How Do You Know the Inverse Is a Function

Composition of Functions

In mathematics, it is frequently the instance that the result of 1 function is evaluated past applying a second function. For example, consider the functions divers by $f\left(\mathrm{ten}\right)=x^{\mathrm{two}}$ and $\mathrm{grand}\left(x\right)=2x+5.$ First, g is evaluated where $x=-\mathrm{one}$ and then the result is squared using the second function, f.

This sequential adding results in 9. We tin streamline this process past creating a new function divers by $f\left(g\left(x\right)\right)$ , which is explicitly obtained past substituting $g\left(x\right)$ into $f\left(x\right).$

$$\begin{array}{ccc}\hfill f\left(\mathrm{one\; thousand}\left(x\right)\right)& =& f\left(2x+5\right)\hfill \\ & =& {\left(\mathrm{ii}x+5\right)}^2\hfill \\ & =& \mathrm{four}{x}^2+20\mathrm{10}+25\hfill \end{array}$$

Therefore, $f\left(m\left(x\right)\right)=\mathrm{iv}{x}^{\mathrm{two}}+20x+25$ and nosotros can verify that when $x=-1$ the effect is ix.

$$\begin{array}{ccc}\hfill f\left(g\left(-1\right)\right)& =& 4{\left(-\mathrm{one}\right)}^{\mathrm{ii}}+20\left(-1\right)+25\hfill \\ & =& \mathrm{four}-20+25\hfill \\ & =& 9\hfill \end{array}$$

The adding above describes composition of functionsApplying a function to the results of another function. , which is indicated using the composition operatorThe open up dot used to indicate the function composition $\left(f○g\right)\left(x\right)=f\left(g\left(x\right)\right).$ ( $○$ ). If given functions f and grand,

$$\left(f○g\right)\left(x\right)=f\left(g\left(x\right)\right)Co\mathrm{grand}positionofFunctio\mathrm{northward}\mathrm{due\; south}$$

The notation $f○\mathrm{1000}$ is read, "f composed with thousand." This operation is only defined for values, ten, in the domain of k such that $\mathrm{yard}\left(x\right)$ is in the domain of f.

Instance 1

Given $f\left(\mathrm{10}\right)=x^2-\mathrm{ten}+\mathrm{three}$ and $g\left(x\right)=2x-\mathrm{i}$ summate:

1. $\left(f○\mathrm{thou}\right)\left(x\right).$
2. $\left(\mathrm{one\; thousand}○f\right)\left(\mathrm{10}\right).$

Solution:

1. Substitute m into f.

   $$\begin{array}{ccc}\hfill \left(f○g\right)\left(x\right)& =& f\left(\mathrm{thousand}\left(\mathrm{ten}\right)\right)\hfill \\ & =& f\left(2x-1\right)\hfill \\ & =& {\left(2x-1\right)}^2-\left(2x-\mathrm{one}\right)+\mathrm{iii}\hfill \\ & =& 4x^{\mathrm{two}}-\mathrm{four}\mathrm{ten}+\mathrm{one}-2\mathrm{ten}+\mathrm{ane}+3\hfill \\ & =& 4x^2-6\mathrm{10}+5\hfill \end{array}$$

2. Substitute f into g.

   $$\begin{array}{ccc}\hfill \left(g○f\right)\left(x\right)& =& g\left(f\left(\mathrm{10}\right)\right)\hfill \\ & =& g\left({\mathrm{ten}}^2-\mathrm{ten}+\mathrm{iii}\right)\hfill \\ & =& 2\left(x^{\mathrm{two}}-\mathrm{ten}+3\right)-1\hfill \\ & =& 2x^2-2\mathrm{ten}+6-\mathrm{ane}\hfill \\ & =& 2{\mathrm{ten}}^2-2\mathrm{ten}+\mathrm{v}\hfill \end{array}$$

Reply:

1. $\left(f○g\right)\left(x\right)=4{\mathrm{10}}^{\mathrm{two}}-\mathrm{vi}x+5$
2. $\left(g○f\right)\left(x\right)=2x^2-\mathrm{two}x+5$

The previous instance shows that composition of functions is not necessarily commutative.

Example two

Given $f\left(\mathrm{10}\right)={\mathrm{10}}^{\mathrm{iii}}+1$ and $g\left(x\right)=\sqrt[3]{\mathrm{three}x-\mathrm{ane}}$ detect $\left(f○g\right)\left(4\right).$

Solution:

Begin by finding $\left(f○\mathrm{1000}\right)\left(x\right).$

$$\begin{array}{ccc}\hfill \left(f○\mathrm{yard}\right)\left(\mathrm{ten}\right)& =& f\left(g\left(\mathrm{10}\right)\right)\hfill \\ & =& f\left(\sqrt[3]{\mathrm{three}\mathrm{10}-\mathrm{ane}}\right)\hfill \\ & =& {\left(\sqrt[3]{3x-1}\right)}^{\mathrm{iii}}+\mathrm{i}\hfill \\ & =& 3x-1+1\hfill \\ & =& 3x\hfill \end{array}$$

Next, substitute 4 in for ten.

$$\begin{array}{ccc}\hfill \left(f○\mathrm{grand}\right)\left(\mathrm{10}\right)& =& \mathrm{three}\mathrm{10}\hfill \\ \hfill \left(f○\mathrm{yard}\right)\left(4\right)& =& 3\left(4\right)\hfill \\ & =& 12\hfill \end{array}$$

Answer: $\left(f○\mathrm{thou}\right)\left(4\right)=12$

Functions can exist composed with themselves.

