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Composition of Functions

Combine functions using algebraic operations.

Composition of Functions

Find the value of a function (IA 3.5.3), (CA 3.1.2)

A function is a relation that assigns to each element in its domain exactly one element in the range. For each ordered pair in the relation, each \(x\) -value is matched with only one \(y\) -value. The notation \(y=f(x)\) defines a function named \(f\) . This is read as “ \(y\) is a function of \(x\) .” The letter \(x\) represents the input value, or independent variable. The letter \(y\), or \(f(x)\) , represents the output value, or dependent variable.

Evaluating and Solving a Function Represented in Table Form

Try it.

ⓐ Evaluate \(g(3)\)
ⓑ Solve \(g(n)=6\)

\(n\)	1	2	3	4	5
\(g(n)\)	8	6	7	6	8

Solution

ⓐ Evaluating g(3) means determining the output value of the function g for the input value of n=3. The table output value corresponding to n=3 is 7, so g(3)=7.
ⓑ Solving g(n)=6 means identifying the input values, n, that produce an output value of 6. The table shows two values where g(n) = 6 at x=2 and 4.

Find the value of a function.

Try it.

Given the function \(k(t)=2t-1\):

ⓐ Evaluate \(k(2)\)
ⓑ Solve \(k(t)=7\)

Try it.

Given the function \(f(x)=\sqrt{x+2}\) :

ⓐ Evaluate \(f(7)\)
ⓑ Solve \(f(x)=4\)

Try it.

For the function \(f(x)=2{x}^{2}+3x+1\) , find

ⓐ \(f(3)\)
ⓑ \(f(-2)\)
ⓒ \(f(t)\)
ⓓ The value(s) of x that make \(f(x)=1\)

Try it.

Use the table below to help answer the following:

x	1	2	3	4	5
\(g(x)\)	5	12	21	32	45

ⓐ Evaluate \(g(4)\)
ⓑ Solve \(g(x)=32\)

Condensed — the full section is in OpenStax College Algebra 2e.

Combining Functions Using Algebraic Operations

Function composition is only one way to combine existing functions. Another way is to carry out the usual algebraic operations on functions, such as addition, subtraction, multiplication and division. We do this by performing the operations with the function outputs, defining the result as the output of our new function.

Suppose we need to add two columns of numbers that represent a husband and wife’s separate annual incomes over a period of years, with the result being their total household income. We want to do this for every year, adding only that year’s incomes and then collecting all the data in a new column. If \(w(y)\) is the wife’s income and \(h(y)\) is the husband’s income in year \(y,\) and we want \(T\) to represent the total income, then we can define a new function.

\[T(y)=h(y)+w(y)\]

If this holds true for every year, then we can focus on the relation between the functions without reference to a year and write

\[T=h+w\]

Just as for this sum of two functions, we can define difference, product, and ratio functions for any pair of functions that have the same kinds of inputs (not necessarily numbers) and also the same kinds of outputs (which do have to be numbers so that the usual operations of algebra can apply to them, and which also must have the same units or no units when we add and subtract). In this way, we can think of adding, subtracting, multiplying, and dividing functions.

For two functions \(f(x)\) and \(g(x)\) with real number outputs, we define new functions \(f+g,\ f-g,\ fg,\) and \(\frac{f}{g}\) by the relations

\[\begin{array}{llll}(f+g)(x) & = & f(x)+g(x) & \\ (f-g)(x) & = & f(x)-g(x) & \\ (fg)(x) & = & f(x)g(x) & \\ (\frac{f}{g})(x) & = & \frac{f(x)}{g(x)} & \text{where}\ g(x)\ \ne \ 0\end{array}\]

Example

Try it.

Find and simplify the functions \((g-f)(x)\) and \((\frac{g}{f})(x),\) given \(f(x)=x-1\) and \(g(x)={x}^{2}-1.\) Are they the same function?

