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Section 4.1 An Introduction to Functions


Calculus studies the relationships between variables. These relationships are most often expressed in terms of functions. In mathematics, a function is a predictive relationship between an independent and a dependent variable such that knowing the value of the independent variable allows us to predict the value of the dependent variable.

We have previously learned to think of these relationships as maps. One of the easiest ways to establish a map is with an equation where a dependent variable, say \(y\text{,}\) is equal to an expression involving only the independent variable, say \(x\text{.}\) The equation defines the map \(x \mapsto y\text{.}\)

In this section, we will review and extend these concepts relating to functions. We use linear functions to review the concepts of maps and to introduce the idea of rate of change. We discuss composition of functions as a way of applying a sequence of operations. Finally, we look at inverse functions in terms of function composition.

Subsection 4.1.1 Linear Functions

A linear function describes a relationship between variables that has a constant rate of change. Let us review this in the context of maps between variables \(x\) and \(y\text{.}\) Suppose we know two states for the variables \((x,y)\text{,}\) say \((x_1,y_1)\) and \((x_2,y_2)\text{.}\) From the perspective of a map, we want a function that maps \(x_1 \mapsto y_1\) and \(x_2 \mapsto y_2\text{.}\) These two points states have corresponding increments in \(x\) and \(y\) given by \(\Delta x = x_2 - x_1\) and \(\Delta y = y_2 - y_1\text{.}\)

A constant rate of change is a constant ratio of the increments. The ratio of the increment in the output to the increment in the input,

\begin{equation*} m = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1}\text{,} \end{equation*}

will always have the same value regardless of which two states are selected. The rate of change is called the slope of the line.

Below is an interactive figure illustrating a linear function with states \((x,y)=(1,2)\) and \((x,y)=(3,8)\text{.}\) The increment of the input is \(\Delta x = x_2 -x_1 = 3-1 = 2\text{.}\) The increment of the output is \(\Delta y = y_2 - y_1 = 8-2 = 6\text{.}\) The rate of change is \(m=\frac{6}{2} = 3\) because the increment of the output is three times the increment of the input. You can adjust a slider controlling the value of \(x\) as input. Notice that the increment of output relative to \((x,y)=(1,2)\) is always \(m=3\) times the increment of the output.

Figure 4.1.1 The linear function that maps \(x=1 \mapsto y=2\) and \(x=3 \mapsto y=8\) has rate of change \(m=2\text{.}\)
Example 4.1.2

For each of the following data tables, determine if the data could come from a linear relation. If so, give an equation of the line.

  1. \(x\) 1 4 10
    \(y\) 5 3 -1
  2. \(x\) 2 6 12
    \(y\) 11 19 27
  1. For the first data set, we measure the slope if we compare the first point with each of the other two points. If those are the same, then the data are consistent with a linear relation. Let us label the states \((x_1,y_1)=(1,5)\text{,}\) \((x_2,y_2)=(4,3)\text{,}\) and \((x_3,y_3)=(10,-1)\text{.}\) Compute the slopes:

    \begin{align*} \frac{y_2-y_1}{x_2-x_1} &= \frac{3-5}{4-1} = -\frac{2}{3} \\ \frac{y_3-y_1}{x_3-x_1} &= \frac{-1-5}{10-1} = -\frac{6}{9} = -\frac{2}{3} \end{align*}

    Because the slopes are the same value \(m=-\frac{2}{3}\text{,}\) there is a linear relation with these states.

    We find the equation \(y=mx+b\) by computing \(b=y-mx\) using one of the given states and \(m=-\frac{2}{3}\text{,}\) such as \((x_1,y_1)=(1,5)\text{:}\)

    \begin{equation*} \textstyle b = 5-(-\frac{2}{3})(1) = 5+\frac{2}{3} = \frac{17}{3}. \end{equation*}

    The linear relation is characterized by the equation

    \begin{equation*} \textstyle y=-\frac{2}{3}x+\frac{17}{3}. \end{equation*}

    If we multiply both sides by 3, we can find an equivalent equation involving only integers \(3y = -2x+17\text{,}\) which is equivalent to an equation \(2x+3y=17\text{.}\) The graph of this line along with the data are shown below.