Example 3

Given $f\left(x\right)={\mathrm{ten}}^2-2$ detect $\left(f○f\right)\left(x\right).$

Solution:

$$\begin{array}{ccc}\hfill \left(f○f\right)\left(\mathrm{ten}\right)& =& f\left(f\left(x\right)\right)\hfill \\ & =& f\left(x^2-2\right)\hfill \\ & =& {\left({\mathrm{10}}^2-\mathrm{two}\right)}^2-\mathrm{ii}\hfill \\ & =& x^{\mathrm{four}}-\mathrm{four}{\mathrm{ten}}^2+\mathrm{four}-2\hfill \\ & =& x^4-\mathrm{iv}{x}^2+\mathrm{two}\hfill \end{array}$$

Reply: $\left(f○f\right)\left(\mathrm{ten}\right)=x^{\mathrm{four}}-\mathrm{iv}{\mathrm{10}}^{\mathrm{ii}}+\mathrm{two}$

Try this! Given $f\left(\mathrm{10}\right)=2x+3$ and $g\left(x\right)=\sqrt{\mathrm{10}-1}$ find $\left(f○g\right)\left(\mathrm{five}\right).$

Answer: 7

Inverse Functions

Consider the function that converts degrees Fahrenheit to degrees Celsius: $C\left(x\right)=\frac{\mathrm{five}}{9}\left(x-32\right).$ We tin employ this office to catechumen 77°F to degrees Celsius as follows.

$$\begin{array}{ccc}\hfill C\left(77\right)& =& \frac{5}{9}\left(77-32\right)\hfill \\ & =& \frac{5}{9}\left(45\right)\hfill \\ & =& 25\hfill \end{array}$$

Therefore, 77°F is equivalent to 25°C. If we wish to convert 25°C dorsum to degrees Fahrenheit we would use the formula: $F\left(x\right)=\frac{9}{\mathrm{v}}\mathrm{ten}+32.$

$$\begin{array}{ccc}\hfill F\left(25\right)& =& \frac{\mathrm{ix}}{\mathrm{five}}\left(25\right)+32\hfill \\ & =& 45+32\hfill \\ & =& 77\hfill \end{array}$$

Find that the ii functions $C$ and $F$ each opposite the effect of the other.

This describes an changed relationship. In general, f and g are inverse functions if,

$$\begin{array}{ccc}\hfill \left(f○\mathrm{yard}\right)\left(\mathrm{ten}\right)& =& f\left(g\left(x\right)\right)=\mathrm{10}forallxinthedomainofgand\hfill \\ \hfill \left(g○f\right)\left(x\right)& =& \mathrm{one\; thousand}\left(f\left(\mathrm{10}\right)\right)=xforal\mathrm{fifty}xinthedokainoff.\hfill \end{array}$$

In this case,

$$\begin{array}{ccc}\hfill C\left(F\left(25\right)\right)& =& C\left(77\right)=25\hfill \\ \hfill F\left(C\left(77\right)\right)& =& F\left(25\right)=77\hfill \end{array}$$

Example 4

Verify algebraically that the functions divers by $f\left(x\right)=\frac{1}{2}\mathrm{ten}-\mathrm{v}$ and $g\left(x\right)=2\mathrm{ten}+\mathrm{x}$ are inverses.

Solution:

Compose the functions both means and verify that the result is 10.

$\begin{array}{ccc}\hfill \left(f○g\right)\left(x\right)& =& f\left(g\left(x\right)\right)\hfill \\ & =& f\left(2\mathrm{ten}+\mathrm{ten}\right)\hfill \\ & =& \frac{1}{2}\left(2x+10\right)-5\hfill \\ & =& \mathrm{ten}+5-\mathrm{five}\hfill \\ & =& x\text{✓}\hfill \end{array}$

$\begin{array}{ccc}\hfill \left(k○f\right)\left(x\right)& =& g\left(f\left(\mathrm{10}\right)\right)\hfill \\ & =& g\left(\frac{\mathrm{ane}}{2}\mathrm{10}-5\right)\hfill \\ & =& 2\left(\frac{\mathrm{ane}}{\mathrm{two}}x-5\right)+10\hfill \\ & =& x-\mathrm{x}+\mathrm{ten}\hfill \\ & =& \mathrm{10}\text{✓}\hfill \end{array}$

Answer: Both $\left(f○\mathrm{one\; thousand}\right)\left(x\right)=\left(g○f\right)\left(\mathrm{10}\right)=\mathrm{10}$ ; therefore, they are inverses.

Next we explore the geometry associated with inverse functions. The graphs of both functions in the previous example are provided on the aforementioned ready of axes beneath.

Note that there is symmetry about the line $y=x$ ; the graphs of f and grand are mirror images most this line. Also discover that the point (20, v) is on the graph of f and that (5, twenty) is on the graph of g. Both of these observations are true in full general and nosotros take the following properties of changed functions:

1. The graphs of inverse functions are symmetric about the line $y=x.$
2. If $\left(a,b\right)$ is on the graph of a function, then $\left(b,a\right)$ is on the graph of its changed.