Solution

Begin by writing the general form, and then substitute the given functions.

\[\begin{array}{llll}(g-f)(x) & = & g(x)-f(x) & \\ (g-f)(x) & = & {x}^{2}-1-(x-1) & \\ (g-f)(x) & = & {x}^{2}-x & \\ (g-f)(x) & = & x(x-1) & \\ & & & \\ & & & \\ (\frac{g}{f})(x) & = & \frac{g(x)}{f(x)} & \\ (\frac{g}{f})(x) & = & \frac{{x}^{2}-1}{x-1} & \\ (\frac{g}{f})(x) & = & \frac{(x+1)(x-1)}{x-1} & \text{where }x\ne 1 \\ (\frac{g}{f})(x) & = & x+1 & \end{array}\]

No, the functions are not the same.

Note: For \((\frac{g}{f})(x),\) the condition \(x\ne 1\) is necessary because when \(x=1,\) the denominator is equal to 0, which makes the function undefined.

Condensed — the full section is in OpenStax College Algebra 2e.

Create a Function by Composition of Functions

Performing algebraic operations on functions combines them into a new function, but we can also create functions by composing functions. When we wanted to compute a heating cost from a day of the year, we created a new function that takes a day as input and yields a cost as output. The process of combining functions so that the output of one function becomes the input of another is known as a composition of functions. The resulting function is known as a composite function. We represent this combination by the following notation:

\[(f∘g)(x)=f(g(x))\]

We read the left-hand side as \(“f\) composed with \(g\) at \(x,”\) and the right-hand side as \(“f\) of \(g\) of \(x.”\) The two sides of the equation have the same mathematical meaning and are equal. The open circle symbol \(∘\) is called the composition operator. We use this operator mainly when we wish to emphasize the relationship between the functions themselves without referring to any particular input value. Composition is a binary operation that takes two functions and forms a new function, much as addition or multiplication takes two numbers and gives a new number. However, it is important not to confuse function composition with multiplication because, as we learned above, in most cases \(f(g(x))\ne f(x)g(x).\)

It is also important to understand the order of operations in evaluating a composite function. We follow the usual convention with parentheses by starting with the innermost parentheses first, and then working to the outside. In the equation above, the function \(g\) takes the input \(x\) first and yields an output \(g(x).\) Then the function \(f\) takes \(g(x)\) as an input and yields an output \(f(g(x)).\)

In general, \(f∘g\) and \(g∘f\) are different functions. In other words, in many cases \(f(g(x))\ne g(f(x))\) for all \(x.\) We will also see that sometimes two functions can be composed only in one specific order.

For example, if \(f(x)={x}^{2}\) and \(g(x)=x+2,\) then

\[\begin{array}{lll}f(g(x)) & = & f(x+2) \\ & = & {(x+2)}^{2} \\ & = & {x}^{2}+4x+4\end{array}\]

but

\[\begin{array}{lll}g(f(x)) & = & g({x}^{2}) \\ & = & {x}^{2}+2\end{array}\]

These expressions are not equal for all values of \(x,\) so the two functions are not equal. It is irrelevant that the expressions happen to be equal for the single input value \(x=-\frac{1}{2}.\)

Condensed — the full section is in OpenStax College Algebra 2e.

Evaluating Composite Functions

Once we compose a new function from two existing functions, we need to be able to evaluate it for any input in its domain. We will do this with specific numerical inputs for functions expressed as tables, graphs, and formulas and with variables as inputs to functions expressed as formulas. In each case, we evaluate the inner function using the starting input and then use the inner function’s output as the input for the outer function.

When working with functions given as tables, we read input and output values from the table entries and always work from the inside to the outside. We evaluate the inside function first and then use the output of the inside function as the input to the outside function.

Example

Try it.

