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Finding Composite and Inverse Functions

Find and evaluate composite functions

Find and Evaluate Composite Functions

Before we introduce the functions, we need to look at another operation on functions called composition. In composition, the output of one function is the input of a second function. For functions \(f\) and \(g,\) the composition is written \(f∘g\) and is defined by \((f∘g)(x)=f(g(x)).\)

We read \(f(g(x))\) as \(\text{“}f\) of \(g\) of \(x\text{.”}\)

To do a composition, the output of the first function, \(g(x),\) becomes the input of the second function, f, and so we must be sure that it is part of the domain of f.

We have actually used composition without using the notation many times before. When we graphed quadratic functions using translations, we were composing functions. For example, if we first graphed \(g(x)={x}^{2}\) as a parabola and then shifted it down vertically four units, we were using the composition defined by \((f∘g)(x)=f(g(x))\) where \(f(x)=x-4.\)

The next example will demonstrate that \((f∘g)(x),\) \((g∘f)(x)\) and \((f\cdot g)(x)\) usually result in different outputs.

Example

Try it.

For functions \(f(x)=4x-5\) and \(g(x)=2x+3,\) find: ⓐ \((f∘g)(x),\) ⓑ \((g∘f)(x),\) and ⓒ \((f\cdot g)(x).\)

Solution

ⓐ

Use the definition of \((f∘g)(x).\)


Distribute.
Simplify.

ⓑ

Use the definition of \((f∘g)(x).\)


Distribute.
Simplify.

Notice the difference in the result in part ⓐ and part ⓑ.

ⓒ Notice that \((f\cdot g)(x)\) is different than \((f∘g)(x).\) In part ⓐ we did the composition of the functions. Now in part ⓒ we are not composing them, we are multiplying them.

\(\begin{array}{llllll}\text{Use the definition of}\ (f\cdot g)(x). & & & & & (f\cdot g)(x)=f(x)\cdot g(x) \\ \text{Substitute}\ f(x)=4x-5\ \text{and}\ g(x)=2x+3. & & & & & (f\cdot g)(x)=(4x-5)\cdot (2x+3) \\ \text{Multiply.} & & & & & (f\cdot g)(x)=8{x}^{2}+2x-15\end{array}\)

In the next example we will evaluate a composition for a specific value.

Example

Try it.

For functions \(f(x)={x}^{2}-4,\) and \(g(x)=3x+2,\) find: ⓐ \((f∘g)(-3),\) ⓑ \((g∘f)(-1),\) and ⓒ \((f∘f)(2).\)

Solution

ⓐ

Use the definition of \((f∘g)(-3).\)

Simplify.

Simplify.

ⓑ

Use the definition of \((g∘f)(-1).\)

Simplify.

Simplify.

ⓒ

Use the definition of \((f∘f)(2).\)

Simplify.

Simplify.

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

Determine Whether a Function is One-to-One

When we first introduced functions, we said 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.

We used the birthday example to help us understand the definition. Every person has a birthday, but no one has two birthdays and it is okay for two people to share a birthday. Since each person has exactly one birthday, that relation is a function.

A function is one-to-one if each value in the range has exactly one element in the domain. For each ordered pair in the function, each y-value is matched with only one x-value.

Our example of the birthday relation is not a one-to-one function. Two people can share the same birthday. The range value August 2 is the birthday of Liz and June, and so one range value has two domain values. Therefore, the function is not one-to-one.

Example

Try it.

For each set of ordered pairs, determine if it represents a function and, if so, if the function is one-to-one.

ⓐ \(\{(-3,27),(-2,8),(-1,1),(0,0),(1,1),(2,8),(3,27)\}\) and ⓑ \(\{(0,0),(1,1),(4,2),(9,3),(16,4)\}.\)

Solution

ⓐ
\(\begin{array}{lllll} & & & & \ \{(-3,27),(-2,8),(-1,1),(0,0),(1,1),(2,8),(3,27)\}\end{array}\)

Each x-value is matched with only one y-value. So this relation is a function.

But each y-value is not paired with only one x-value, \((-3,27)\) and \((3,27),\) for example. So this function is not one-to-one.

ⓑ
\(\begin{array}{lllll} & & & & \ \{(0,0),(1,1),(4,2),(9,3),(16,4)\}\end{array}\)

Each x-value is matched with only one y-value. So this relation is a function.

