{.}\)

Example 4.2.20

Write down the polynomial \(f(x)\) of degree \(n=4\) with coefficients \((c_0,c_1,c_2,c_3,c_4) = (16, 0, -8, 0, 1)\text{.}\)

Solution

Because \(c_1=0\) and \(c_3=0\text{,}\) we skip the terms with powers \(x^1=x\) and \(x^3\text{.}\) We usually write polynomials in decreasing powers, so we have

\begin{align*} f(x) &= 1x^4 + 0x^3 + -8x^2 + 0x^1 + 16 x^0 \\ &= x^4-8x^2+16. \end{align*}

Subsection 4.2.3 Parametrized Functions

When functions have the same structure but use different constants, we say they are in the same parametrized (or parametric) family. For example, linear functions can always be written in the form \(f(x)=mx+b\) with different values of \(m\) and \(b\text{.}\) The constants are called the parameters of the family. Some examples of parametrized functions are:

  1. Linear functions \(f(x)=mx+b\text{,}\) parameters \(m\) and \(b\text{;}\)
  2. General power functions \(f(x)=Ax^p\text{,}\) parameters \(A\) and \(p\text{;}\)
  3. General exponential functions \(f(x)=Ab^x\text{,}\) parameters \(A\) and \(b\) with \(b \gt 0\) and \(b \ne 1\text{.}\) These can always be written (see Theorem 2.5.10) in terms of the natural base \(f(x) = Ae^{kx}\text{,}\) with parameters \(A\) and \(k\text{.}\)

In addition, polynomials could also be considered a parametrized function where the coefficients are the parameters.

  1. Quadratic functions \(f(x)=ax^2+bx+c\text{,}\) parameters \(a\text{,}\) \(b\text{,}\) and \(c\text{;}\)
  2. Cubic functions \(f(x)=ax^3+bx^2+cx+d\text{,}\) parameters \(a\text{,}\) \(b\text{,}\) \(c\text{,}\) and \(d\text{;}\)

To find the values of the parameters for a function, we use given states for the relation to create equations. The equation associated with each state is called a constraint. We generally need as many constraints as parameters in the function. We solve the system of equations simultaneously to find the parameters. This usually requires a method called solving by substitution.

The method of substitution for a system of equations is to take one equation and solve for one of the unknowns. We then substitute that formula in place of that unknown in the other equation. This allows us to find the value of the second unknown. We then back-substitute that value into our first equation to solve for the first unknown.

Example 4.2.21

Use parameter constraint equations to find a linear function \(f(x)=mx+b\) satisfying \(f(2)=5\) and \(f(5)=1\text{.}\)

Solution

We could find an equation of the function by computing the slope and \(y\)-intercept, but we want to illustrate parameter constraints. The data will give us equations for \(m\) and \(b\text{.}\) We use the parametrized formula \(f(x)=mx+b\) for the two given values \(x=2\) and \(x=5\) to give us two constraint equations.

\begin{gather*} f(2)=5 \quad \Rightarrow \quad \left\{ \begin{matrix} x=2 \\ f(x)=5 \end{matrix} \right. \quad \Rightarrow \quad 5 = m(2)+b \\ f(5)=1 \quad \Rightarrow \quad \left\{ \begin{matrix} x=5 \\ f(x)=1 \end{matrix} \right. \quad \Rightarrow \quad 1 = m(5)+b \end{gather*}

The system of equations that we need to solve is given by

\begin{equation*} \left\{ \begin{matrix} 2m+b = 5, \\ 5m+b= 1. \end{matrix}\right. \end{equation*}

We solve for \(b\) in the first equation.

\begin{gather*} 2m+b = 5\\ b=5-2m \end{gather*}

Now substitute this formula into the second equation.

\begin{gather*} 5m+b=1\\ 5m+(5-2m)=1\\ 3m+5=1\\ \textstyle m=-\frac{4}{3} \end{gather*}

We finish by back-substituting this value into the first equation to find the value for \(b\text{,}\)

\begin{equation*} \textstyle b=5-2m=5-2(\frac{4}{3})=5-\frac{8}{3}=\frac{7}{3}. \end{equation*}

The function is \(f(x)=-\frac{4}{3}x+\frac{7}{3}\text{.}\)

Solving the parameter equations for a parametric model provides us a way of identifying a function from a particular family that passes through given points.

