Study Island

Generation Date: 09/26/2012
Generated By: Afiya Thomas

1. Consider the two functions shown below.

$\begin{eqnarray} f(x) &=& 5x\,-\,3 \\ \vspace*{, \hspace*{-10}} \\ g(x) &=& \frac{1}{5}x\,+\,\frac{3}{5} \end{eqnarray}$

A. The functions $f(x)$ and $g(x)$ are inverses because $f(g(x))\,\neq\,g(f(x))$ .

B. The functions $f(x)$ and $g(x)$ are not inverses because $f(g(x))\,=\,g(f(x))\,=\,x$ .

C. The functions $f(x)$ and $g(x)$ are not inverses because $f(g(x))\,\neq\,x$ .

D. The functions $f(x)$ and $g(x)$ are inverses because $f(g(x))\,=\,g(f(x))\,=\,x$ .

2. Consider the two functions shown below.

$\begin{eqnarray} f(x) &=& 2^x\,+\,7 \\ \vspace*{, \hspace*{-10}} \\ g(x) &=& \frac{1}{2^x\,-\,7} \end{eqnarray}$

A. The functions $f(x)$ and $g(x)$ are inverses because $f(g(x))\,\neq\,x$ .

B. The functions $f(x)$ and $g(x)$ are inverses because $f(g(x))\,=\,g(f(x))\,=\,x$ .

C. The functions $f(x)$ and $g(x)$ are not inverses because $f(g(x))\,\neq\,x$ .

D. The functions $f(x)$ and $g(x)$ are not inverses because $f(g(x))\,=\,g(f(x))\,=\,x$ .

3. Consider the two functions shown below.

$\begin{eqnarray} f(x) &=& \frac{1}{7}x\,-\,4 \\ \vspace*{, \hspace*{-10}} \\ g(x) &=& 7x\,+\,28 \end{eqnarray}$

A. The functions $f(x)$ and $g(x)$ are inverses because $f(g(x))\,\neq\,x$ .

B. The functions $f(x)$ and $g(x)$ are not inverses because $f(g(x))\,\neq\,x$ .

C. The functions $f(x)$ and $g(x)$ are inverses because $f(g(x))\,=\,g(f(x))\,=\,x$ .

D. The functions $f(x)$ and $g(x)$ are not inverses because $f(g(x))\,=\,g(f(x))\,=\,x$ .

4. Consider the two functions shown below.

$\begin{eqnarray} f(x) &=& 3x\,-\,8 \\ \vspace*{, \hspace*{-10}} \\ g(x) &=& \frac{1}{3}x\,+\,8 \end{eqnarray}$

A. The functions $f(x)$ and $g(x)$ are not inverses because $f(g(x))\,=\,g(f(x))\,=\,x$ .

B. The functions $f(x)$ and $g(x)$ are not inverses because $f(g(x))\,\neq\,x$ .

C. The functions $f(x)$ and $g(x)$ are inverses because $f(g(x))\,\neq\,x$ .

D. The functions $f(x)$ and $g(x)$ are inverses because $f(g(x))\,=\,g(f(x))\,=\,x$ .

5. Consider the two functions shown below over the interval $[7\,,\,\infty)$ .

$\begin{eqnarray} f(x) &=& x^2\,-\,8x\,+\,23 \\ \vspace*{, \hspace*{-10}} \\ g(x) &=& \sqrt{x\,-\,7}\,+\,4 \end{eqnarray}$

A. The functions $f(x)$ and $g(x)$ are inverses because $f(g(x))\,=\,g(f(x))\,=\,x$ .

B. The functions $f(x)$ and $g(x)$ are not inverses because $f(g(x))\,\neq\,x$ .

C. The functions $f(x)$ and $g(x)$ are inverses because $f(g(x))\,\neq\,x$ .

D. The functions $f(x)$ and $g(x)$ are not inverses because $f(g(x))\,=\,g(f(x))\,=\,x$ .

6. Which of the following domain restrictions should be used for
the function f(x) = x² - 10x + 25 so that it will have an inverse function?

A. - < x < 6

B. 5 < x <

C. 2 < x < 8

D. 4 < x <

7. The graph of f(x) is shown below.

(Hint: A function has an inverse if it is one to one. How do we test for one to oneness?)