Since each y-value is paired with only one x-value, this function is one-to-one.

The vertical line is representing an x-value and we check that it intersects the graph in only one y-value. Then it is a function.

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

Find the Inverse of a Function

Let’s look at a one-to one function, \(f\), represented by the ordered pairs \(\{(0,5),(1,6),(2,7),(3,8)\}.\) For each \(x\)-value, \(f\) adds 5 to get the \(y\)-value. To ‘undo’ the addition of 5, we subtract 5 from each \(y\)-value and get back to the original \(x\)-value. We can call this “taking the inverse of \(f\)” and name the function \({f}^{-1}.\)

Notice that that the ordered pairs of \(f\) and \({f}^{-1}\) have their \(x\)-values and \(y\)-values reversed. The domain of \(f\) is the range of \({f}^{-1}\) and the domain of \({f}^{-1}\) is the range of \(f.\)

In the next example we will find the inverse of a function defined by ordered pairs.

Example

Try it.

Find the inverse of the function \(\{(0,3),(1,5),(2,7),(3,9)\}.\) Determine the domain and range of the inverse function.

Solution

This function is one-to-one since every \(x\)-value is paired with exactly one \(y\)-value.

To find the inverse we reverse the \(x\)-values and \(y\)-values in the ordered pairs of the function.

Function	\(\{(0,3),(1,5),(2,7),(3,9)\}\)
Inverse Function	\(\{(3,0),(5,1),(7,2),(9,3)\}\)
Domain of Inverse Function	\(\{3,5,7,9\}\)
Range of Inverse Function	\(\{0,1,2,3\}\)

We just noted that if \(f(x)\) is a one-to-one function whose ordered pairs are of the form \((x,y),\) then its inverse function \({f}^{-1}(x)\) is the set of ordered pairs \((y,x).\)

So if a point \((a,b)\) is on the graph of a function \(f(x),\) then the ordered pair \((b,a)\) is on the graph of \({f}^{-1}(x).\) See .

The distance between any two pairs \((a,b)\) and \((b,a)\) is cut in half by the line \(y=x.\) So we say the points are mirror images of each other through the line \(y=x.\)

Since every point on the graph of a function \(f(x)\) is a mirror image of a point on the graph of \({f}^{-1}(x),\) we say the graphs are mirror images of each other through the line \(y=x.\) We will use this concept to graph the inverse of a function in the next example.

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

Key Concepts

Composition of Functions: The composition of functions \(f\) and \(g,\) is written \(f∘g\) and is defined by
\[(f∘g)(x)=f(g(x))\]
We read \(f(g(x))\) as \(f\) of \(g\) of \(x.\)
Horizontal Line Test: If every horizontal line, intersects the graph of a function in at most one point, it is a one-to-one function.
Inverse of a Function Defined by Ordered Pairs: If \(f(x)\) is a one-to-one function whose ordered pairs are of the form \((x,y),\) then its inverse function \({f}^{-1}(x)\) is the set of ordered pairs \((y,x).\)
Inverse Functions: For every \(x\) in the domain of one-to-one function \(f\) and \({f}^{-1},\)
\[\begin{array}{lll}{f}^{-1}(f(x)) & = & x \\ f({f}^{-1}(x)) & = & x\end{array}\]
How to Find the Inverse of a One-to-One Function:
1. Substitute y for \(f(x).\)
2. Interchange the variables x and y.
3. Solve for y.
4. Substitute \({f}^{-1}(x)\) for \(y.\)
5. Verify that the functions are inverses.

Finding Composite and Inverse Functions

Find and Evaluate Composite Functions

In the following exercises, find ⓐ (f ∘ g)(x), ⓑ (g ∘ f)(x), and ⓒ (f · g)(x).

Try it.

\(f(x)=4x+3\) and \(g(x)=2x+5\)

Solution

ⓐ \(8x+23\) ⓑ \(8x+11\) ⓒ
\(8{x}^{2}+26x+15\)

Try it.

\(f(x)=3x-1\) and \(g(x)=5x-3\)

Try it.

\(f(x)=6x-5\) and \(g(x)=4x+1\)

Solution

ⓐ \(24x+1\) ⓑ \(24x-19\)
ⓒ \(24{x}^{2}-14x-5\)

Try it.