Example 4.2.22

Find a power function \(f(x)=Ax^p\) satisfying \(f(2)=5\) and \(f(5)=1\text{.}\)

Solution

Start by creating the system of constraint equations.

\begin{equation*} \left\{ \begin{matrix} A \cdot 2^p = 5 \\ A \cdot 5^p = 1 \end{matrix} \right. \end{equation*}

To use the method of substitution, it is easiest to solve for \(A\text{:}\)

\begin{gather*} A \cdot 2^p = 5, \\ A = \frac{5}{2^p}. \end{gather*}

Substitute this into the second equation to get an equation for \(p\text{:}\)

\begin{gather*} A \cdot 5^p = 1 \\ \Big( \frac{5}{\displaystyle 2^p} \Big) \cdot 5^p = 1 \\ 5 \cdot \Big( \frac{5^p}{2^p} \Big) = 1 \\ 5 \cdot (\textstyle \frac{5}{2})^p = 1 \end{gather*}

To solve this, we isolate the exponential with base \(\frac{5}{2}\) and then use a logarithm as the inverse operation.

\begin{gather*} 5 \cdot (\textstyle \frac{5}{2})^p = 1 \\ (\textstyle \frac{5}{2})^p = \frac{1}{5} \\ \textstyle \log_{\frac{5}{2}}(\exp_{\frac{5}{2}}(p)) = \log_{\frac{5}{2}}(\frac{1}{5})\\ \textstyle p = \log_{\frac{5}{2}}(\frac{1}{5})\\ p = \frac{\log(\frac{1}{5})}{\log(\frac{5}{2})} \approx -1.7565 \end{gather*}

This allows us to find the value of \(A\text{:}\)

\begin{equation*} A = \frac{5}{2^p} \approx \frac{5}{2^{-1.7565}} \approx 16.894. \end{equation*}

The exact function is defined by

\begin{equation*} f(x) = \frac{5}{2^p} \cdot x^p, \qquad p = \log_{\frac{5}{2}}({\textstyle \frac{1}{5}}), \end{equation*}

with an approximate form

\begin{equation*} f(x) \approx 16.894 \cdot x^{-1.7565}. \end{equation*}
Example 4.2.23

Find an exponential function \(f(x)=Ab^x\) satisfying \(f(2)=5\) and \(f(5)=1\text{.}\)

Solution

Start by creating the system of constraint equations.

\begin{equation*} \left\{ \begin{matrix} A \cdot b^2 = 5 \\ A \cdot b^5 = 1 \end{matrix} \right. \end{equation*}

We solve for \(A\text{:}\)

\begin{gather*} A \cdot b^2 = 5, \\ A = \frac{5}{b^2}. \end{gather*}

Substitute this into the second equation to get an equation for \(p\text{:}\)

\begin{gather*} A \cdot b^5 = 1 \\ \Big( \frac{5}{\displaystyle b^2} \Big) \cdot b^5 = 1 \\ 5 \cdot \Big( \frac{b^5}{b^2} \Big) = 1 \\ 5 \cdot b^3 = 1 \\ b^3 = \textstyle{\frac{1}{5}}\\ b = \sqrt[3]{\textstyle \frac{1}{5}} = \frac{1}{5^{\frac{1}{3}}} = 5^{-\frac{1}{3}} \end{gather*}

The value of \(A\) can be found:

\begin{equation*} A = \frac{5}{b^2} = \frac{5}{(5^{-\frac{1}{3}})^2} = \frac{5}{5^{-\frac{2}{3}}} = 5 \cdot 5^{\frac{2}{3}} = 5^{\frac{5}{3}}. \end{equation*}

The exact exponential function is defined by

\begin{equation*} f(x) =5^{\frac{5}{3}} \cdot (5^{-\frac{1}{3}})^x\text{,} \end{equation*}

and is approximated by

\begin{equation*} f(x) \approx 14.620 \cdot 0.58480^x\text{.} \end{equation*}

When we know how to find models for functions, we can use those models to answer questions about physical systems.

Example 4.2.24

A chemical reaction is tracked with concentrations measured every five minutes. The concentration of one reactant is predicted to follow an exponential decay. Let \(R\) be the concentration of that reactant in moles per liter as a function of time \(t\) in minutes. When \(t=0\text{,}\) the concentration was \(R=2.5\text{.}\) When \(t=5\text{,}\) the concentration was \(R=2.1\text{.}\) At what time will the concentration be \(R=1.0\text{?}\)

Solution

This question is a bit more challenging as it will involve several stages. We begin by looking at the question. We want to find the time \(t\) knowing the concentration \(R\text{.}\) That is, we would like to apply a map \(R \mapsto t\) which would be the inverse of \(t \mapsto R\text{.}\) To find our answer, we will first need to write \(R\) as a dependent variable with \(t\) as the independent variable.