$f(x)\ =\ 2^{0.5x}\ -\ 3$

Which of the following shows the graph of the inverse function of f(x)?

W.	X.
Y.	Z.

A. Y

B. Z

C. X

D. W

8. The graph of f(x) is shown below.

$f(x)\ =\ \sqrt{x\ +\ 1}$

Which of the following shows the graph of the inverse function of f(x)?

W.	X.
Y.	Z.

A. W

B. Z

C. X

D. Y

9. The graph of f(x) is shown below.

$f(x)\ =\ \frac{2}{9}x\ -\ 1$

Which of the following shows the graph of the inverse function of f(x)?

W.	X.
Y.	Z.

A. W

B. Z

C. Y

D. X

10. $\text{Which of the following is the inverse of }f(x)\ =\ \frac{3}{6x\ -\ 7}?$

A. $f^{-1}(x)\ =\ \frac{1}{2x}\ -\ \frac{7}{6}$

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

C. $f^{-1}(x)\ =\ \frac{2}{x}\ +\ \frac{7}{6}$

D. $f^{-1}(x)\ =\ \frac{1}{2x}\ +\ \frac{7}{6}$

11. $\text{Which of the following is the inverse of }f(x)\ =\ 6x\ -\ 4?$

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

B. $f^{-1}(x)\ =\ \frac{x\ +\ 4}{6}$

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

D. $f^{-1}(x)\ =\ -\ \frac{x\ +\ 4}{6}$

12. $\text{Which of the following is the inverse of }f(x)\ =\ -1\ +\ \frac{5}{x}?$

A. $f^{-1}(x)\ =\ \frac{5}{x\ +\ 1}$

B. $f^{-1}(x)\ =\ \frac{1}{x\ +\ 5}$

C. $f^{-1}(x)\ =\ \frac{5}{x\ -\ 1}$

D. $f^{-1}(x)\ =\ -\ \frac{5}{x\ +\ 1}$

13. $\text{Which of the following is the inverse of }f(x)\ =\ (x\ -\ 3)^{3}\ +\ 8?$

A. $f^{-1}(x)\ =\ \sqrt[3]{x\ -\ 8}\ +\ 3$

B. $f^{-1}(x)\ =\ \sqrt[3]{x\ +\ 8}\ -\ 3$

C. $f^{-1}(x)\ =\ \sqrt[3]{x\ +\ 3}\ -\ 8$

D. $f^{-1}(x)\ =\ \sqrt[3]{x\ -\ 3}\ +\ 8$

14. $\text{Which of the following is the inverse of }f(x)\ =\ \sqrt{x\ -\ 4}?$

A. $f^{-1}(x)\ =\ x^{2}\ -\ 4$

B. $f^{-1}(x)\ =\ (x\ -\ 4)^{2}$

C. $f^{-1}(x)\ =\ x^{2}\ +\ 4$

D. $f^{-1}(x)\ =\ (x\ +\ 4)^{2}$

15. $\text{Which of the following is the inverse of }f(x)\ =\ -x^{3}\ +\ 3?$

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

B. $f^{-1}(x)\ =\ -\sqrt[3]{x\ -\ 3}$

C. $f^{-1}(x)\ =\ \sqrt[3]{-x\,+\ 3}$

D. $f^{-1}(x)\ =\ \sqrt[3]{x\ -\ 3}$

16.

$f(x)\ =\ \sqrt{x\ +\ 1}\ -\ 2$

The graph of the function above is shown below.

Which of the following statements about f(x) is true?

A. The function f(x) has an inverse because it is one-to-one.

B. The function f(x) does not have an inverse because it is not one-to-one.

C. The function f(x) does not have an inverse because its domain is not $(-\infty,\ \infty)$ .

D. The function f(x) has an inverse because it passes the vertical line test.

17.

$f(x)\ =\ (x\ +\ 2)^{2}\ -\ 3$

The graph of the function above is shown below.

Which of the following statements about f(x) is true?

A. The function f(x) has an inverse because it is one-to-one.

B. The function f(x) does not have an inverse because it is not one-to-one.

C. The function f(x) has an inverse because it passes the vertical line test.

D. The function f(x) does not have an inverse because its domain is $(-\infty,\ \infty)$ .

18.

$f(x)\ =\ \left(\frac{5}{4}\right)^{x}\ +\ 1$

The graph of the function above is shown below.