  2. For the second data set, we measure the slopes using the states \((x_1,y_1)=(2,11)\text{,}\) \((x_2,y_2)=(6,19)\text{,}\) and \((x_3,y_3)=(12,27)\text{:}\)

    \begin{align*} \frac{y_2-y_1}{x_2-x_1} &= \frac{19-11}{6-2} = \frac{8}{4} = 2 \\ \frac{y_3-y_1}{x_3-x_1} &= \frac{27-11}{12-2} = \frac{16}{10} = \frac{8}{5} \end{align*}

    Because the slopes have different values, these data are not consistent with a linear relation. The figure below illustrates the data along with the line passing through the first two states. The third data point is not on the line.

In the context of functions, the equation of a line solved for \(y\) defines a function \(x \mapsto y\text{.}\) Most algebra courses introduce the slope-intercept form A.2.6 of a line,

\begin{equation*} y=mx+b\text{,} \end{equation*}

where \(m\) is the slope and \(b\) is the \(y\)-value of the \(y\)-intercept. However, the point-slope form of a line is actually more useful to remember,

\begin{equation*} y=m(x-h) + k\text{,} \end{equation*}

where \(m\) is still the slope but \((h,k)\) is any point included on the line. Rewriting this slightly, the equation is equivalent to

\begin{equation*} y-k = m (x-h)\text{.} \end{equation*}

The point-slope form of a line guarantees that \(x=h \mapsto y=k\) and the change in output from that point \(\Delta y = y-k\) is exactly equal to the slope \(m\) times the change in input \(\Delta x = x-h\text{.}\)

Example 4.1.3

In 1990, the population of Harrisonburg, Virginia, was 30,707. In 2010, the population was 48,914. If the rate of change of the population increased at a constant rate, find a model for the population of Harrisonburg as a function of the year.


A model with a constant rate of change is a linear function, \(y \mapsto P\text{,}\) where \(y\) is the year and \(P\) is the population. The given data show \(y=1990 \mapsto P=30707\) and \(y=2010 \mapsto P=48914\text{.}\) We find the slope as the ratio of \(\Delta P\) (change in output) to \(\Delta y\) (change in input).

\begin{align*} m &= \frac{\Delta P}{\Delta y} \\ &= \frac{48914 - 30707}{2010-1990} \\ &= \frac{18207}{20} = 910.35 \end{align*}

Using the point-slope form of the line, we start with the known mapping \(1990 \mapsto 30707\) and the constant rate of change to give

\begin{equation*} P = 30707 + 910.35(y - 1990)\text{.} \end{equation*}

Subsection 4.1.2 Function Composition

In Subsection 2.2.2, we introduced the idea that a function is often assigned a name, which might be a letter, a word, or any symbol. In math textbooks, the most common name for a function is \(f\text{.}\) We write \(x \overset{f}{\mapsto} y\) or \(f : x \mapsto y\) as a symbolic statement that \(f\) is the name of the function relating an independent variable \(x\) and a dependent variable \(y\text{.}\) When the function is defined by an equation, as in our parametric models, we often define the function by saying \(f(x)\) is equal to the expression that defines the map.

Imagine we had a function \(t \mapsto B\) where \(t\) measures the time in years since we opened a bank account and \(B\) measures the account balance. This account hypothetically grows according to the model

\begin{equation*} B = 500 (1.02)^t\text{.} \end{equation*}

Consider how we would communicate calculating the balance after two, five, and ten years. We would use a statement like, “When \(t=2\text{,}\) we have \(B=500 (1.02)^{2}\text{,}\)” and then we could compute the value. For each value of time, we have to specify the value we substitute for \(t\) in the equation.

Function notation provides a concise way of specifying which value we are using in substitution. If we would introduce a function \(f\) to represent the map,

\begin{equation*} B = f(t) = 500 (1.02)^t\text{,} \end{equation*}

then simply writing \(f(2)\) immediately implies \(t=2\) without having to say this separately. The three values of interest could be written,

\begin{gather*} f(2) = 500 (1.02)^2 = 520.20,\\ f(5) = 500 (1.02)^5 \approx 552.04,\\ f(10) = 500 (1.02)^{10} \approx 609.50. \end{gather*}
Note 4.1.4

Although writing parentheses in mathematics next to a number or a variable means multiplication, a function is not a variable. The parentheses after the function do not mean multiplication but evaluation. It is unfortunate that the same symbols have different meanings, so you will need to pay close attention to the context.

The next few examples illustrate how evaluation notation relates to substitution.

Example 4.1.5

For \(f(x)=-\frac{2}{3}x+2\text{,}\) simplify each of the following expressions.

