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Composition of Functions

Combine functions using algebraic operations.

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

\[\begin{array}{l}(f+g)(x)=f(x)+g(x) \\ (f-g)(x)=f(x)-g(x) \\ (fg)(x)=f(x)g(x) \\ (\frac{f}{g})(x)=\frac{f(x)}{g(x)}\end{array}\]

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 Precalculus 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

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

but

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

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 Precalculus 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 \(f(2)=8.\)

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

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 Precalculus 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 \(g(x)=1.\)

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

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 Precalculus 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.

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?

Жауап

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?

Жауап

\(\begin{array}{l}(fg)(x)=f(x)g(x)=(x-1)({x}^{2}-1)={x}^{3}-{x}^{2}-x+1 \\ (f-g)(x)=f(x)-g(x)=(x-1)-({x}^{2}-1)=x-{x}^{2}\end{array}\)
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.
\[f(x)=2x+1\ g(x)=3-x\]

Жауап

Let’s begin by substituting \(g(x)\) into \(f(x).\)
\[\begin{array}{l}f(g(x))=2(3-x)+1 \\ =6-2x+1 \\ =7-2x\end{array}\]
Now we can substitute \(f(x)\) into \(g(x).\)
\[\begin{array}{l}g(f(x))=3-(2x+1) \\ =3-2x-1 \\ =-2x+2\end{array}\]
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 \(c(s(3)).\)

Жауап

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 \(g(f(x))?\)

Жауап

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.

Жауап

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

Жауап

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 \(f(2)=8.\)
\[\begin{array}{l}g(3)=2 \\ f(g(3))=f(2)=8\end{array}\]
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 \(g(f(4)).\)

Жауап

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

Жауап

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 \(g(f(2)).\)

Жауап

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

Жауап

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

Жауап

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 \(g(x)=1.\)
\[\begin{array}{l}\frac{4}{3x-2}=1 \\ 4=3x-2 \\ 6=3x \\ x=2\end{array}\]
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
\[(f∘g)(x)\text{ where}\ f(x)=\sqrt{x+2}\text{ and}\ g(x)=\sqrt{3-x}\]

Жауап

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

Жауап

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

Жауап

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.

Жауап

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

Жауап

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.

Жауап

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 \(\frac{f}{g}.\) Determine the domain for each function in interval notation.

Жауап

\((f+g)(x)=2x+6,\) 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 \(\frac{f}{g}.\) Determine the domain for each function in interval notation.
Given \(f(x)=2{x}^{2}+4x\) and \(g(x)=\frac{1}{2x},\) find \(f+g,\ f-g,\ fg,\) and \(\frac{f}{g}.\) Determine the domain for each function in interval notation.

Жауап

\((f+g)(x)=\frac{4{x}^{3}+8{x}^{2}+1}{2x},\) 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 \(\frac{f}{g}.\) Determine the domain for each function in interval notation.
Given \(f(x)=3{x}^{2}\) and \(g(x)=\sqrt{x-5},\) find \(f+g,\ f-g,\ fg,\) and \(\frac{f}{g}.\) Determine the domain for each function in interval notation.

Жауап

\((f+g)(x)=3{x}^{2}+\sqrt{x-5},\) 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 \(\frac{g}{f}.\) Determine the domain of the function in interval notation.
Given \(f(x)=2{x}^{2}+1\) and \(g(x)=3x-5,\) find the following:
1. ⓐ \(f(g(2))\)
2. ⓑ \(f(g(x))\)
3. ⓒ \(g(f(x))\)
4. ⓓ \((g∘g)(x)\)
5. ⓔ \((f∘f)(-2)\)
Жауап
1. ⓐ 3
2. ⓑ \(f(g(x))=2{(3x-5)}^{2}+1;\)
3. ⓒ \(g(f)(x))=6{x}^{2}-2;\)
4. ⓓ \((g∘g)(x)=3(3x-5)-5=9x-20;\)
5. ⓔ \((f∘f)(-2)=163\)
\(f(x)={x}^{2}+1,\ g(x)=\sqrt{x+2}\)
\(f(x)=\sqrt{x}+2,\ g(x)={x}^{2}+3\)

Жауап

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

Жауап

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

Жауап

\((f∘g)(x)=\frac{1}{\frac{2}{x}+4-4}=\frac{x}{2},\ (g∘f)(x)=2x-4\)
\(f(x)={x}^{4}+6,\) \(g(x)=x-6,\) and \(h(x)=\sqrt{x}\)
\(f(x)={x}^{2}+1,\) \(g(x)=\frac{1}{x},\) and \(h(x)=x+3\)

Жауап

\(f(g(h(x)))={(\frac{1}{x+3})}^{2}+1\)
Given \(f(x)=\frac{1}{x}\) and \(g(x)=x-3,\) find the following:
1. ⓐ \((f∘g)(x)\)
2. ⓑ the domain of \((f∘g)(x)\) in interval notation
3. ⓒ \((g∘f)(x)\)
4. ⓓ the domain of \((g∘f)(x)\)
5. ⓔ \((\frac{f}{g})x\)
Given \(f(x)=\sqrt{2-4x}\) and \(g(x)=-\frac{3}{x},\) find the following:
1. ⓐ \((g∘f)(x)\)
2. ⓑ the domain of \((g∘f)(x)\) in interval notation
Жауап
- ⓐ Text \((g∘f)(x)=-\frac{3}{\sqrt{2-4x}};\)
- ⓑ\((-\infty ,\frac{1}{2})\)
Given the functions \(f(x)=\frac{1-x}{x}\ \text{and}\ g(x)=\frac{1}{1+{x}^{2}},\) find the following:
1. ⓐ \((g∘f)(x)\)
2. ⓑ \((g∘f)(\text{2})\)
Given functions \(p(x)=\frac{1}{\sqrt{x}}\) and \(m(x)={x}^{2}-4,\) state the domain of each of the following functions using interval notation:
1. ⓐ \(\frac{p(x)}{m(x)}\)
2. ⓑ \(p(m(x))\)
3. ⓒ \(m(p(x))\)
Жауап
1. ⓐ \((0,2)\cup (2,\infty );\)
2. ⓑ \((-\infty ,-2)\cup (2,\infty );\) c. \((0,\infty )\)

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.

|x|

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

\binom{n}{k}

binomial coefficient, "n choose k"
Number of k-element subsets of n things: n!/(k!(n−k)!).

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

Euler's number
2.71828…, the base whose exponential is its own derivative.

imaginary unit
i² = −1.

\theta

theta
The usual name for an angle.

\log_b x,\ \ln x

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

\mathbf{v},\ \vec{v}

vector
A quantity with magnitude and direction; a column of 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.
Locate the given input to the inner function on the
Read off the output of the inner function from the
Locate the inner function output on the

Questions people ask

What is a function, really?

A rule that assigns exactly one output to each input. The vertical-line test on a graph is the same idea: no input may have two outputs.

Why do we need complex numbers?

Because x² + 1 = 0 has no real solution, and allowing one new number i with i² = −1 makes every polynomial equation solvable. They then turn out to describe rotation, waves and alternating current more naturally than real numbers do.

Өзүңүздүн аракетиңизди көрүңүз

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

Кээ бирлери Precalculus

Complex numbers Polynomial functions Rational functions Sequences and series The binomial theorem Conic sections Vectors Exponential and logarithmic functions Polynomial division and the remainder theorem Parametric equations and polar coordinates