(Hint: A function has an inverse if it is one to one. How do we test for one to oneness?)

Which of the following statements about f(x) is true?

A. The function f(x) has an inverse because it passes the horizontal line test.

B. The function f(x) does not have an inverse because it never crosses the x-axis.

C. The function f(x) does not have an inverse because it does not pass the horizontal line test.

D. The function f(x) has an inverse because it passes the vertical line test.

Answers

1. D
2. C
3. C
4. B
5. A
6. B
7. D
8. A
9. C
10. D
11. B
12. A
13. A
14. C
15. C
16. A
17. B
18. A

Explanations

1. Two functions, $f(x)$ and $g(x)$ , will be inverses of each other if $f(g(x))\,=\,g(f(x))\,=\,x$ . To determine if the given functions are inverses of each other, find if the compositions both equal $x$ .

$\begin{eqnarray} f(g(x)) &=& 5\left(\frac{1}{5}x\,+\,\frac{3}{5}\right)\,-\,3 \\ \vspace*{, \hspace*{-10}} \\ &=& x\,+\,3\,-\,3 \\ \vspace*{, \hspace*{-10}} \\ &=& x \end{eqnarray}$

$\begin{eqnarray} g(f(x)) &=& \frac{1}{5}(5x\,-\,3)\,+\,\frac{3}{5} \\ \vspace*{, \hspace*{-10}} \\ &=& x\,-\,\frac{3}{5}\,+\,\frac{3}{5} \\ \vspace*{, \hspace*{-10}} \\ &=& x \end{eqnarray}$

Since $f(g(x))\,=\,g(f(x))\,=\,x$ , $f(x)$ and $g(x)$ are inverses.

2. Two functions, $f(x)$ and $g(x)$ , will be inverses of each other if $f(g(x))\,=\,g(f(x))\,=\,x$ . To determine if the given functions are inverses of each other, find if the compositions both equal $x$ .

$\begin{eqnarray} f(g(x)) &=& \Large{2^{\left(\frac{1}{2^x\,-\,7}\right)}}\,+\,7 \\ \vspace*{, \hspace*{-10}} \\ &\neq& x \end{eqnarray}$

Since $f(g(x))\,\neq\,x$ , $f(x)$ and $g(x)$ are not inverses.

3. Two functions, $f(x)$ and $g(x)$ , will be inverses of each other if $f(g(x))\,=\,g(f(x))\,=\,x$ . To determine if the given functions are inverses of each other, find if the compositions both equal $x$ .

$\begin{eqnarray} f(g(x)) &=& \frac{1}{7}(7x\,+\,28)\,-\,4 \\ \vspace*{, \hspace*{-10}} \\ &=& x\,+\,4\,-\,4 \\ \vspace*{, \hspace*{-10}} \\ &=& x \\ \vspace*{, \hspace*{-10}} \\ \vspace*{, \hspace*{-10}} \\ g(f(x)) &=& 7\left(\frac{1}{7}x\,-\,4\right)\,+\,28 \\ \vspace*{, \hspace*{-10}} \\ &=& x\,-\,28\,+\,28 \\ \vspace*{, \hspace*{-10}} \\ &=& x \end{eqnarray}$

Since $f(g(x))\,=\,g(f(x))\,=\,x$ , $f(x)$ and $g(x)$ are inverses.

4. Two functions, $f(x)$ and $g(x)$ , will be inverses of each other if $f(g(x))\,=\,g(f(x))\,=\,x$ . To determine if the given functions are inverses of each other, find if the compositions both equal $x$ .

$\begin{eqnarray} f(g(x)) &=& 3\left(\frac{1}{3}x\,+\,8\right)\,-\,8 \\ \vspace*{, \hspace*{-10}} \\ &=& x\,+\,24\,-\,8 \\ \vspace*{, \hspace*{-10}} \\ &=& x\,+\,16 \\ \vspace*{, \hspace*{-10}} \\ &\neq& x \end{eqnarray}$

Since $f(g(x))\,\neq\,x$ , $f(x)$ and $g(x)$ are not inverses.

5. Two functions, $f(x)$ and $g(x)$ , will be inverses of each other if $f(g(x))\,=\,g(f(x))\,=\,x$ . To determine if the given functions are inverses of each other, find if the compositions both equal $x$ .