  1. \(f(4)\)
  2. \(f(t)\)
  3. \(f(3x+1)\)
  1. To simplify \(f(4)\text{,}\) we see that the value inside parentheses is 4. So we substitute the independent variable with the value 4.

    \begin{align*} f(x) &= -{\textstyle \frac{2}{3}}x+2\\ f(4) &= -{\textstyle \frac{2}{3}}(4)+2 = {-\textstyle \frac{8}{3} + \frac{6}{3}} = {\textstyle -\frac{2}{3}} \end{align*}
  2. The expression \(f(t)\) substitutes the variable \(t\) in place of \(x\text{:}\)

    \begin{equation*} f(t) = -{\textstyle \frac{2}{3}}t+2. \end{equation*}
  3. The expression \(f(3x+1)\) substitutes the expression \(3x+1\) for the independent variable \(x\text{.}\) Although the original independent variable of \(f(x)=-\frac{2}{3}x+2\) is \(x\text{,}\) we should think of this substitution in terms of a placeholder \(f(\square)=-\frac{2}{3}\square+2\) and substitution \(\square = 3x+1\text{.}\)

    \begin{align*} f(x) &= -{\textstyle \frac{2}{3}}x+2\\ f(\square) &= -{\textstyle \frac{2}{3}}\square+2\\ f(3x+1) &= -{\textstyle \frac{2}{3}}(3x+1)+2\\ &= \textstyle -\frac{2}{3}(3x) + -\frac{2}{3}(1) + 2 \\ &= \textstyle -2x + \frac{4}{3} \end{align*}

    Introducing the placeholder symbol \(\square\) is not necessary but might help emphasize that we are not just including things with the \(x\text{.}\) We are performing an actual substitution, \(\square = 3x+1\text{.}\)

Note 4.1.6

When an expression is inside the parentheses of function evaluation notation, you should use parentheses at the substitution step as well to ensure that the entire expression is being used. Once substitution occurs, you should then apply standard algebraic rules appropriately.

Example 4.1.7

For \(g(y)=-\frac{3}{2}y+3\text{,}\) simplify each of the following expressions.

  1. \(g(x)\)
  2. \(g(2t)\)
  3. \(3g(2x)+1\)
  1. To find \(g(x)\) requires a simple substitution.
    \begin{align*} g(y) &= \textstyle -\frac{3}{2}y+3\\ g(x) &= \textstyle -\frac{3}{2}x+3 \end{align*}
  2. To find \(g(2t)\) requires a substitution of the expression \(x=2t\text{.}\)
    \begin{align*} g(2t) &= \textstyle -\frac{3}{2}(2t)+3\\ &= -3t+3 \end{align*}
  3. To find \(3g(2x)+1\) will require substitution of \(g(2x)\) by a formula that itself involves substitution. Note the use of parentheses.
    \begin{align*} 3g(2x)+1 &= 3\Big({\textstyle -\frac{3}{2}(2x)+3}\Big) + 1\\ &= 3(-3x+3)+1 = -9x+9+1\\ &=-9x+10 \end{align*}

Evaluating a function using an expression as the input is called composition. In the examples above, \(f(3x+1)\) and \(g(2t)\) were composition of functions. First, \(f(3x+1)\) is the composition of \(f\) with an input expression of \(3x+1\text{.}\) Second, \(g(2t)\) is the composition of the function \(g\) with an input expression \(2t\text{.}\) In each case, the input of the function is itself a dependent variable (expression) involving another variable.

Composition corresponds to linking functions together, with the output of one function becoming the input to another function. In the context of a physical system, we are considering where the state involves multiple variables, say \((A,B,C,\ldots)\text{.}\) Suppose we know one function, \(f: A \mapsto B\text{,}\) that determines the value of \(B\) knowing the value of \(A\text{.}\) Then suppose know another function, \(g: B \mapsto C\text{,}\) that predicts the value of \(C\) from the value of \(B\text{.}\) If we link these together in a chain, we can start with a value of \(A\text{,}\) compute the value of \(B=f(A)\text{,}\) and then use that value of \(B\) to compute the value of \(C=g(B)\text{.}\) Together, this creates a map \(A \mapsto C\) which is the composition of the individual maps. Using substitution, we have \(C= g\big(f(A)\big)\text{,}\) the output of \(f\) becoming the input to \(g\text{.}\)

Example 4.1.8

The radius \(r\text{,}\) circumference \(C\text{,}\) and area \(A\) of a circle are all related. The equation \(C=2 \pi r\) defines \(C\) as a function of \(r\) and the equation \(A = \pi r^2\) defines \(A\) as a function of \(r\text{.}\) Use composition to define the function \(C \mapsto A\text{.}\)