Furthermore, if thousand is the inverse of f nosotros use the notation $\mathrm{chiliad}=f^{-1}.$ Here $f^{-1}$ is read, "f inverse," and should not be confused with negative exponents. In other words, $f^{-1}\left(x\right)\ne \frac{1}{f\left(\mathrm{ten}\right)}$ and we have,

$$\begin{array}{cccc}\hfill \left(f○f^{-1}\right)\left(\mathrm{ten}\right)& =& f\left(f^{-\mathrm{i}}\left(\mathrm{ten}\right)\right)=x\hfill & \text{and}\hfill \\ \hfill \left(f^{-1}○f\right)\left(x\right)& =& f^{-1}\left(f\left(\mathrm{ten}\right)\right)=\mathrm{10}\hfill & \end{array}$$

Example 5

Verify algebraically that the functions defined past $f\left(x\right)=\frac{1}{x}-2$ and $f^{-1}\left(\mathrm{10}\right)=\frac{\mathrm{i}}{x+\mathrm{ii}}$ are inverses.

Solution:

Compose the functions both ways to verify that the result is x.

$\begin{array}{ccc}\hfill \left(f○f^{-\mathrm{i}}\right)\left(\mathrm{10}\right)& =& f\left(f^{-\mathrm{ane}}\left(x\right)\right)\hfill \\ & =& f\left(\frac{\mathrm{one}}{\mathrm{ten}+\mathrm{two}}\right)\hfill \\ & =& \frac{1}{\left(\frac{1}{\mathrm{ten}+\mathrm{two}}\right)}-2\hfill \\ & =& \frac{x+2}{1}-\mathrm{two}\hfill \\ & =& \mathrm{10}+\mathrm{ii}-\mathrm{two}\hfill \\ & =& x\text{✓}\hfill \end{array}$

$\begin{array}{ccc}\hfill \left(f^{-1}○f\right)\left(x\right)& =& f^{-1}\left(f\left(x\right)\right)\hfill \\ & =& f^{-1}\left(\frac{1}{\mathrm{10}}-2\right)\hfill \\ & =& \frac{1}{\left(\frac{1}{x}-2\right)+\mathrm{two}}\hfill \\ & =& \frac{1}{\frac{\mathrm{ane}}{x}}\hfill \\ & =& \mathrm{10}\text{✓}\hfill \end{array}$

Answer: Since $\left(f○f^{-\mathrm{one}}\right)\left(\mathrm{10}\right)=\left(f^{-1}○f\right)\left(x\right)=x$ they are inverses.

Call back that a function is a relation where each element in the domain corresponds to exactly one chemical element in the range. Nosotros use the vertical line test to determine if a graph represents a office or not. Functions can exist further classified using an inverse relationship. Ane-to-one functionsFunctions where each value in the range corresponds to exactly 1 value in the domain. are functions where each value in the range corresponds to exactly 1 element in the domain. The horizontal line examinationIf a horizontal line intersects the graph of a office more than in one case, and then it is not one-to-one. is used to determine whether or not a graph represents a one-to-one function. If a horizontal line intersects a graph more than than in one case, then it does not represent a one-to-ane function.

The horizontal line represents a value in the range and the number of intersections with the graph represents the number of values it corresponds to in the domain. The role defined by $f\left(x\right)=x^3$ is one-to-one and the role defined by $f\left(x\right)=\left|x\right|$ is not. Determining whether or not a office is ane-to-one is important because a function has an changed if and merely if it is one-to-i. In other words, a function has an changed if it passes the horizontal line test.

Note: In this text, when nosotros say "a function has an inverse," nosotros mean that in that location is another part, $f^{-1}$ , such that $\left(f○f^{-\mathrm{i}}\right)\left(x\right)=\left(f^{-\mathrm{i}}○f\right)\left(x\right)=x.$

Example 6

Make up one's mind whether or not the given function is one-to-one.

Solution:

Answer: The given office passes the horizontal line test and thus is 1-to-one.

In fact, whatever linear function of the form $f\left(x\right)=mx+b$ where $\mathrm{thousand}\ne 0$ , is one-to-one and thus has an inverse. The steps for finding the inverse of a i-to-one role are outlined in the following example.

Example vii

Discover the changed of the office defined past $f\left(x\right)=\frac{3}{2}\mathrm{ten}-\mathrm{five}.$

Solution:

Earlier beginning this process, you should verify that the function is one-to-ane. In this case, we have a linear function where $\mathrm{chiliad}\ne 0$ and thus it is ane-to-one.