Using , evaluate \(f(g(3))\) and \(g(f(3)).\)

\(x\)	\(f(x)\)	\(g(x)\)
1	6	3
2	8	5
3	3	2
4	1	7

Solution

To evaluate \(f(g(3)),\) we start from the inside with the input value 3. We then evaluate the inside expression \(g(3)\) using the table that defines the function \(g:\) \(g(3)=2.\) We can then use that result as the input to the function \(f,\) so \(g(3)\) is replaced by 2 and we get \(f(2).\) Then, using the table that defines the function \(f,\) we find that \(f(2)=8.\)

\[\begin{array}{lll}g(3) & = & 2 \\ f(g(3)) & = & f(2)=8\end{array}\]

To evaluate \(g(f(3)),\) we first evaluate the inside expression \(f(3)\) using the first table: \(f(3)=3.\) Then, using the table for \(g\text{,\,}\) we can evaluate

\[g(f(3))=g(3)=2\]

shows the composite functions \(f∘g\) and \(g∘f\) as tables.

\(x\)	\(g(x)\)	\(f(g(x))\)	\(f(x)\)	\(g(f(x))\)
3	2	8	3	2

Condensed — the full section is in OpenStax College Algebra 2e.

Finding the Domain of a Composite Function

As we discussed previously, the domain of a composite function such as \(f∘g\) is dependent on the domain of \(g\) and the domain of \(f.\) It is important to know when we can apply a composite function and when we cannot, that is, to know the domain of a function such as \(f∘g.\) Let us assume we know the domains of the functions \(f\) and \(g\) separately. If we write the composite function for an input \(x\) as \(f(g(x)),\) we can see right away that \(x\) must be a member of the domain of \(g\) in order for the expression to be meaningful, because otherwise we cannot complete the inner function evaluation. However, we also see that \(g(x)\) must be a member of the domain of \(f,\) otherwise the second function evaluation in \(f(g(x))\) cannot be completed, and the expression is still undefined. Thus the domain of \(f∘g\) consists of only those inputs in the domain of \(g\) that produce outputs from \(g\) belonging to the domain of \(f.\) Note that the domain of \(f\) composed with \(g\) is the set of all \(x\) such that \(x\) is in the domain of \(g\) and \(g(x)\) is in the domain of \(f.\)

Example

Try it.

Find the domain of

\[(f∘g)(x)\ \text{where}\ \ \ f(x)=\frac{5}{x-1}\ \text{and}\ g(x)=\frac{4}{3x-2}\]

Solution

The domain of \(g(x)\) consists of all real numbers except \(x=\frac{2}{3},\) since that input value would cause us to divide by 0. Likewise, the domain of \(f\) consists of all real numbers except 1. So we need to exclude from the domain of \(g(x)\) that value of \(x\) for which \(g(x)=1.\)

\[\begin{array}{lll}\frac{4}{3x-2} & = & 1 \\ 4 & = & 3x-2 \\ 6 & = & 3x \\ x & = & 2\end{array}\]

So the domain of \(f∘g\) is the set of all real numbers except \(\frac{2}{3}\) and \(2.\) This means that

\[x\ne \frac{2}{3}\ \text{or}\ x\ne 2\]

We can write this in interval notation as

\[(-\infty ,\frac{2}{3})\cup (\frac{2}{3},2)\cup (2,\infty )\]

Condensed — the full section is in OpenStax College Algebra 2e.

Decomposing a Composite Function into its Component Functions

In some cases, it is necessary to decompose a complicated function. In other words, we can write it as a composition of two simpler functions. There may be more than one way to decompose a composite function, so we may choose the decomposition that appears to be most expedient.

Example

Try it.

Write \(f(x)=\sqrt{5-{x}^{2}}\) as the composition of two functions.

Solution

We are looking for two functions, \(g\) and \(h,\) so \(f(x)=g(h(x)).\) To do this, we look for a function inside a function in the formula for \(f(x).\) As one possibility, we might notice that the expression \(5-{x}^{2}\) is the inside of the square root. We could then decompose the function as

\[h(x)=5-{x}^{2}\ \text{and }g(x)=\sqrt{x}\]

We can check our answer by recomposing the functions.

\[g(h(x))=g(5-{x}^{2})=\sqrt{5-{x}^{2}}\]