The final output should be area \(A\text{,}\) which we can compute if we know \(r\text{.}\) We can use the relation between \(C\) and \(r\) to solve for \(r\) as a function of \(C\text{.}\)

\begin{gather*} C = 2 \pi r\\ \frac{C}{2 \pi} = r \end{gather*}

This equation defines \(C \mapsto r\text{,}\) so that we have a chain

\begin{align*} r &= \frac{C}{2 \pi},\\ A &= \pi r^2. \end{align*}

Composition uses substitution to replace \(r\) by its formula,

\begin{equation*} A = \pi r^2 = \pi \Big(\frac{C}{2\pi}\Big)^2 = \frac{\pi C^2}{2^2 \pi^2} = \frac{C^2}{4 \pi}. \end{equation*}

We use circle notation to represent the composition of functions. We write \(f \circ g(x)\) to mean that \(g(x)\) is the input expression to the function \(f\text{:}\)

\begin{equation*} f \circ g(x) = f\big( g(x) \big). \end{equation*}

We can remember the order if we notice that the input to a function is always on the right of the function name. So \(f \circ g(x)\) has \(x\) as the input to \(g\text{,}\) which has an output as input to \(f\text{.}\)

Even when a function is introduced without the context of mapping between variables, we can think of a composition as a chain of relations. The composition \(f \circ g(x) = f\big(g(x)\big)\) has an inner expression \(g(x)\) as the input to \(f\text{.}\) We treat \(g(x)\) as an intermediate variable, say \(u\text{.}\) There are two functions: \(x \overset{g}\mapsto u\) and \(u \overset{f} \mapsto y\text{.}\) The composition \(f \circ g\) puts these together as a chain \(x \overset{f \circ g} \mapsto y = x \overset{g}\mapsto u \overset{f}\mapsto y\) to find the value of \(y\) starting from \(x\text{.}\)

Example 4.1.9

Suppose \(f(x)=x^2\) and \(g(x)=x+3\text{.}\) Compute and simplify the compositions \(f \circ g(x)\) and \(g \circ f(x)\text{.}\)


The composition \(f \circ g(x)=f(g(x))\) corresponds a chain. The inner expression defines an intermediate variable \(x \overset{g}\mapsto u=x+3\text{,}\) while the outer function squares the result \(u \overset{f}\mapsto y=u^2\text{.}\) The composition \(f \circ g(x) = (x+3)^2\) combines the operations,

\begin{equation*} x \overset{g}\mapsto x+3 \overset{f}\mapsto (x+3)^2\text{.} \end{equation*}

The chain is illustrated in the figure below.

The composition \(g \circ f(x)=g(f(x))\) reverses the order in which the chain is applied. The inner expression \(f(x)\) defines a different intermediate variable \(x \overset{f}\mapsto w=w^2\text{,}\) while the outer function adds three to the result \(w \overset{g}\mapsto y=w+3\text{.}\) The composition \(g \circ f(x) = x^2+3\) combines the operations,

\begin{equation*} x \overset{f}\mapsto x^2 \overset{g}\mapsto x^2+3\text{.} \end{equation*}

This chain is illustrated in the next figure.

The previous example clearly illustrated that the order of composition matters. Compositions \(f \circ g\) and \(g \circ f\) represent different functions. We can see this visually by considering the chains represented by interactive maps.

Figure 4.1.10 \(f \circ g(x) = (x+3)^2\)
Figure 4.1.11 \(g\circ f(x) = x^2+3\)

We now consider some additional examples of using substitution to find a simplified formula for a composition.

Example 4.1.12

Suppose \(f(x)=2x-3\) and \(g(x)=3x+1\text{.}\) Simplify each of the following expressions.

  1. \(f(x^2-1)\)
  2. \(f \circ g(x)\)
  3. \(g \circ f(x-2)\)
  1. \(f(x^2-1)\) is a composition of \(f\) with input \(x^2-1\text{.}\) Since \(f(x)=2x-3\) or \(f(\square)=2\,\square - 3\text{,}\) using substitution, we find

    \begin{equation*} f(x^2-1) = 2(x^2-1)-3 = 2x^2-2-3 = 2x^2-5\text{.} \end{equation*}
  2. \(f\circ g(x)\) is a composition of \(f\) with input \(g(x)\text{.}\) Since \(f(x)=2x-3\) and \(g(x)=3x+1\text{,}\) using substitution, we find

    \begin{equation*} f\circ g(x) = f(g(x)) = 2(3x+1)-3 = 6x+2-3 = 6x-1\text{.} \end{equation*}
  3. \(g\circ f(x-2)\) is a composition of a composition. The outermost function is \(g\) has an input coming from the output of \(f\text{,}\) and \(f\) has an input expression \(x-2\text{.}\) Careful substitution one step at a time will give the value.