• Step one: Replace the function notation $f\left(\mathrm{ten}\right)$ with y.

  $$\begin{array}{ccc}\hfill f\left(x\right)& =& \frac{3}{\mathrm{two}}\mathrm{10}-5\hfill \\ \hfill y& =& \frac{3}{\mathrm{ii}}x-5\hfill \end{array}$$

• Step 2: Interchange x and y. Nosotros employ the fact that if $\left(x,y\right)$ is a betoken on the graph of a office, then $\left(y,x\right)$ is a point on the graph of its changed.

  $$\mathrm{ten}=\frac{\mathrm{three}}{2}y-\mathrm{v}$$

• Pace 3: Solve for y.

  $$\begin{array}{ccc}\hfill x& =& \frac{3}{2}y-5\hfill \\ \hfill \mathrm{ten}+5& =& \frac{3}{\mathrm{two}}y\hfill \\ \hfill \frac{2}{3}\cdot \left(\mathrm{10}+5\right)& =& \frac{2}{3}\cdot \frac{\mathrm{iii}}{2}y\hfill \\ \hfill \frac{2}{3}x+\frac{\mathrm{ten}}{3}& =& y\hfill \end{array}$$

• Step 4: The resulting function is the inverse of f. Replace y with $f^{-1}\left(x\right).$

  $$f^{-1}\left(x\right)=\frac{\mathrm{ii}}{\mathrm{iii}}x+\frac{\mathrm{x}}{3}$$

• Pace 5: Cheque.

  $\begin{array}{c}\left(f○f^{-\mathrm{i}}\right)\left(\mathrm{10}\right)\hfill \\ =f\left(f^{-\mathrm{ane}}\left(\mathrm{10}\right)\right)\hfill \\ =f\left(\frac{\mathrm{two}}{\mathrm{three}}x+\frac{\mathrm{x}}{\mathrm{iii}}\right)\hfill \\ =\frac{\mathrm{three}}{\mathrm{ii}}\left(\frac{\mathrm{ii}}{3}x+\frac{10}{\mathrm{three}}\right)-5\hfill \\ =x+5-5\hfill \\ =x\text{✓}\hfill \end{array}$

  $\begin{array}{c}\left(f^{-1}○f\right)\left(x\right)\hfill \\ =f^{-1}\left(f\left(x\right)\right)\hfill \\ =f^{-\mathrm{i}}\left(\frac{3}{2}\mathrm{10}-5\right)\hfill \\ =\frac{\mathrm{ii}}{3}\left(\frac{3}{2}\mathrm{ten}-5\right)+\frac{10}{\mathrm{iii}}\hfill \\ =\mathrm{ten}-\frac{10}{3}+\frac{10}{\mathrm{iii}}\hfill \\ =x\text{✓}\hfill \end{array}$

Answer: $f^{-\mathrm{ane}}\left(\mathrm{ten}\right)=\frac{2}{3}\mathrm{10}+\frac{\mathrm{x}}{3}$

If a function is non one-to-ane, it is often the case that we can restrict the domain in such a manner that the resulting graph is one-to-one. For example, consider the squaring function shifted upward one unit, $g\left(x\right)=x^2+\mathrm{ane}.$ Note that it does not pass the horizontal line exam and thus is non 1-to-i. However, if nosotros restrict the domain to nonnegative values, $x\ge 0$ , and so the graph does pass the horizontal line test.

On the restricted domain, g is i-to-1 and nosotros can find its inverse.

Case eight

Find the inverse of the office defined by $m\left(\mathrm{ten}\right)=x^{\mathrm{ii}}+1$ where $x\ge 0.$

Solution:

Begin by replacing the function annotation $\mathrm{one\; thousand}\left(\mathrm{ten}\right)$ with y.

$$\begin{array}{ccc}\hfill g\left(\mathrm{10}\right)& =& {\mathrm{ten}}^2+\mathrm{one}\hfill \\ \hfill y& =& x^2+\mathrm{ane}\hfill & \mathrm{due\; west}h\mathrm{east}re\hfill & x\ge 0\hfill \end{array}$$

Interchange x and y.

$$\begin{array}{ccc}x=y^2+1& where& y\ge 0\end{array}$$

Solve for y.

$$\begin{array}{ccc}\hfill x& =& y^2+1\hfill \\ \hfill x-1& =& y^2\hfill \\ \hfill \pm \sqrt{\mathrm{ten}-1}& =& y\hfill \end{array}$$

Since $y\ge 0$ nosotros only consider the positive effect.

$$\begin{array}{ccc}\hfill y& =& \sqrt{x-\mathrm{one}}\hfill \\ \hfill k^{-\mathrm{one}}\left(\mathrm{10}\right)& =& \sqrt{\mathrm{10}-1}\hfill \end{array}$$

Answer: $g^{-1}\left(\mathrm{10}\right)=\sqrt{x-1}.$ The bank check is left to the reader.

The graphs in the previous instance are shown on the aforementioned set of axes beneath. Take annotation of the symmetry most the line $y=x.$

Instance 9

Find the inverse of the role divers by $f\left(x\right)=\frac{2x+1}{x-3}.$

Solution:

Use a graphing utility to verify that this function is ane-to-1. Begin by replacing the function notation $f\left(\mathrm{ten}\right)$ with y.

$$\begin{array}{ccc}\hfill f\left(x\right)& =& \frac{2x+\mathrm{ane}}{x-3}\hfill \\ \hfill y& =& \frac{\mathrm{ii}\mathrm{10}+\mathrm{i}}{x-3}\hfill \end{array}$$

Interchange x and y.

$$x=\frac{\mathrm{ii}y+\mathrm{i}}{y-3}$$

Solve for y.

$$\begin{array}{ccc}\hfill x& =& \frac{\mathrm{ii}y+1}{y-3}\hfill \\ \hfill x\left(y-3\right)& =& \mathrm{ii}y+1\hfill \\ \hfill xy-3x& =& 2y+\mathrm{i}\hfill \end{array}$$

Obtain all terms with the variable y on one side of the equation and everything else on the other. This will enable us to treat y equally a GCF.

$$\begin{array}{ccc}\hfill xy-3x& =& 2y+1\hfill \\ \hfill \mathrm{10}y-2y& =& 3x+\mathrm{ane}\hfill \\ \hfill y\left(\mathrm{ten}-\mathrm{ii}\right)& =& \mathrm{three}x+1\hfill \\ \hfill y& =& \frac{3\mathrm{ten}+\mathrm{one}}{\mathrm{ten}-2}\hfill \end{array}$$

Answer: $f^{-1}\left(x\right)=\frac{\mathrm{iii}x+\mathrm{one}}{x-2}.$ The check is left to the reader.