Key Concepts

We can perform algebraic operations on functions. See .
When functions are composed, the output of the first (inner) function becomes the input of the second (outer) function.
The function produced by composing two functions is a composite function. See and .
The order of function composition must be considered when interpreting the meaning of composite functions. See .
A composite function can be evaluated by evaluating the inner function using the given input value and then evaluating the outer function taking as its input the output of the inner function.
A composite function can be evaluated from a table. See .
A composite function can be evaluated from a graph. See .
A composite function can be evaluated from a formula. See .
The domain of a composite function consists of those inputs in the domain of the inner function that correspond to outputs of the inner function that are in the domain of the outer function. See and .
Just as functions can be combined to form a composite function, composite functions can be decomposed into simpler functions.
Functions can often be decomposed in more than one way. See .

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

1. ⓐ Evaluate \(g(3)\)
2. ⓑ Solve \(g(n)=6\)
\(n\) 1 2 3 4 5
\(g(n)\) 8 6 7 6 8
Bonisa impendulo
1. ⓐ Evaluating g(3) means determining the output value of the function g for the input value of n=3. The table output value corresponding to n=3 is 7, so g(3)=7.
2. ⓑ Solving g(n)=6 means identifying the input values, n, that produce an output value of 6. The table shows two values where g(n) = 6 at x=2 and 4.
Given the function \(k(t)=2t-1\):
1. ⓐ Evaluate \(k(2)\)
2. ⓑ Solve \(k(t)=7\)
Given the function \(f(x)=\sqrt{x+2}\) :
1. ⓐ Evaluate \(f(7)\)
2. ⓑ Solve \(f(x)=4\)
For the function \(f(x)=2{x}^{2}+3x+1\) , find
1. ⓐ \(f(3)\)
2. ⓑ \(f(-2)\)
3. ⓒ \(f(t)\)
4. ⓓ The value(s) of x that make \(f(x)=1\)
Use the table below to help answer the following:
x 1 2 3 4 5
\(g(x)\) 5 12 21 32 45
1. ⓐ Evaluate \(g(4)\)
2. ⓑ Solve \(g(x)=32\)
Use the table below showing values of \(f(x)\) and \(g(x)\) to find each of the following. Remember when working with composition functions we always evaluate the inner function first.
\(x\) –3 –2 –1 0 1 2 3
\(f(x)\) 11 9 7 5 3 1 –1
\(g(x)\) –8 –3 0 1 0 –3 –8
1. ⓐ \(f(1)=\)
2. ⓑ \(g(f(1))=\)
3. ⓒ \(g(0)=\)
4. ⓓ \(f(g(0))=\)
5. ⓔ \(f(g(2))=\)
6. ⓕ \(f(f(3))=\)
Find and simplify the functions \((g-f)(x)\) and \((\frac{g}{f})(x),\) given \(f(x)=x-1\) and \(g(x)={x}^{2}-1.\) Are they the same function?

Bonisa impendulo

Begin by writing the general form, and then substitute the given functions.
\[\begin{array}{llll}(g-f)(x) & = & g(x)-f(x) & \\ (g-f)(x) & = & {x}^{2}-1-(x-1) & \\ (g-f)(x) & = & {x}^{2}-x & \\ (g-f)(x) & = & x(x-1) & \\ & & & \\ & & & \\ (\frac{g}{f})(x) & = & \frac{g(x)}{f(x)} & \\ (\frac{g}{f})(x) & = & \frac{{x}^{2}-1}{x-1} & \\ (\frac{g}{f})(x) & = & \frac{(x+1)(x-1)}{x-1} & \text{where }x\ne 1 \\ (\frac{g}{f})(x) & = & x+1 & \end{array}\]
No, the functions are not the same.
Note: For \((\frac{g}{f})(x),\) the condition \(x\ne 1\) is necessary because when \(x=1,\) the denominator is equal to 0, which makes the function undefined.
Find and simplify the functions \((fg)(x)\) and \((f-g)(x).\)
\[f(x)=x-1\ \text{ and }g(x)={x}^{2}-1\]
Are they the same function?