\(f(x)=2x+7\) and \(g(x)=3x-4\)

Try it.

\(f(x)=3x\) and \(g(x)=2{x}^{2}-3x\)

Solution

ⓐ \(6{x}^{2}-9x\) ⓑ \(18{x}^{2}-9x\)
ⓒ \(6{x}^{3}-9{x}^{2}\)

Try it.

\(f(x)=2x\) and \(g(x)=3{x}^{2}-1\)

Try it.

\(f(x)=2x-1\) and \(g(x)={x}^{2}+2\)

Solution

ⓐ \(2{x}^{2}+3\) ⓑ \(4{x}^{2}-4x+3\)
ⓒ \(2{x}^{3}-{x}^{2}+4x-2\)

Try it.

\(f(x)=4x+3\) and \(g(x)={x}^{2}-4\)

In the following exercises, find the values described.

Try it.

For functions \(f(x)=2{x}^{2}+3\) and \(g(x)=5x-1,\) find ⓐ \((f∘g)(-2)\) ⓑ \((g∘f)(-3)\) ⓒ \((f∘f)(-1)\)

Solution

ⓐ 245 ⓑ 104 ⓒ 53

Try it.

For functions \(f(x)=5{x}^{2}-1\) and \(g(x)=4x-1,\) find ⓐ \((f∘g)(1)\) ⓑ \((g∘f)(-1)\) ⓒ \((f∘f)(2)\)

Try it.

For functions \(f(x)=2{x}^{3}\) and \(g(x)=3{x}^{2}+2,\) find ⓐ \((f∘g)(-1)\) ⓑ \((g∘f)(1)\) ⓒ \((g∘g)(1)\)

Solution

ⓐ 250 ⓑ 14 ⓒ 77

Try it.

For functions \(f(x)=3{x}^{3}+1\) and \(g(x)=2{x}^{2}-3,\) find ⓐ \((f∘g)(-2)\) ⓑ \((g∘f)(-1)\) ⓒ \((g∘g)(1)\)

Determine Whether a Function is One-to-One

In the following exercises, determine if the set of ordered pairs represents a function and if so, is the function one-to-one.

Try it.

\(\{(-3,9),(-2,4),(-1,1),(0,0)\),
\((1,1),(2,4),(3,9)\}\)

Solution

Function; not one-to-one

Try it.

\(\{(9,-3),(4,-2),(1,-1),(0,0)\),
\((1,1),(4,2),(9,3)\}\)

Try it.

\(\{(-3,-5),(-2,-3),(-1,-1)\),
\((0,1),(1,3),(2,5),(3,7)\}\)

Solution

One-to-one function

Try it.

\(\{(5,3),(4,2),(3,1),(2,0)\),
\((1,-1),(0,-2),(-1,-3)\}\)

In the following exercises, determine whether each graph is the graph of a function and if so, is it one-to-one.

Try it.

ⓐ

ⓑ

Solution

ⓐ Not a function ⓑ Function; not one-to-one

Try it.

ⓐ

ⓑ

Try it.

ⓐ

ⓑ

Solution

ⓐ One-to-one function
ⓑ Function; not one-to-one

Try it.

ⓐ

ⓑ

In the following exercises, find the inverse of each function. Determine the domain and range of the inverse function.

Try it.

\(\{(2,1),(4,2),(6,3),(8,4)\}\)

Solution

Inverse function: \(\{(1,2),(2,4),(3,6),(4,8)\}.\) Domain: \(\{1,2,3,4\}.\) Range: \(\{2,4,6,8\}.\)

Try it.

\(\{(6,2),(9,5),(12,8),(15,11)\}\)

Try it.

\(\{(0,-2),(1,3),(2,7),(3,12)\}\)

Solution

Inverse function: \(\{(-2,0),(3,1),(7,2),(12,3)\}.\) Domain: \(\{-2,3,7,12\}.\) Range: \(\{0,1,2,3\}.\)

Try it.

\(\{(0,0),(1,1),(2,4),(3,9)\}\)

Try it.