    \begin{align*} g \circ f(x-2) & = g\big(f(x-2)\big)\\ & = g\big(2(x-2)-3\big) & \hbox{substitute $f(x-2)$}\\ & = g\big(2x-7\big) & \text{simplify}\\ & = 3(2x-7)+1 & \hbox{substitute $g(2x-7)$}\\ & = 6x-20 \end{align*}

Subsection 4.1.3 Inverse Functions and Composition

When we discussed inverse functions earlier in Subsection 2.4.3, we thought of them as inverse maps. Given an equation defining the map \(x \mapsto y\text{,}\) if we could solve the equation for the input \(x\) as a single expression involving \(y\text{,}\) then this new equation defined the inverse function. Inverse functions undo one another's operations.

Example 4.1.13

Consider the function

\begin{equation*} a \overset{f}{\mapsto} b = 3a+2\text{.} \end{equation*}

The inverse function is found by solving for \(a\) in the equation \(b=3a+2\) which gives \(\displaystyle a = \frac{b-2}{3}\text{.}\) We have the inverse function

\begin{equation*} b \overset{f^{-1}}{\mapsto} a = \frac{b-2}{3}\text{.} \end{equation*}

Using a placeholder variable, like \(x\text{,}\) these can be written in function notation:

\begin{gather*} f(x) = 3x+2,\\ f^{-1}(x) = \frac{x-2}{3}. \end{gather*}

Let us consider the calculations involved in the previous example. The function \(f\) took an input and performed the following operations in order:

  • Multiply by 3.
  • Add 2.

The inverse function \(f^{-1}\) took an input and performed related operations:

  • Subtract 2.
  • Divide by 3.

The functions are inverse because they will exactly undo one another's operations.

Consider what happens if you create a chain and apply \(f^{-1}\) immediately after \(f\text{:}\)

  • Multiply by 3.
  • Add 2.
  • Subtract 2.
  • Divide by 3.

The middle two steps cancel one another's effects, so this would be the same as the simpler chain of steps:

  • Multiply by 3.
  • Divide by 3.

Again, the operations cancel each other out. The output will always be the same as the original input,

\begin{equation*} f^{-1} \circ f(x) = x\text{.} \end{equation*}

The following interactive figure shows this composition as a chain of maps.

Figure 4.1.14 Composition \(y=f^{-1} \circ f(a)\text{,}\) corresponding to chain \(a \overset{f}\mapsto b \overset{f^{-1}}\mapsto y\text{.}\) As the functions are inverses, this always yields \(y=a\text{.}\)

A composition in the reverse order, \(f \circ f^{-1}(x)\text{,}\) also results in exact cancellation.

Figure 4.1.15 Composition \(y=f \circ g(b)\text{,}\) corresponding to chain \(b \overset{g}\mapsto a \overset{f}\mapsto y\text{.}\) As the functions are inverses, this always yields \(y=b\text{.}\)

Simplifying the composition of inverse functions algebraically reveals the cancellation directly.

Example 4.1.16

For \(f(x)=3x+2\) and \(\displaystyle f^{-1}(x)=\frac{x-2}{3}\) compute and simplify \(f \circ g(x)\) and \(g \circ f(x)\text{.}\)


Using substitution and algebraic simplification, we find the values requested.

\begin{align*} f \circ g(x) & = f\big(g(x)\big) \\ &= f\Big( \frac{x-2}{3} \Big) & \hbox{substitute $g(x)$} \\ &= 3\Big( \frac{x-2}{3} \Big)+2 & \hbox{substitute $f(\square)$} \\ &= x-2+2 = x \end{align*}
\begin{align*} g \circ f(x) & = g\big(f(x)\big) \\ &= g\big( 3x+2 \big) & \hbox{substitute $f(x)$} \\ &= \frac{(3x+2)-2}{3} & \hbox{substitute $g(\square)$} \\ &= \frac{3x}{3} = x \end{align*}

Inverse functions will always simplify in this way: the composition of inverse functions cancel to just leave the input. Functions are not always defined by an equation, so we shouldn't define inverses through solving equations. Mathematicians actually define inverse functions in terms of the property of composition.