Next, we look at the given information to see what it says about the relation \(f: t \mapsto R\text{.}\) Because the problem states that the reactant follows exponential decay, we interpret this as telling us that \(R = f(t)\) is an exponential function. The general exponential function has the form

\begin{equation*} f(t) = A \cdot b^{t} \end{equation*}

where \(A\) and \(b\) are the model parameters (constants).

We use the given information to determine the values of \(A\) and \(b\text{.}\) Each state will give us an equation that can be used to constrain the values of the parameters. The state \((t,R) = (0,2.5)\) corresponds to the equation \(f(0) = 2.5\text{.}\) The exponential function applied to this equation gives

\begin{equation*} f(0) = A \cdot b^0 = A \qquad \Rightarrow \qquad A = 2.5\text{.} \end{equation*}

The state \((t,R) = (5,2.1)\) corresponds to the equation \(f(5)=2.1\text{,}\) which gives an equation

\begin{equation*} f(5) = A \cdot b^5 \qquad \Rightarrow \qquad A \cdot b^5 = 2.1\text{.} \end{equation*}

Substituting \(A=2.5\) into the second equation, we can then solve for \(b\text{.}\)

\begin{gather*} 2.5 \cdot b^5 = 2.1\\ b^5 = \frac{2.1}{2.5} = 0.84 \end{gather*}

The inverse operation of a power \(p=5\) is a fifth-root or a power \(p=\frac{1}{5}\text{.}\)

\begin{equation*} b = \sqrt[5]{0.84} = 0.84^{1/5} \approx 0.96573 \end{equation*}

We will use the exact value.

Knowing \(A=2.5\) and \(b = 0.84^{1/5}\text{,}\) we have our function \(f : t \mapsto R\text{,}\)

\begin{equation*} R = f(t) = 2.5 \cdot (0.84^{1/5})^{t} = 2.5 \cdot 0.84^{t/5}. \end{equation*}

The inverse function \(f^{-1} : R \mapsto t\) can be used to take a concentration and report back a time. We need to solve the equation for \(t\) as the dependent variable. When choosing what operation to apply an inverse, we first need to isolate the exponential.

\begin{gather*} R = 2.5 \cdot 0.84^{t/5}\\ \frac{R}{2.5} = 0.84^{t/5} \end{gather*}

We now have an exponential with base \(b=0.84\) on the right. The balanced inverse operation would be to use a logarithm with \(b=0.84\) on both sides.

\begin{gather*} \log_{0.84}(\frac{R}{2.5}) = \frac{t}{5}\\ 5 \log_{0.84}(\frac{R}{2.5}) = t \end{gather*}

We finish by using the change of base formula for logarithms:

\begin{equation*} t = 5 \cdot \frac{\log(R/2.5)}{\log(0.84)}\text{.} \end{equation*}

To answer the question, we use the inverse relation with \(R=1.0\) and compute \(t\text{.}\)

\begin{align*} t &= 5 \cdot \frac{\log(1.0/2.5)}{\log(0.84)}\\ &= 5 \cdot \frac{\log(0.4)}{\log(0.84)}\\ &\approx 26.277 \end{align*}

The concentration will be \(R=1.0\) approximately 26.277 minutes after the experiment begins.

Subsection 4.2.4 Summary

  • Functions defined by formulas are typically constructed from elementary functions: constant functions, the identity function, power functions, exponential functions, logarithms, and trigonometric functions.
  • Combinations of expressions can be arithmetic (sum, difference, product, or quotient) or the composition of functions.
  • An expression is classified by the last operation used to construct that expression.
  • Binary operations involving a constant operand are special cases. They can be constructed using only constant sums, constant multiples, and reciprocals.
  • A parametrized family of functions is a set of functions that have the same structure with different constants. The constants that can change are called parameters.
  • Common parametrized families of functions are linear, exponential, and power functions.