\(\{(-2,-3),(-1,-1),(0,1),(1,3)\}\)

Solution

Inverse function: \(\{(-3,\text{-}2),(-1,-1),(1,0),(3,1)\}.\) Domain: \(\{-3,\text{-}1,1,3\}.\) Range: \(\{-2,-1,0,1\}.\)

Try it.

\(\{(5,3),(4,2),(3,1),(2,0)\}\)

In the following exercises, graph, on the same coordinate system, the inverse of the one-to-one function shown.

Try it.

Solution

Try it.

Solution

Try it.

In the following exercises, determine whether or not the given functions are inverses.

Try it.

\(f(x)=x+8\) and \(g(x)=x-8\)

Solution

\(g(f(x))=x,\) and \(f(g(x))=x,\) so they are inverses.

Try it.

\(f(x)=x-9\) and \(g(x)=x+9\)

Try it.

\(f(x)=7x\) and \(g(x)=\frac{x}{7}\)

Solution

\(g(f(x))=x,\) and \(f(g(x))=x,\) so they are inverses.

Try it.

\(f(x)=\frac{x}{11}\) and \(g(x)=11x\)

Try it.

\(f(x)=7x+3\) and \(g(x)=\frac{x-3}{7}\)

Solution

\(g(f(x))=x,\) and \(f(g(x))=x,\) so they are inverses.

Try it.

\(f(x)=5x-4\) and \(g(x)=\frac{x-4}{5}\)

Try it.

\(f(x)=\sqrt{x+2}\) and \(g(x)={x}^{2}-2(x>0)\)

Solution

\(g(f(x))=x,\) and \(f(g(x))=x,\) so they are inverses (for nonnegative \(x).\)

Try it.

\(f(x)=\sqrt[3]{x-4}\) and \(g(x)={x}^{3}+4\)

In the following exercises, find the inverse of each function.

Try it.

\(f(x)=x-12\)

Solution

\({f}^{-1}(x)=x+12\)

Try it.

\(f(x)=x+17\)

Try it.

\(f(x)=9x\)

Solution

\({f}^{-1}(x)=\frac{x}{9}\)

Try it.

\(f(x)=8x\)

Try it.

\(f(x)=\frac{x}{6}\)

Solution

\({f}^{-1}(x)=6x\)

Try it.

\(f(x)=\frac{x}{4}\)

Try it.

\(f(x)=6x-7\)

Solution

\({f}^{-1}(x)=\frac{x+7}{6}\)

Try it.

\(f(x)=7x-1\)

Try it.

\(f(x)=-2x+5\)

Solution

\({f}^{-1}(x)=\frac{x-5}{-2}\)

Try it.

\(f(x)=-5x-4\)

Try it.

\(f(x)={x}^{2}+6,\) \(x\ge 0\)

Solution

\({f}^{-1}(x)=\sqrt{x-6}\)

Try it.

\(f(x)={x}^{2}-9,\) \(x\ge 0\)

Try it.

\(f(x)={x}^{3}-4\)

Solution

\({f}^{-1}(x)=\sqrt[3]{x+4}\)

Try it.

\(f(x)={x}^{3}+6\)

Try it.

\(f(x)=\frac{1}{x+2}\)

Solution

\({f}^{-1}(x)=\frac{1}{x}-2\)

Try it.

\(f(x)=\frac{1}{x-6}\)

Try it.

\(f(x)=\sqrt{x-2},\) \(x\ge 2\)

Solution

\({f}^{-1}(x)={x}^{2}+2\), \(x\ge 0\)

Try it.

\(f(x)=\sqrt{x+8},\) \(x\ge -8\)

Try it.

\(f(x)=\sqrt[3]{x-3}\)

Solution

\({f}^{-1}(x)={x}^{3}+3\)

Try it.

\(f(x)=\sqrt[3]{x+5}\)

Try it.

\(f(x)=\sqrt[4]{9x-5},\) \(x\ge \frac{5}{9}\)

Solution

\({f}^{-1}(x)=\frac{{x}^{4}+5}{9}\), \(x\ge 0\)

Try it.

\(f(x)=\sqrt[4]{8x-3},\) \(x\ge \frac{3}{8}\)

Try it.

\(f(x)=\sqrt[5]{-3x+5}\)

Solution

\({f}^{-1}(x)=\frac{{x}^{5}-5}{-3}\)

Try it.