Try this! Find the inverse of $f\left(x\right)=\sqrt[3]{\mathrm{10}+1}-\mathrm{three}.$

Reply: $f^{-1}\left(x\right)={\left(x+3\right)}^3-1$

Cardinal Takeaways

• The composition operator ( $○$ ) indicates that we should substitute one function into some other. In other words, $\left(f○\mathrm{chiliad}\right)\left(\mathrm{10}\right)=f\left(m\left(\mathrm{10}\right)\right)$ indicates that we substitute $g\left(x\right)$ into $f\left(x\right).$
• If ii functions are inverses, then each will contrary the effect of the other. Using note, $\left(f○g\right)\left(\mathrm{ten}\right)=f\left(g\left(x\right)\right)=x$ and $\left(\mathrm{grand}○f\right)\left(x\right)=g\left(f\left(\mathrm{10}\right)\right)=x.$
• Changed functions have special notation. If $\mathrm{thousand}$ is the changed of $f$ , then nosotros tin write $\mathrm{yard}\left(x\right)=f^{-\mathrm{one}}\left(x\right).$ This notation is often dislocated with negative exponents and does not equal one divided past $f\left(\mathrm{ten}\right).$
• The graphs of inverses are symmetric nigh the line $y=\mathrm{ten}.$ If $\left(a,b\right)$ is a point on the graph of a function, and so $\left(b,a\right)$ is a indicate on the graph of its inverse.
• If each point in the range of a office corresponds to exactly ane value in the domain then the function is one-to-ane. Use the horizontal line test to decide whether or non a function is one-to-one.
• A one-to-one function has an changed, which tin can ofttimes exist found past interchanging x and y, and solving for y. This new function is the changed of the original function.

Topic Exercises

Part A: Composition of Functions

Given the functions defined by f and g find $\left(f○\mathrm{chiliad}\right)\left(x\right)$ and $\left(m○f\right)\left(x\right).$

1. $f\left(x\right)=\mathrm{iv}x-1$ , $g\left(x\right)=3x$

2. $f\left(\mathrm{10}\right)=-\mathrm{ii}x+5$ , $g\left(x\right)=2\mathrm{10}$

3. $f\left(x\right)=3\mathrm{ten}-5$ , $g\left(x\right)=x-4$

4. $f\left(x\right)=5\mathrm{ten}+1$ , $\mathrm{grand}\left(x\right)=2x-3$

5. $f\left(\mathrm{ten}\right)=x^2-x+1$ , $g\left(x\right)=2x-\mathrm{i}$

6. $f\left(x\right)={\mathrm{10}}^{\mathrm{ii}}-3\mathrm{10}-2$ , $\mathrm{yard}\left(x\right)=x-2$

7. $f\left(x\right)=x^{\mathrm{two}}+3$ , $\mathrm{grand}\left(x\right)=x^{\mathrm{two}}-\mathrm{five}$

8. $f\left(\mathrm{10}\right)=\mathrm{ii}{x}^{\mathrm{ii}}$ , $g\left(\mathrm{10}\right)={\mathrm{10}}^2-x$

9. $f\left(x\right)=\mathrm{viii}{x}^3+\mathrm{five}$ , $k\left(\mathrm{ten}\right)=\sqrt[3]{\mathrm{ten}-\mathrm{v}}$

10. $f\left(\mathrm{ten}\right)=27x^{\mathrm{three}}-1$ , $g\left(\mathrm{10}\right)=\sqrt[3]{x+1}$

11. $f\left(\mathrm{10}\right)=\frac{\mathrm{one}}{x+5}$ , $g\left(x\right)=\frac{1}{\mathrm{ten}}$

12. $f\left(x\right)=\frac{1}{x}-3$ , $\mathrm{one\; thousand}\left(\mathrm{10}\right)=\frac{3}{x+3}$

13. $f\left(\mathrm{10}\right)=5\sqrt{\mathrm{ten}}$ , $g\left(x\right)=\mathrm{iii}x-2$

14. $f\left(\mathrm{ten}\right)=\sqrt{2x}$ , $g\left(x\right)=4x+\mathrm{one}$

15. $f\left(\mathrm{ten}\right)=\frac{1}{\mathrm{ii}x}$ , $\mathrm{yard}\left(x\right)=x^2+8$

16. $f\left(x\right)=\mathrm{two}x-\mathrm{i}$ , $\mathrm{yard}\left(x\right)=\frac{\mathrm{ane}}{\mathrm{10}+1}$

17. $f\left(x\right)=\frac{1-\mathrm{ten}}{2x}$ , $g\left(x\right)=\frac{\mathrm{one}}{2x+1}$

18. $f\left(\mathrm{10}\right)=\frac{\mathrm{two}\mathrm{10}}{x+\mathrm{one}}$ , $\mathrm{one\; thousand}\left(x\right)=\frac{\mathrm{ten}+1}{\mathrm{10}}$

Given the functions defined by $f\left(x\right)=\mathrm{iii}{x}^{\mathrm{two}}-2$ , $\mathrm{one\; thousand}\left(x\right)=\mathrm{v}x+\mathrm{ane}$ , and $h\left(x\right)=\sqrt{x}$ , calculate the following.