Bonisa impendulo

\(\begin{array}{l}(fg)(x)=f(x)g(x)=(x-1)({x}^{2}-1)={x}^{3}-{x}^{2}-x+1 \\ (f-g)(x)=f(x)-g(x)=(x-1)-({x}^{2}-1)=x-{x}^{2}\end{array}\)
No, the functions are not the same.
Using the functions provided, find \(f(g(x))\) and \(g(f(x)).\) Determine whether the composition of the functions is commutative.
\[f(x)=2x+1\ \ \ \ \ \ \ g(x)=3-x\]

Bonisa impendulo

Let’s begin by substituting \(g(x)\) into \(f(x).\)
\[\begin{array}{lll}f(g(x)) & = & 2(3-x)+1 \\ & = & 6-2x+1 \\ & = & 7-2x\end{array}\]
Now we can substitute \(f(x)\) into \(g(x).\)
\[\begin{array}{lll}g(f(x)) & = & 3-(2x+1) \\ & = & 3-2x-1 \\ & = & -2x+2\end{array}\]
We find that \(g(f(x))\ne f(g(x)),\) so the operation of function composition is not commutative.
The function \(c(s)\) gives the number of calories burned completing \(s\) sit-ups, and \(s(t)\) gives the number of sit-ups a person can complete in \(t\) minutes. Interpret \(c(s(3)).\)

Bonisa impendulo

The inside expression in the composition is \(s(3).\) Because the input to the s-function is time, \(t=3\) represents 3 minutes, and \(s(3)\) is the number of sit-ups completed in 3 minutes.
Using \(s(3)\) as the input to the function \(c(s)\) gives us the number of calories burned during the number of sit-ups that can be completed in 3 minutes, or simply the number of calories burned in 3 minutes (by doing sit-ups).
Suppose \(f(x)\) gives miles that can be driven in \(x\) hours and \(g(y)\) gives the gallons of gas used in driving \(y\) miles. Which of these expressions is meaningful: \(f(g(y))\) or \(g(f(x))?\)

Bonisa impendulo

The function \(y=f(x)\) is a function whose output is the number of miles driven corresponding to the number of hours driven.
\[\text{number of miles }=f\ (\text{number of hours})\]
The function \(g(y)\) is a function whose output is the number of gallons used corresponding to the number of miles driven. This means:
\[\text{number of gallons }=g\ (\text{number of miles})\]
The expression \(g(y)\) takes miles as the input and a number of gallons as the output. The function \(f(x)\) requires a number of hours as the input. Trying to input a number of gallons does not make sense. The expression \(f(g(y))\) is meaningless.
The expression \(f(x)\) takes hours as input and a number of miles driven as the output. The function \(g(y)\) requires a number of miles as the input. Using \(f(x)\) (miles driven) as an input value for \(g(y),\) where gallons of gas depends on miles driven, does make sense. The expression \(g(f(x))\) makes sense, and will yield the number of gallons of gas used, \(g,\) driving a certain number of miles, \(f(x),\) in \(x\) hours.
The gravitational force on a planet a distance r from the sun is given by the function \(G(r).\) The acceleration of a planet subjected to any force \(F\) is given by the function \(a(F).\) Form a meaningful composition of these two functions, and explain what it means.

Bonisa impendulo

A gravitational force is still a force, so \(a(G(r))\) makes sense as the acceleration of a planet at a distance r from the Sun (due to gravity), but \(G(a(F))\) does not make sense.
Using , evaluate \(f(g(3))\) and \(g(f(3)).\)
\(x\) \(f(x)\) \(g(x)\)
1 6 3
2 8 5
3 3 2
4 1 7