\(f(x)=\sqrt[5]{-4x-3}\)

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

Practice (40)

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

If \(f(x)=2x-3\) and \(g(x)={x}^{2}+2x-3,\) find \(f(4).\)
If you missed this problem, review .

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ⓐ \(f\left(4\right)=5\); ⓑ \(g\left(f\left(4\right)\right)=32\)
Solve for \(x,\) \(3x+2y=12.\)
If you missed this problem, review .

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\(x=-\frac{2}{3}y+4\)
Simplify: \(5\frac{(x+4)}{5}-4.\)
If you missed this problem, review .

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\(x\)
For functions \(f(x)=4x-5\) and \(g(x)=2x+3,\) find: ⓐ \((f∘g)(x),\) ⓑ \((g∘f)(x),\) and ⓒ \((f\cdot g)(x).\)

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ⓐ

Use the definition of \((f∘g)(x).\)
Distribute.
Simplify.

ⓑ

Use the definition of \((f∘g)(x).\)
Distribute.
Simplify.
Notice the difference in the result in part ⓐ and part ⓑ.
ⓒ Notice that \((f\cdot g)(x)\) is different than \((f∘g)(x).\) In part ⓐ we did the composition of the functions. Now in part ⓒ we are not composing them, we are multiplying them.

\(\begin{array}{llllll}\text{Use the definition of}\ (f\cdot g)(x). & & & & & (f\cdot g)(x)=f(x)\cdot g(x) \\ \text{Substitute}\ f(x)=4x-5\ \text{and}\ g(x)=2x+3. & & & & & (f\cdot g)(x)=(4x-5)\cdot (2x+3) \\ \text{Multiply.} & & & & & (f\cdot g)(x)=8{x}^{2}+2x-15\end{array}\)
For functions \(f(x)=3x-2\) and \(g(x)=5x+1,\) find ⓐ \((f∘g)(x)\) ⓑ \((g∘f)(x)\) ⓒ \((f\cdot g)(x)\).

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ⓐ \(15x+1\) ⓑ \(15x-9\)
ⓒ \(15{x}^{2}-7x-2\)
For functions \(f(x)=4x-3,\) and \(g(x)=6x-5,\) find ⓐ \((f∘g)(x),\) ⓑ \((g∘f)(x),\) and ⓒ \((f\cdot g)(x).\)

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ⓐ \(24x-23\) ⓑ \(24x-23\)
ⓒ \(24{x}^{2}-38x+15\)
For functions \(f(x)={x}^{2}-4,\) and \(g(x)=3x+2,\) find: ⓐ \((f∘g)(-3),\) ⓑ \((g∘f)(-1),\) and ⓒ \((f∘f)(2).\)

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ⓐ

Use the definition of \((f∘g)(-3).\)
Simplify.
Simplify.

ⓑ

Use the definition of \((g∘f)(-1).\)
Simplify.
Simplify.

ⓒ

Use the definition of \((f∘f)(2).\)
Simplify.
Simplify.
For functions \(f(x)={x}^{2}-9,\) and \(g(x)=2x+5,\) find ⓐ \((f∘g)(-2),\) ⓑ \((g∘f)(-3),\) and ⓒ \((f∘f)(4).\)

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ⓐ –8 ⓑ 5 ⓒ 40
For functions \(f(x)={x}^{2}+1,\) and \(g(x)=3x-5,\) find ⓐ \((f∘g)(-1),\) ⓑ \((g∘f)(2),\) and ⓒ \((f∘f)(-1).\)

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ⓐ 65 ⓑ 10 ⓒ 5
For each set of ordered pairs, determine if it represents a function and, if so, if the function is one-to-one.
ⓐ \(\{(-3,27),(-2,8),(-1,1),(0,0),(1,1),(2,8),(3,27)\}\) and ⓑ \(\{(0,0),(1,1),(4,2),(9,3),(16,4)\}.\)