Definition 4.1.17

Two functions \(f\) and \(g\) are inverses of one another, and we write \(g=f^{-1}\) and \(f=g^{-1}\text{,}\) if for every \(x\) in the domain of \(g\text{,}\) we have

\begin{equation*} f \circ g(x) = f\big(g(x)\big) = x, \end{equation*}

and for every \(x\) in the domain of \(f\text{,}\) we have

\begin{equation*} g \circ f(x) = g \big(f(x)\big) = x. \end{equation*}

It is time for a comment about real variables. In science, variables represent physical measurements and the variables are the objects of study. These variables can be related by functions. However, in mathematics, it is the function itself that is being studied. For simplicity, mathematics textbooks have adopted an approach where \(x\) is almost universally the independent variable of every function and \(y\) is the dependent variable. This makes it easier to remember the role each variable plays, but it can lead to confusion in actual applications.

Example 4.1.18

An enzyme is a protein that helps catalyze a chemical reaction. For many enzymes, the rate of reaction \(R\) and the concentration of the reactant \(C\) satisfy a relation called Michaelis-Menten kinetics

\begin{equation*} R = \frac{aC}{C+K}, \end{equation*}

where \(a\) and \(K\) are parameters that characterize the particular reaction. Physically, we require \(C \ge 0\text{.}\) In mathematics, this relation might be characterized by a function

\begin{equation*} f(x) = \frac{ax}{x+K}. \end{equation*}

We would then say \(R = f(C)\text{.}\) This is equivalent to mapping notation

\begin{equation*} C \overset{f}{\mapsto} R = \frac{aC}{C+K}. \end{equation*}

To find the inverse function, most mathematics textbooks say to write \(y=f(x)\text{,}\) switch all \(x\) and \(y\) and then solve for \(y\text{.}\) The only reason to switch the variables is to preserve \(x\) as the independent variable of the relation. This is an artificial requirement. We might as well just solve for \(C\) as a function of \(R\text{.}\) Start by cross-multiplying to eliminate the denominator in the equation.

\begin{gather*} R = \frac{aC}{C+K}\\ R(C+K) = aC\\ RC+KR = aC \end{gather*}

Because we are solving for \(C\text{,}\) we need to collect \(C\) terms on one side of the equation and then factor.

\begin{gather*} RC-aC = -KR\\ C(R-a) = -KR\\ C = \frac{-KR}{R-a} \end{gather*}

Multiplying the numerator and denominator each by \(-1\text{,}\) we get an equivalent and simpler explicit function

\begin{equation*} R \overset{g}{\mapsto}C = \frac{KR}{a-R}. \end{equation*}

As the functions come from the same relation, we know \(g = f^{-1}\text{.}\)

This equation shows that \(C\) is the dependent variable and is a function of the independent variable \(R\text{.}\) Mathematically, using \(x\) as the independent variable, we would have written

\begin{equation*} f^{-1}(x) = \frac{Kx}{a-x}\text{.} \end{equation*}

However, this equation loses the context of what the input variable \(x\) and the output value represent. In applications, it is better to include the variables so that their interpretation can be preserved.

Recall that the composition of inverse functions should result in the input of the inner function. Consider how that applies in the context of actual variables. Recall the earlier example relating a reaction rate \(R\) and the reactant concentration \(C\text{.}\) We had inverse functions \(C \overset{f}{\mapsto} R\) and \(R \overset{g}{\mapsto} C\text{.}\) Composition applies these operations one immediately after the other, with the inner function applied first. Composition \(f \circ g\) applies \(g\) to the input followed by \(f\text{,}\) which would be written in mapping notation with the variables as

\begin{equation*} R \overset{g}{\mapsto} C \overset{f}{\mapsto} R\text{.} \end{equation*}

The original input is the value \(R\) and the final output is also the value \(R\text{.}\) So the comoposition is equal to the original input. Algebra should verify that this actually works.

Example 4.1.19

For the inverse functions of Michaelis-Menten kinetics,

\begin{gather*} C \overset{f}{\mapsto} R = \frac{aC}{C+K},\\ R \overset{g}{\mapsto} C = \frac{KR}{a-R}, \end{gather*}

show that the composition of functions cancel.