19. $\left(f○g\right)\left(2\right)$

20. $\left(g○f\right)\left(-1\right)$

21. $\left(m○f\right)\left(0\right)$

22. $\left(f○g\right)\left(0\right)$

23. $\left(f○h\right)\left(3\right)$

24. $\left(\mathrm{one\; thousand}○h\right)\left(\mathrm{xvi}\right)$

25. $\left(h○g\right)\left(\frac{3}{5}\right)$

26. $\left(h○f\right)\left(-3\right)$

Given the functions defined by $f\left(x\right)=\sqrt[3]{x+3}$ , $\mathrm{chiliad}\left(x\right)=\mathrm{eight}{x}^3-3$ , and $h\left(\mathrm{10}\right)=2x-\mathrm{i}$ , calculate the post-obit.

27. $\left(f○g\right)\left(\mathrm{i}\right)$

28. $\left(\mathrm{thou}○f\right)\left(-2\right)$

29. $\left(g○f\right)\left(0\right)$

30. $\left(f○k\right)\left(-2\right)$

31. $\left(f○h\right)\left(-\mathrm{ane}\right)$

32. $$\left(h○g\right)\left(-\frac{\mathrm{i}}{2}\right)$$

33. $\left(h○f\right)\left(24\right)$

34. $\left(g○h\right)\left(0\right)$

Given the function, decide $\left(f○f\right)\left(x\right).$

35. $f\left(\mathrm{10}\right)=\mathrm{iii}x-1$

36. $f\left(\mathrm{ten}\right)=\frac{2}{5}x+1$

37. $f\left(x\right)=x^2+5$

38. $f\left(\mathrm{10}\right)={\mathrm{ten}}^2-x+\mathrm{half-dozen}$

39. $f\left(x\right)={\mathrm{ten}}^{\mathrm{three}}+2$

40. $f\left(\mathrm{ten}\right)=x^3-x$

41. $f\left(\mathrm{ten}\right)=\frac{1}{\mathrm{ten}+1}$

42. $f\left(x\right)=\frac{x+1}{2x}$

Part B: Inverse Functions

Are the given functions one-to-one? Explain.

43. 

44. 

45. 

46. 

47. $f\left(x\right)=x+\mathrm{one}$

48. $\mathrm{grand}\left(x\right)={\mathrm{10}}^2+1$

49. $h\left(\mathrm{10}\right)=\left|\mathrm{10}\right|+1$

50. $r\left(x\right)={\mathrm{10}}^{\mathrm{iii}}+\mathrm{ane}$

51. $f\left(x\right)=\sqrt{\mathrm{ten}+1}$

52. $\mathrm{thou}\left(x\right)=3$

Given the graph of a one-to-one function, graph its inverse.

53. 

54. 

55. 

56. 

57. 

58. 

Verify algebraically that the ii given functions are inverses. In other words, show that $\left(f○f^{-\mathrm{i}}\right)\left(x\right)=x$ and $\left(f^{-\mathrm{ane}}○f\right)\left(x\right)=x.$

59. $f\left(x\right)=3x-\mathrm{iv}$ , $f^{-1}\left(x\right)=\frac{\mathrm{10}+4}{3}$

60. $f\left(\mathrm{ten}\right)=-\mathrm{five}x+1$ , $f^{-1}\left(\mathrm{ten}\right)=\frac{1-\mathrm{ten}}{\mathrm{v}}$

61. $f\left(x\right)=-\frac{2}{\mathrm{three}}\mathrm{ten}+1$ , $f^{-1}\left(\mathrm{10}\right)=-\frac{\mathrm{iii}}{\mathrm{ii}}\mathrm{10}+\frac{\mathrm{iii}}{2}$

62. $f\left(x\right)=4x-\frac{\mathrm{i}}{3}$ , $f^{-1}\left(x\right)=\frac{\mathrm{i}}{4}x+\frac{1}{12}$

63. $f\left(x\right)=\sqrt{x-8}$ , $f^{-1}\left(x\right)={\mathrm{ten}}^{\mathrm{ii}}+\mathrm{viii}$ , $x\ge 0$

64. $f\left(x\right)=\sqrt[3]{6x}-3$ , $f^{-1}\left(x\right)=\frac{{\left(x+3\right)}^{\mathrm{iii}}}{6}$

65. $f\left(x\right)=\frac{x}{x+1}$ , $f^{-\mathrm{ane}}\left(x\right)=\frac{x}{1-x}$

66. $f\left(\mathrm{10}\right)=\frac{x-3}{3x}$ , $f^{-1}\left(\mathrm{10}\right)=\frac{\mathrm{iii}}{1-3\mathrm{10}}$

67. $f\left(x\right)=2{\left(\mathrm{10}-1\right)}^3+\mathrm{iii}$ , $f^{-\mathrm{ane}}\left(\mathrm{10}\right)=1+\sqrt[3]{\frac{x-3}{\mathrm{ii}}}$

68. $f\left(x\right)=\sqrt[\mathrm{iii}]{5\mathrm{ten}-\mathrm{one}}+4$ , $f^{-1}\left(\mathrm{10}\right)=\frac{{\left(\mathrm{10}-4\right)}^3+1}{5}$

Find the inverses of the following functions.