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ⓐ
\(\begin{array}{lllll} & & & & \ \{(-3,27),(-2,8),(-1,1),(0,0),(1,1),(2,8),(3,27)\}\end{array}\)
Each x-value is matched with only one y-value. So this relation is a function.
But each y-value is not paired with only one x-value, \((-3,27)\) and \((3,27),\) for example. So this function is not one-to-one.
ⓑ
\(\begin{array}{lllll} & & & & \ \{(0,0),(1,1),(4,2),(9,3),(16,4)\}\end{array}\)
Each x-value is matched with only one y-value. So this relation is a function.
Since each y-value is paired with only one x-value, this function is one-to-one.
For each set of ordered pairs, determine if it represents a function and if so, is the function one-to-one.
ⓐ \(\{(-3,-6),(-2,-4),(-1,-2),(0,0),(1,2),(2,4),(3,6)\}\) ⓑ \(\{(-4,8),(-2,4),(-1,2),(0,0),(1,2),(2,4),(4,8)\}\)

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ⓐ One-to-one function
ⓑ Function; not one-to-one
For each set of ordered pairs, determine if it represents a function and if so, is the function one-to-one.
ⓐ \(\{(27,-3),(8,-2),(1,-1),(0,0),(1,1),(8,2),(27,3)\}\) ⓑ \(\{(7,-3),(-5,-4),(8,0),(0,0),(-6,4),(-2,2),(-1,3)\}\)

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ⓐ Not a function
ⓑ Function; not one-to-one
Determine ⓐ whether each graph is the graph of a function and, if so, ⓑ whether it is one-to-one.

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ⓐ

Since any vertical line intersects the graph in at most one point, the graph is the graph of a function. Since any horizontal line intersects the graph in at most one point, the graph is the graph of a one-to-one function.
ⓑ

Since any vertical line intersects the graph in at most one point, the graph is the graph of a function. The horizontal line shown on the graph intersects it in two points. This graph does not represent a one-to-one function.
Determine whether each graph is the graph of a function and, if so, whether it is one-to-one.

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ⓐ Not a function ⓑ One-to-one function
Determine whether each graph is the graph of a function and, if so, whether it is one-to-one.

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ⓐ Function; not one-to-one ⓑ One-to-one function
Find the inverse of the function \(\{(0,3),(1,5),(2,7),(3,9)\}.\) Determine the domain and range of the inverse function.

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This function is one-to-one since every \(x\)-value is paired with exactly one \(y\)-value.
To find the inverse we reverse the \(x\)-values and \(y\)-values in the ordered pairs of the function.

Function \(\{(0,3),(1,5),(2,7),(3,9)\}\)
Inverse Function \(\{(3,0),(5,1),(7,2),(9,3)\}\)
Domain of Inverse Function \(\{3,5,7,9\}\)
Range of Inverse Function \(\{0,1,2,3\}\)
Find the inverse of \(\{(0,4),(1,7),(2,10),(3,13)\}.\) Determine the domain and range of the inverse function.

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Inverse function: \(\{(4,0),(7,1),(10,2),(13,3)\}.\) Domain: \(\{4,7,10,13\}.\) Range: \(\{0,1,2,3\}.\)
Find the inverse of \(\{(-1,4),(-2,1),(-3,0),(-4,2)\}.\) Determine the domain and range of the inverse function.

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Inverse function: \(\{(4,-1),(1,-2),(0,-3),(2,-4)\}.\) Domain: \(\{0,1,2,4\}.\) Range: \(\{-4,-3,-2,-1\}.\)
Graph, on the same coordinate system, the inverse of the one-to one function shown.

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We can use points on the graph to find points on the inverse graph. Some points on the graph are: \((-5,-3),(-3,-1),(-1,0),(0,2),(3,4)\).
So, the inverse function will contain the points: \((-3,-5),(-1,-3),(0,-1),(2,0),(4,3)\).

Notice how the graph of the original function and the graph of the inverse functions are mirror images through the line \(y=x.\)
Graph, on the same coordinate system, the inverse of the one-to one function.

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Graph, on the same coordinate system, the inverse of the one-to one function.

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Verify that \(f(x)=5x-1\) and \(g(x)=\frac{x+1}{5}\) are inverse functions.