To compute \(f \circ g(x)\text{,}\) we use \(g(x)\) as the input to \(f\text{.}\) Using meaningful variables, \(g\) takes a reaction rate as input, so we compute \(f(g(R))\) and simplify. Recall that function evaluation is just substitution of the input in a formula.

\begin{align*} f\big(g(R)\big) &= f\big( \frac{KR}{a-R}\big) \\ &= \frac{a\big( \frac{KR}{a-R}\big)}{\big( \frac{KR}{a-R}\big)+K} \end{align*}

We replaced the \(C\) as input to \(f\) with the formula for \(g(R)\text{.}\) To simplify this, we can clear the denominator of the fractions inside the fraction by multiplying numerator and denominator by \((a-R)\text{.}\)

\begin{align*} f\big(g(R)\big) &= \frac{a\big( \frac{KR}{a-R}\big)(a-R)}{\big( \frac{KR}{a-R}+K\big)(a-R)}\\ &= \frac{aKR}{KR + K(a-R)}\\ &= \frac{aKR}{KR + Ka-KR}\\ &= \frac{aKR}{Ka} \\ &= R \end{align*}

Using the placeholder variable \(x\text{,}\) we have \(f \circ g(x)=x\text{,}\) as required for inverse functions.

The algebraic verification that \(g\) undoes the evaluation of \(f\text{,}\)

\begin{equation*} C \overset{f}{\mapsto} R \overset{g}{\mapsto} C, \end{equation*}

follows a similar calculation. To compute \(g \circ f(x)\text{,}\) we use \(f(x)\) as the input to \(g\text{.}\) In context, \(f\) takes a reactant concentration \(C\) as input, so we compute \(g(f(C))\) and simplify.

\begin{align*} g\big(f(C)\big) &= g\big(\frac{aC}{C+K}\big) \\ &= \frac{K\big( \frac{aC}{C+K}\big)}{a-\big( \frac{aC}{C+K}\big)}\\ &= \frac{K\big( \frac{aC}{C+K}\big)(C+K)}{\big(a- \frac{aC}{C+K}\big)(C+K)}\\ &= \frac{aKC}{a(C+K) - aC}\\ &= \frac{aKC}{aC+aK -aC}\\ &= \frac{aKC}{aK} \\ &= C \end{align*}

Subsection 4.1.4 Summary

  • A function is a relation between an independent variable (input) and a dependent variable (output) such that for each value of the input, there is exactly one value for the output.
  • An equation in two variables defines a relation. When we can solve the equation for one variable (dependent) as a single expression of the other variable (independent), the expression defines an explicit function.
  • A linear function \(x \mapsto y\) is a relationship between variables that have a constant rate of change. The rate of change equals the slope between two states \((x_1,y_1)\) and \((x_2,y_2)\) and is the ratio of the change in the output to the change in the input:
    \begin{equation*} m = \frac{\Delta y}{\Delta x} = \frac{y_2-y_1}{x_2-x_1}\text{.} \end{equation*}
  • Function mapping notation \(x \overset{f}{\mapsto} y\) indicates that \(y\) is a function of \(x\) and \(f\) is the name of the function.
  • Function evaluation notation \(f(\square)\) uses substitution of whatever appears between the parentheses \((\square)\) in place of the independent variable.
  • Composition \(f \circ g\) is evaluation of the outer function \(f\) with an input using the output of the inner function \(g\text{,}\)
    \begin{equation*} f \circ g(x) = f\big(g(x)\big)\text{.} \end{equation*}
    As maps, if \(g:x \mapsto u\) and \(f:u \mapsto y\text{,}\) then
    \begin{equation*} x \overset{f \circ g}\mapsto y \quad = \quad x \overset{g}\mapsto u \overset{f}\mapsto y. \end{equation*}
  • Two functions \(f\) and \(g\) are inverses of one another if \(f \circ g(x) = x\) for all \(x\) in the domain of \(g\) and \(g \circ f(x) = x\) for all \(x\) in the domain of \(f\text{.}\) This means that inverse functions cancel one another when applied in a chain:

    \begin{equation*} f \circ f^{-1}(x) = x \quad \text{and} \quad f^{-1}\circ f(x) = x\text{.} \end{equation*}
  • If an equation can be solved for each variable in terms of the other (e.g., \(x \mapsto y\) and \(y \mapsto x\)), the relation is one-to-one. The two resulting functions are inverse functions.

Subsection 4.1.5 Exercises

For each equation, determine if the relation defines functions \(x \mapsto y\) and \(y \mapsto x\) by solving the equation for the dependent variable.


For the equation \(3x-5y=10\text{,}\) do the following.

  1. Determine if \(x \mapsto y\text{.}\)
  2. Determine if \(y \mapsto x\text{.}\)

For the equation \(2xy - 6 = 4x - 3y\text{,}\) do the following.