Bonisa impendulo

To evaluate \(f(g(3)),\) we start from the inside with the input value 3. We then evaluate the inside expression \(g(3)\) using the table that defines the function \(g:\) \(g(3)=2.\) We can then use that result as the input to the function \(f,\) so \(g(3)\) is replaced by 2 and we get \(f(2).\) Then, using the table that defines the function \(f,\) we find that \(f(2)=8.\)
\[\begin{array}{lll}g(3) & = & 2 \\ f(g(3)) & = & f(2)=8\end{array}\]
To evaluate \(g(f(3)),\) we first evaluate the inside expression \(f(3)\) using the first table: \(f(3)=3.\) Then, using the table for \(g\text{,\,}\) we can evaluate
\[g(f(3))=g(3)=2\]
shows the composite functions \(f∘g\) and \(g∘f\) as tables.
\(x\) \(g(x)\) \(f(g(x))\) \(f(x)\) \(g(f(x))\)
3 2 8 3 2
Using , evaluate \(f(g(1))\) and \(g(f(4)).\)

Bonisa impendulo

\(f(g(1))=f(3)=3\) and \(g(f(4))=g(1)=3\)
Using , evaluate \(f(g(1)).\)

Bonisa impendulo

To evaluate \(f(g(1)),\) we start with the inside evaluation. See .
We evaluate \(g(1)\) using the graph of \(g(x),\) finding the input of 1 on the \(x\text{-}\) axis and finding the output value of the graph at that input. Here, \(g(1)=3.\) We use this value as the input to the function \(f.\)
\[f(g(1))=f(3)\]
We can then evaluate the composite function by looking to the graph of \(f(x),\) finding the input of 3 on the \(x\text{-}\) axis and reading the output value of the graph at this input. Here, \(f(3)=6,\) so \(f(g(1))=6.\)
Using , evaluate \(g(f(2)).\)

Bonisa impendulo

\(g(f(2))=g(5)=3\)
Given \(f(t)={t}^{2}-t\) and \(h(x)=3x+2,\) evaluate \(f(h(1)).\)

Bonisa impendulo

Because the inside expression is \(h(1),\) we start by evaluating \(h(x)\) at 1.
\[\begin{array}{lll}h(1) & = & 3(1)+2 \\ h(1) & = & 5\end{array}\]
Then \(f(h(1))=f(5),\) so we evaluate \(f(t)\) at an input of 5.
\[\begin{array}{lll}f(h(1)) & = & f(5) \\ f(h(1)) & = & {5}^{2}-5 \\ f(h(1)) & = & 20\end{array}\]
Given \(f(t)={t}^{2}-t\) and \(h(x)=3x+2,\) evaluate
1. ⓐ \(h(f(2))\)
2. ⓑ \(h(f(-2))\)
Bonisa impendulo
1. ⓐ 8
2. ⓑ 20
Find the domain of
\[(f∘g)(x)\ \text{where}\ \ \ f(x)=\frac{5}{x-1}\ \text{and}\ g(x)=\frac{4}{3x-2}\]

Bonisa impendulo

The domain of \(g(x)\) consists of all real numbers except \(x=\frac{2}{3},\) since that input value would cause us to divide by 0. Likewise, the domain of \(f\) consists of all real numbers except 1. So we need to exclude from the domain of \(g(x)\) that value of \(x\) for which \(g(x)=1.\)
\[\begin{array}{lll}\frac{4}{3x-2} & = & 1 \\ 4 & = & 3x-2 \\ 6 & = & 3x \\ x & = & 2\end{array}\]
So the domain of \(f∘g\) is the set of all real numbers except \(\frac{2}{3}\) and \(2.\) This means that
\[x\ne \frac{2}{3}\ \text{or}\ x\ne 2\]
We can write this in interval notation as
\[(-\infty ,\frac{2}{3})\cup (\frac{2}{3},2)\cup (2,\infty )\]
Find the domain of
\[(f∘g)(x)\ \text{ where}\ \ \ f(x)=\sqrt{x+2}\ \text{ and }\ g(x)=\sqrt{3-x}\]

Bonisa impendulo

Because we cannot take the square root of a negative number, the domain of \(g\) is \((-\infty ,3].\) Now we check the domain of the composite function
\[(f∘g)(x)=\sqrt{\sqrt{3-x}+2}\]
For \((f∘g)(x)=\sqrt{\sqrt{3-x}+2},\sqrt{3-x}+2\ge 0,\) since the radicand of a square root must be positive. Since square roots are positive, \(\ \sqrt{3-x}\ge 0,\\) or, \(3-x\ge 0,\\) which gives a domain of \((-\infty ,3]\) .
Find the domain of
\[(f∘g)(x)\ \text{ where}\ \ \ f(x)=\frac{1}{x-2}\ \text{ and }\ g(x)=\sqrt{x+4}\]

Bonisa impendulo

\([-4,0)\cup (0,\infty )\)
Write \(f(x)=\sqrt{5-{x}^{2}}\) as the composition of two functions.