69. $f\left(\mathrm{10}\right)=\mathrm{v}x$

70. $f\left(\mathrm{10}\right)=\frac{1}{2}x$

71. $f\left(x\right)=\mathrm{ii}\mathrm{ten}+5$

72. $f\left(x\right)=-\mathrm{four}\mathrm{ten}+\mathrm{three}$

73. $f\left(x\right)=-\frac{2}{3}x+\frac{\mathrm{ane}}{3}$

74. $f\left(\mathrm{10}\right)=-\frac{1}{\mathrm{ii}}\mathrm{10}+\frac{3}{4}$

75. $m\left(x\right)={\mathrm{10}}^2+\mathrm{v}$ , $\mathrm{10}\ge 0$

76. $g\left(x\right)=x^2-7$ , $x\ge 0$

77. $f\left(\mathrm{ten}\right)={\left(\mathrm{ten}-5\right)}^2$ , $\mathrm{ten}\ge 5$

78. $f\left(\mathrm{ten}\right)={\left(x+\mathrm{i}\right)}^{\mathrm{ii}}$ , $x\ge -1$

79. $h\left(\mathrm{10}\right)=\mathrm{three}{x}^3+\mathrm{v}$

80. $h\left(x\right)=\mathrm{two}{\mathrm{10}}^{\mathrm{three}}-\mathrm{i}$

81. $f\left(x\right)={\left(2x-3\right)}^3$

82. $f\left(x\right)={\left(x+\mathrm{iv}\right)}^3-1$

83. $$g\left(x\right)=\frac{\mathrm{ii}}{{\mathrm{ten}}^{\mathrm{three}}+1}$$

84. $$\mathrm{1000}\left(x\right)=\frac{1}{x^{\mathrm{three}}}-\mathrm{ii}$$

85. $$f\left(x\right)=\frac{\mathrm{v}}{x+1}$$

86. $$f\left(x\right)=\frac{1}{\mathrm{ii}x-9}$$

87. $$f\left(\mathrm{10}\right)=\frac{x+5}{\mathrm{10}-5}$$

88. $$f\left(x\right)=\frac{\mathrm{three}x-\mathrm{four}}{\mathrm{ii}x-1}$$

89. $$h\left(x\right)=\frac{x-5}{10x}$$

90. $$h\left(\mathrm{10}\right)=\frac{\mathrm{ix}\mathrm{10}+\mathrm{one}}{\mathrm{iii}\mathrm{ten}}$$

91. $g\left(x\right)=\sqrt[3]{\mathrm{v}x+2}$

92. $g\left(x\right)=\sqrt[3]{4x-3}$

93. $f\left(x\right)=\sqrt[3]{x-\mathrm{vi}}-4$

94. $f\left(\mathrm{10}\right)=2\sqrt[3]{x+\mathrm{two}}+5$

95. $h\left(\mathrm{ten}\right)=\sqrt[5]{x+\mathrm{one}}-3$

96. $h\left(\mathrm{10}\right)=\sqrt[5]{\mathrm{10}-\mathrm{eight}}+1$

97. $f\left(\mathrm{ten}\right)=mx+b$ , $m\ne 0$

98. $f\left(\mathrm{10}\right)=a{\mathrm{ten}}^2+c$ , $x\ge 0$

99. $f\left(x\right)=a{x}^{\mathrm{iii}}+d$

100. $f\left(\mathrm{10}\right)=a{\left(x-h\right)}^{\mathrm{two}}+k$ , $x\ge h$

Graph the function and its inverse on the same set of axes.

101. $f\left(x\right)=\mathrm{10}+\mathrm{two}$

102. $f\left(x\right)=\frac{2}{3}\mathrm{10}-\mathrm{four}$

103. $f\left(x\right)=-\mathrm{ii}x+2$

104. $f\left(x\right)=-\frac{1}{3}x+4$

105. $g\left(\mathrm{ten}\right)=x^{\mathrm{two}}-\mathrm{two}$ , $x\ge 0$

106. $\mathrm{thou}\left(\mathrm{10}\right)={\left(\mathrm{10}-2\right)}^{\mathrm{two}}$ , $x\ge 2$

107. $h\left(x\right)=x^{\mathrm{three}}+1$

108. $h\left(x\right)={\left(x+\mathrm{two}\right)}^3-2$

109. $f\left(x\right)=2-\sqrt{x}$

110. $f\left(\mathrm{ten}\right)=\sqrt{-x}+1$

Part C: Discussion Board

111. Is composition of functions associative? Explain.

112. Explain why $C\left(\mathrm{ten}\right)=\frac{\mathrm{v}}{9}\left(\mathrm{10}-32\right)$ and $F\left(x\right)=\frac{9}{\mathrm{five}}\mathrm{ten}+32$ define changed functions. Show it algebraically.