  1. Determine if \(x \mapsto y\text{.}\)
  2. Determine if \(y \mapsto x\text{.}\)

For the equation \(6x+4y-3xy=0\text{,}\) do the following.

  1. Determine if \(x \mapsto y\text{.}\)
  2. Determine if \(y \mapsto x\text{.}\)

For the equation \(x^2+3y=25\text{,}\) do the following.

  1. Determine if \(x \mapsto y\text{.}\)
  2. Determine if \(y \mapsto x\text{.}\)

Given a function, compute and simplify the expressions listed.


Suppose \(f(x)=\frac{2}{3}x+4\text{.}\) Simplify each of the following expressions.

  1. \(f(5)\)
  2. \(f(t)\)
  3. \(f(t^2-1)\)
  4. \(3f(2x)-8\)

Suppose \(\displaystyle g(x)=\frac{4}{x+1}\text{.}\) Simplify each of the following.

  1. \(g(1)\)
  2. \(g(\frac{1}{x})\)
  3. \(\displaystyle \frac{1}{g(x)}\)
  4. \(g\big(\frac{1}{x}-1\big)\)

Suppose \(f(x)=2x-5\text{,}\) \(g(x)=\frac{1}{2}x+5\text{,}\) and \(h(x)=\frac{1}{2}(x+5)\text{.}\) Simplify each of the following.

  1. \(f \circ g(x)\)
  2. \(f \circ h(x)\)
  3. \(g \circ f(x)\)
  4. \(g \circ h(x)\)

What conclusion can be drawn?


Suppose \(\displaystyle f(x)=\frac{3}{x+2}\) and \(\displaystyle g(x)=\frac{3}{x}-2\text{.}\) Simplify each of the following.

  1. \(f(x-2)\)
  2. \(g(\frac{1}{x})\)
  3. \(f \circ g(x)\)
  4. \(g \circ f(x)\)

Is \(g = f^{-1}\text{?}\)



Let \(C\) be the temperature measured in degrees Celsius, and let \(F\) be the temperature measured in degrees Fahrenheit. The function \(g(x) = \frac{9}{5}x + 32\) defines the map \(g : C \mapsto F\text{,}\) and \(h(x) = \frac{5}{9}(x-32)\) defines \(h : F \mapsto C\text{.}\)

  1. Use algebra to verify that \(g\) and \(h\) are inverse functions.
  2. What is the value and interpretation of \(g(30)\text{?}\)
  3. What is the value and interpretation of \(g \circ h(30)\text{?}\)

A spring force scale uses the distance a spring is stretched to determine the force that is applied to the spring. We calibrate the scale by using known forces (e.g., weights) and record the corresponding location of the tip on a ruler. Let \(F\) be the force (Newtons) applied to the spring and let \(L\) be the corresponding location (centimeters). The following table is used for calibration.

\(F\) (N) 0 10.0
\(L\) (cm) 20.0 42.5
  1. Find a linear equation relating the variables \(F\) and \(L\text{.}\)
  2. Determine functions \(g\) and \(h\) so that \(F \overset{g}{\mapsto} L\) and \(L \overset{h}{\mapsto} F\text{.}\) What are the corresponding equations using evaluation notation?
  3. Suppose a force of 5 N is applied to the spring. What will be the location of the tip of the ruler? Which function was used?
  4. Suppose a force is applied that results in the tip having a location of 28.7 cm. What was the force? Which function was used?

The perimeter \(P\) and area \(A\) of a square are each functions of the length of the sides \(s\) by \(P=4s\) and \(A=s^2\text{.}\) Find perimeter as a function of area, \(P \mapsto A\text{.}\)


The volume of a sphere is related to the radius of the sphere by the equation \(V = \frac{4}{3} \pi r^3\text{.}\) Suppose the radius is a function of time defined by \(r = 1+2t\text{.}\) Find the volume as a function of time, \(t \mapsto V\text{.}\)


The cost \(C\) of materials for a project depends on the required area \(A\) of materials needed. The unit price is $3.50 per m2. The project involves making two squares, each of them having sides with length \(s\) (meters).

  1. Find \(A \overset{f}{\mapsto} C\text{.}\)
  2. Find \(s \overset{g}{\mapsto} A\text{.}\)
  3. Use composition to find \(s \mapsto C\text{.}\) Is this \(f \circ g\) or \(g \circ f\text{?}\)

  4. How much would a project with \(s=4\) cost? How much area of materials will be required? What function is used for each calculation?