Bonisa impendulo

We are looking for two functions, \(g\) and \(h,\) so \(f(x)=g(h(x)).\) To do this, we look for a function inside a function in the formula for \(f(x).\) As one possibility, we might notice that the expression \(5-{x}^{2}\) is the inside of the square root. We could then decompose the function as
\[h(x)=5-{x}^{2}\ \text{and }g(x)=\sqrt{x}\]
We can check our answer by recomposing the functions.
\[g(h(x))=g(5-{x}^{2})=\sqrt{5-{x}^{2}}\]
Write \(f(x)=\frac{4}{3-\sqrt{4+{x}^{2}}}\) as the composition of two functions.

Bonisa impendulo

Possible answer:
\(\begin{array}{l}g(x)=\sqrt{4+{x}^{2}} \\ h(x)=\frac{4}{3-x} \\ f=h∘g\end{array}\)
How does one find the domain of the quotient of two functions, \(\frac{f}{g}?\)

Bonisa impendulo

Find the numbers that make the function in the denominator \(g\) equal to zero, and check for any other domain restrictions on \(f\) and \(g,\) such as an even-indexed root or zeros in the denominator.
What is the composition of two functions, \(f∘g?\)
If the order is reversed when composing two functions, can the result ever be the same as the answer in the original order of the composition? If yes, give an example. If no, explain why not.

Bonisa impendulo

Yes. Sample answer: Let \(f(x)=x+1\ \text{and }g(x)=x-1.\) Then \(f(g(x))=f(x-1)=(x-1)+1=x\) and \(g(f(x))=g(x+1)=(x+1)-1=x.\) So \(f∘g=g∘f.\)
How do you find the domain for the composition of two functions, \(f∘g?\)
Given \(f(x)={x}^{2}+2x\) and \(g(x)=6-{x}^{2},\) find \(f+g,\ f-g,\ fg,\) and \(\ \frac{f}{g}.\)

Bonisa impendulo

\((f+g)(x)=2x+6,\) domain: \((-\infty ,\infty )\)
\((f-g)(x)=2{x}^{2}+2x-6,\) domain: \((-\infty ,\infty )\)
\((fg)(x)=-{x}^{4}-2{x}^{3}+6{x}^{2}+12x,\) domain: \((-\infty ,\infty )\)
\((\frac{f}{g})(x)=\frac{{x}^{2}+2x}{6-{x}^{2}},\) domain: \((-\infty ,-\sqrt{6})\cup (-\sqrt{6},\sqrt{6})\cup (\sqrt{6},\infty )\)
Given \(f(x)=-3{x}^{2}+x\) and \(\ g(x)=5,\) find \(f+g,\ f-g,\ fg,\) and \(\ \frac{f}{g}.\)
Given \(f(x)=2{x}^{2}+4x\) and \(\ g(x)=\frac{1}{2x},\) find \(f+g,\ f-g,\ fg,\) and \(\ \frac{f}{g}.\)