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The functions are inverses of each other if \(g(f(x))=x\) and \(f(g(x))=x.\)
Substitute \(5x-1\) for \(f(x).\)
Simplify.
Simplify.
Substitute \(\frac{x+1}{5}\) for \(g(x).\)
Simplify.
Simplify.
Since both \(g(f(x))=x\) and \(f(g(x))=x\) are true, the functions \(f(x)=5x-1\) and \(g(x)=\frac{x+1}{5}\) are inverse functions. That is, they are inverses of each other.
Verify that the functions are inverse functions.
\(f(x)=4x-3\) and \(g(x)=\frac{x+3}{4}.\)

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\(g(f(x))=x,\) and \(f(g(x))=x,\) so they are inverses.
Verify that the functions are inverse functions.
\(f(x)=2x+6\) and \(g(x)=\frac{x-6}{2}.\)

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\(g(f(x))=x,\) and \(f(g(x))=x,\) so they are inverses.
Find the inverse of \(f(x)=4x+7.\)
Find the inverse of the function \(f(x)=5x-3.\)

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\({f}^{-1}(x)=\frac{x+3}{5}\)
Find the inverse of the function \(f(x)=8x+5.\)

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\({f}^{-1}(x)=\frac{x-5}{8}\)

Find the inverse of \(f(x)=\sqrt[5]{2x-3}.\)

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	\(f(x)=\sqrt[5]{2x-3}\\)
Substitute \(y\) for \(f(x)\).	\(y=\sqrt[5]{2x-3}\\)
Interchange the variables \(x\) and \(y\).	\(x=\sqrt[5]{2y-3}\\)
Solve for \(y\).	\({(x)}^{5}={(\sqrt[5]{2y-3})}^{5}\\)
	\({x}^{5}=2y-3\\)
	\({x}^{5}+3=2y\\)
	\(\frac{{x}^{5}+3}{2}=y\\)
Substitute \({f}^{-1}(x)\) for \(y\).	\({f}^{-1}(x)=\frac{{x}^{5}+3}{2}\\)
Verify that the functions are inverses.
\({f}^{-1}(f(x))\overset{?}{=}x\\)	\(f({f}^{-1}(x))\overset{?}{=}x\\)
\({f}^{-1}(\sqrt[5]{2x-3})\overset{?}{=}x\\)	\(f(\frac{{x}^{5}+3}{2})\overset{?}{=}x\\)
\(\frac{{(\sqrt[5]{2x-3})}^{5}+3}{2}\overset{?}{=}x\\)	\(\sqrt[5]{2(\frac{{x}^{5}+3}{2})-3}\overset{?}{=}x\\)
\(\frac{2x-3+3}{2}\overset{?}{=}x\\)	\(\sqrt[5]{{x}^{5}+3-3}\overset{?}{=}x\\)
\(\frac{2x}{2}\overset{?}{=}x\\)	\(\sqrt[5]{{x}^{5}}\overset{?}{=}x\\)
\(x=x✓\\)	\(x=x✓\\)

Find the inverse of the function \(f(x)=\sqrt[5]{3x-2}.\)

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\({f}^{-1}(x)=\frac{{x}^{5}+2}{3}\)
Find the inverse of the function \(f(x)=\sqrt[4]{6x-7}.\)

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\({f}^{-1}(x)=\frac{{x}^{4}+7}{6}\)
\(f(x)=3x-1\) and \(g(x)=5x-3\)
\(f(x)=2x+7\) and \(g(x)=3x-4\)
\(f(x)=2x\) and \(g(x)=3{x}^{2}-1\)
\(f(x)=4x+3\) and \(g(x)={x}^{2}-4\)
For functions \(f(x)=2{x}^{2}+3\) and \(g(x)=5x-1,\) find ⓐ \((f∘g)(-2)\) ⓑ \((g∘f)(-3)\) ⓒ \((f∘f)(-1)\)

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ⓐ 245 ⓑ 104 ⓒ 53

Symbols used here

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

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

\leq,\ \geq

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

\pm

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

\neq

not equal
The two sides are different.

|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: Finding Composite and Inverse Functions

Find and evaluate composite functions
Determine whether a function is one-to-one
Find the inverse of a function
Substitute
Interchange the variables
Solve for
Substitute
Verify that the functions are inverses.

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.

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

بیشتر در Algebra

Linear equations Quadratic equations Systems of equations Inequalities Factoring Expanding Simplifying expressions Functions and graphs Exponential and logarithmic equations Polynomial equations Absolute value


Substitute \(5x-1\) for \(f(x).\)

Simplify.
Simplify.


Substitute \(\frac{x+1}{5}\) for \(g(x).\)

Simplify.
Simplify.