113. Do the graphs of all directly lines stand for one-to-one functions? Explain.

114. If the graphs of inverse functions intersect, then how can nosotros discover the betoken of intersection? Explain.

Answers

1. $\left(f○\mathrm{1000}\right)\left(\mathrm{ten}\right)=12x-1$ ; $\left(\mathrm{thou}○f\right)\left(\mathrm{10}\right)=12\mathrm{10}-\mathrm{three}$

3. $\left(f○\mathrm{thou}\right)\left(x\right)=3x-17$ ; $\left(m○f\right)\left(x\right)=3x-9$

5. $\left(f○g\right)\left(x\right)=4{\mathrm{ten}}^2-6x+3$ ; $\left(g○f\right)\left(x\right)=\mathrm{two}{x}^2-2\mathrm{10}+1$

7. $\left(f○g\right)\left(x\right)=x^4-10{\mathrm{10}}^2+28$ ; $\left(g○f\right)\left(x\right)=x^4+6{\mathrm{ten}}^{\mathrm{two}}+4$

9. $\left(f○\mathrm{1000}\right)\left(x\right)=8\mathrm{10}-35$ ; $\left(\mathrm{grand}○f\right)\left(x\right)=2x$

11. $\left(f○\mathrm{1000}\right)\left(x\right)=\frac{x}{5\mathrm{10}+1}$ ; $\left(\mathrm{one\; thousand}○f\right)\left(\mathrm{ten}\right)=\mathrm{ten}+5$

13. $\left(f○g\right)\left(\mathrm{10}\right)=5\sqrt{3x-2}$ ; $\left(g○f\right)\left(x\right)=\mathrm{fifteen}\sqrt{x}-2$

15. $$\left(f○k\right)\left(x\right)=\frac{1}{2x^2+16};$$ $$\left(\mathrm{yard}○f\right)\left(\mathrm{10}\right)=\frac{1+32x^2}{4{\mathrm{10}}^2}$$

17. $\left(f○g\right)\left(\mathrm{ten}\right)=\mathrm{10}$ ; $\left(\mathrm{thousand}○f\right)\left(x\right)=x$

19. 361

21. −9

23. 7

25. ii

27. 2

29. 21

31. 0

33. 5

35. $\left(f○f\right)\left(x\right)=9\mathrm{10}-4$

37. $\left(f○f\right)\left(\mathrm{ten}\right)={\mathrm{10}}^4+10x^2+30$

39. $\left(f○f\right)\left(x\right)=x^9+\mathrm{half\; dozen}{\mathrm{10}}^6+12x^3+\mathrm{x}$

41. $\left(f○f\right)\left(x\right)=\frac{x+\mathrm{one}}{x+2}$

43. No, fails the HLT

45. Yeah, passes the HLT

47. Yep, its graph passes the HLT.

49. No, its graph fails the HLT.

51. Yes, its graph passes the HLT.

53. 

55. 

57. 

59. Proof

61. Proof

63. Proof

65. Proof

67. Proof

69. $f^{-\mathrm{one}}\left(\mathrm{ten}\right)=\frac{\mathrm{ten}}{\mathrm{v}}$

71. $$f^{-\mathrm{ane}}\left(\mathrm{10}\right)=\frac{1}{2}x-\frac{5}{2}$$

73. $$f^{-1}\left(\mathrm{10}\right)=-\frac{3}{\mathrm{two}}x+\frac{1}{\mathrm{ii}}$$

75. $g^{-1}\left(\mathrm{10}\right)=\sqrt{\mathrm{ten}-\mathrm{five}}$

77. $f^{-\mathrm{one}}\left(\mathrm{10}\right)=\sqrt{x}+5$

79. $$h^{-1}\left(x\right)=\sqrt[3]{\frac{\mathrm{10}-\mathrm{v}}{3}}$$

81. $$f^{-\mathrm{i}}\left(x\right)=\frac{\sqrt[\mathrm{three}]{x}+3}{2}$$

83. $$g^{-\mathrm{one}}\left(x\right)=\sqrt[3]{\frac{\mathrm{ii}-x}{\mathrm{10}}}$$

85. $$f^{-\mathrm{i}}\left(x\right)=\frac{\mathrm{v}-x}{\mathrm{10}}$$

87. $$f^{-1}\left(x\right)=\frac{\mathrm{v}\left(x+\mathrm{one}\right)}{x-1}$$

89. $$h^{-1}\left(x\right)=-\frac{5}{10x-\mathrm{ane}}$$

91. $${\mathrm{chiliad}}^{-1}\left(\mathrm{10}\right)=\frac{x^3-2}{\mathrm{five}}$$

93. $f^{-1}\left(x\right)={\left(x+4\right)}^3+6$

95. $h^{-\mathrm{one}}\left(x\right)={\left(x+3\right)}^5-1$

97. $$f^{-1}\left(x\right)=\frac{\mathrm{10}-b}{m}$$

99. $$f^{-1}\left(\mathrm{10}\right)=\sqrt[3]{\frac{x-d}{a}}$$

101. 

103. 

105. 

107. 

109. 

111. Answer may vary

113. Reply may vary

How Do You Know the Inverse Is a Function

Source: https://saylordotorg.github.io/text_intermediate-algebra/s10-01-composition-and-inverse-functi.html

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