Bonisa impendulo

\((f+g)(x)=\frac{4{x}^{3}+8{x}^{2}+1}{2x},\) domain: \((-\infty ,0)\cup (0,\infty )\)
\((f-g)(x)=\frac{4{x}^{3}+8{x}^{2}-1}{2x},\) domain: \((-\infty ,0)\cup (0,\infty )\)
\((fg)(x)=x+2,\) domain: \((-\infty ,0)\cup (0,\infty )\)
\((\frac{f}{g})(x)=4{x}^{3}+8{x}^{2},\) domain: \((-\infty ,0)\cup (0,\infty )\)
Given \(f(x)=\frac{1}{x-4}\) and \(g(x)=\frac{1}{6-x},\) find \(f+g,\ f-g,\ fg,\) and \(\ \frac{f}{g}.\)
Given \(f(x)=3{x}^{2}\) and \(g(x)=\sqrt{x-5},\) find \(f+g,\ f-g,\ fg,\) and \(\ \frac{f}{g}.\)

Bonisa impendulo

\((f+g)(x)=3{x}^{2}+\sqrt{x-5},\) domain: \([5,\infty )\)
\((f-g)(x)=3{x}^{2}-\sqrt{x-5},\) domain: \([5,\infty )\)
\((fg)(x)=3{x}^{2}\sqrt{x-5},\) domain: \([5,\infty )\)
\((\frac{f}{g})(x)=\frac{3{x}^{2}}{\sqrt{x-5}},\) domain: \((5,\infty )\)
Given \(f(x)=\sqrt{x}\) and \(g(x)=|x-3|,\) find \(\frac{g}{f}.\)
For the following exercise, find the indicated function given \(f(x)=2{x}^{2}+1\) and \(g(x)=3x-5.\)
1. ⓐ \(f(g(2))\)
2. ⓑ \(f(g(x))\)
3. ⓒ \(g(f(x))\)
4. ⓓ \((g∘g)(x)\)
5. ⓔ \((f∘f)(-2)\)
Bonisa impendulo
1. ⓐ \(3=2(9{x}^{2}-30x+25)+1=18{x}^{2}-60x+51\)
2. ⓑ \(f(g(x))=2{(3x-5)}^{2}+1\)
3. ⓒ \(g(f(x))=3(2{x}^{2}+1)-5=6{x}^{2}-2\)
4. ⓓ\((g∘g)(x)=3(3x-5)-5=9x-20\)
5. ⓔ \((f∘f)(-2)=163\)
\(f(x)={x}^{2}+1,\ g(x)=\sqrt{x+2}\)
\(f(x)=\sqrt{x}+2,\ g(x)={x}^{2}+3\)

Bonisa impendulo

\(f(g(x))=\sqrt{{x}^{2}+3}+2,\ g(f(x))=x+4\sqrt{x}+7\)
\(f(x)=|x|,\ g(x)=5x+1\)
\(f(x)=\sqrt[3]{x},\ g(x)=\frac{x+1}{{x}^{3}}\)

Bonisa impendulo

\(f(g(x))=\sqrt[3]{\frac{x+1}{{x}^{3}}}=\frac{\sqrt[3]{x+1}}{x},\ g(f(x))=\frac{\sqrt[3]{x}+1}{x}\)
\(f(x)=\frac{1}{x-6},\ g(x)=\frac{7}{x}+6\)
\(f(x)=\frac{1}{x-4},\ g(x)=\frac{2}{x}+4\)

Bonisa impendulo

\((f∘g)(x)=\frac{1}{\frac{2}{x}+4-4}=\frac{x}{2},\ (g∘f)(x)=2x-4\)

Symbols used here

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

A \cup B,\ A \cap B,\ A \setminus B

union, intersection, difference
In either; in both; in A but not B.

\neq

not equal
The two sides are different.

\sup,\ \inf

supremum, infimum
Least upper bound, greatest lower bound.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

imaginary unit
i² = −1.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Composition of Functions

Combine functions using algebraic operations.
Create a new function by composition of functions.
Evaluate composite functions.
Find the domain of a composite function.
Decompose a composite function into its component functions.
Find the value of a function (IA 3.5.3), (CA 3.1.2)

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

Zama ngokwakho

Parts of this page are adapted from OpenStax College Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

IiNkqubo Algebra

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\(x\)	–3	–2	–1	0	1	2	3
\(f(x)\)	11	9	7	5	3	1	–1
\(g(x)\)	–8	–3	0	1	0	–3	–8