    Parametric Formula Description
    \(f(x)=mx+b\) linear, slope-intercept
    \(f(x)=A \, x^p\) power
    \(f(x)=A \, b^x\) exponential, general base \(b\)
    \(f(x)=A \, e^{kx}\) exponential, natural base \(e\)
  • A polynomial is a linear combination of simple powers \((1,x,x^2,\ldots,x^n)\text{,}\) or, in other words, a sum of constant multiples of these powers,
    \begin{equation*} f(x)=a_n x^n + \cdots + a_2 x^2 + a_1 x + a_0\text{.} \end{equation*}
    The constant multiples \((a_0,a_1,\ldots,a_n)\) are called the coefficients. The term \(a_n x^n\) is called the leading term.
  • Knowing the value of a parametrized function for a state establishes a constraint equation for the parameters. With enough constraints, we can often solve for the parameter values using the method of substitution.

Subsection 4.2.5 Exercises

1

Classify each elementary function.

  1. \(f(x)=\pi\)
  2. \(g(x)=x\)
  3. \(h(x)=x^\pi\)
  4. \(j(x)=\pi^x\)
  5. \(k(x)=\sin(x)\)
2

Classify each function according to the last operation. Then classify the component expressions. Make note if the operation involves a constant expression.

  1. \(f(x)=4x^4\)
  2. \(g(x)=2^{3x}+5\)
  3. \(h(x)=3^{5x-1}\)
  4. \(\displaystyle j(x)=3\sqrt{x} + \frac{1}{x^2}\)
  5. \(k(x)=4x^2e^{3x}\)
  6. \(\displaystyle m(x)=\frac{x^2(3x-1)}{(x^2+1)^4}\)
3

For each polynomial, determine the degree and list the coefficients.

  1. \(f(x)=3x^2+5x-1\)
  2. \(f(x)=x^3-2x+8\)
  3. \(f(x)=x^4-1\)
  4. \(f(x)=x^4+4x^3+6x^2+4x+1\)

Find the equation of the function \(x \mapsto y\text{,}\) if possible, for each of the following parametric models satisfying the states \((x,y)=(0,3)\) and \((x,y)=(5,9)\text{.}\)

4

linear function

5

power function

6

exponential function

7

quadratic function of the form \(y=a+bx^2\)

8

quadratic function of the form \(y=ax+bx^2\)

Find the equation of the function \(x \mapsto y\text{,}\) if possible, for each of the following parametric models satisfying the states \((x,y)=(1,3)\) and \((x,y)=(4,6)\text{.}\)

9

linear function

10

power function

11

exponential function

12

quadratic function of the form \(y=a+bx^2\)

13

quadratic function of the form \(y=ax+bx^2\)

14

a function the form \(\displaystyle y=\frac{ax}{x+b}\)

Parametrized functions are often used to model real-life scenarios. In particular, exponential functions are often used to model problems involving radioactive decay. A half-life is a length of time during which the amount of the radioactive substance loses exactly half of its mass. We think of the mass \(M\) as a function of time \(t\text{,}\) \(M=A\cdot b^t\text{,}\) with parameters \(A\) and \(b\text{.}\)

Use the information given in each problem to find a parametrized exponential function satisfying the given data and use it to answer the question.

15

P-32 is a radioactive isotope of phosphorus used in labeling biological molecules. P-32 has a half-life of 14.29 days. Suppose an experiment begins with 10 µg of P-32. Find a parametrized model for the mass (in µg) as a function of time (in days) measured from the start of the experiment, \(t \mapsto M\text{,}\) in order to determine how much P-32 remains after 10 days and after 100 days.

Hint

Create two constraints using \(t=0\) and \(t=14.29\text{.}\)

16

C-14 is a radioactive isotope of carbon used in dating biological samples. C-14 has a half-life of 5730 years. Due to cosmic rays on atmospheric nitrogen, C-14 has a steady concentration in the atmosphere. The tissue of living plants match the atmospheric concentration until the time of death. Measuring how much the fraction of carbon that is C-14 allows us to estimate the time the plant tissue died.

Suppose a sample is measured as having 3.5 mg. How old is the sample if the living sample is estimated to have had 5.0 mg?

Hint

The current measurement gives a constraint for \(t=0\text{;}\) the half-life gives a second constraint. Solve for \(t\) corresponding to the living sample.

17

Radioactive masses decay according to an exponential function of time. The isotope of plutonium Pu-239 has a half-life of 24,110 years, which is the time after which half of the mass has decayed. For an initial mass of 1 kg, how much plutonium remains after 100 years?

18

An exponentially growing population that doubles in size every 5 years currently has 1000 individuals.

  1. What will the population be in 4 years?
  2. How long does it take for the population to triple?