$\begin{eqnarray} f(g(x)) &=& \left(\sqrt{x\,-\,7}\,+\,4\right)^2\,-\,8\left(\sqrt{x\,-\,7}\,+\,4\right)\,+\,23 \\ \vspace*{, \hspace*{-10}} \\ &=& x\,-\,7\,+\,8\sqrt{x\,-\,7}\,+\,16\,-\,8\sqrt{x\,-\,7}\,-\,32\,+\,23 \\ \vspace*{, \hspace*{-10}} \\ &=& x \\ \vspace*{, \hspace*{-10}} \\ \vspace*{, \hspace*{-10}} \\ g(f(x)) &=& \sqrt{\left(x^2\,-\,8x\,+\,23\right)\,-\,7}\,+\,4 \\ \vspace*{, \hspace*{-10}} \\ &=& \sqrt{(x\,-\,4)^2}\,+\,4 \\ \vspace*{, \hspace*{-10}} \\ &=& x\,-\,4\,+\,4 \\ \vspace*{, \hspace*{-10}} \\ &=& x \end{eqnarray}$

Since $f(g(x))\,=\,g(f(x))\,=\,x$ , $f(x)$ and $g(x)$ are inverses.

6. In order for a function to have a function inverse, it must be
one-to-one. This means that it must pass the horizontal line test. When a function does not pass the horizontal line test, its domain can be restricted so that it will be one-to-one and will therefore have a function inverse.

The graph of the function given in the question is a parabola which is not one-to-one. However, if its domain were restricted so that it would pass the horizontal line test, it would then have a function inverse. The best way to restrict the domain would be to include the x-value for the vertex of the parabola and all values greater than it. So, find the
x-value for the vertex of the parabola using the formula x = -^b/_2a.

x = -^-10/₂₍₁₎ = 5

Therefore, the best way to restrict the domain is 5 < x < .

7. The graph of the inverse of a function can be found by reflecting the graph of the function about the line y = x.

The graph above shows that W is the correct answer.

8. The graph of the inverse of a function can be found by reflecting the graph of the function about the line y = x.

The graph above shows that W is the correct answer.

9. The graph of the inverse of a function can be found by reflecting the graph of the function about the line y = x.

The graph above shows that Y is the correct answer.

10. $\text{To find the inverse of a function, first replace }f(x)\text{ with }y.\\ \vspace*{a \hspace*{-10}}\\ \begin{eqnarray} &\hspace{500}&\\ &y\ =\ \frac{3}{6x\ -\ 7}&\end{eqnarray}\\ \vspace*{9 \hspace*{-10}}\\ \text{Next, switch all the }x\text{'s and }y\text{'s, and then solve for }y.\\ \vspace*{a \hspace*{-10}}\\$
$\begin{eqnarray} \hspace{250}& &\\ x\ &=&\ \frac{3}{6y\ -\ 7}\\ \vspace*{a \hspace*{-10}}\\ (x)(6y\ -\ 7)\ &=&\ 3\\ \vspace*{a \hspace*{-10}}\\ 6y\ -\ 7\ &=&\ \frac{3}{x}\\ \vspace*{a \hspace*{-10}}\\ 6y\ &=&\ \frac{3}{x}\ +\ 7\\ \vspace*{a \hspace*{-10}}\\ y\ &=&\ \frac{3}{6x}\ +\ \frac{7}{6}\\ \vspace*{a \hspace*{-10}}\\ y\ &=&\ \frac{1}{2x}\ +\ \frac{7}{6}\end{eqnarray}\\ \vspace*{9 \hspace*{-10}}\\$
$\text{Finally, replace }y\text{ with }f^{-1}(x).\\ \vspace*{9 \hspace*{-10}}\\ \begin{eqnarray} &\hspace{500}&\\ &f^{-1}(x)\ =\ \frac{1}{2x}\ +\ \frac{7}{6}&\end{eqnarray}$

11. $\text{To find the inverse of a function, first replace }f(x)\text{ with }y.\\ \vspace*{a \hspace*{-10}}\\ \begin{eqnarray} &\hspace{500}&\\ &y\ =\ 6x\ -\ 4&\end{eqnarray}\\ \vspace*{9 \hspace*{-10}}\\ \text{Next, switch all the }x\text{'s and }y\text{'s, and then solve for }y.\\ \vspace*{a \hspace*{-10}}\\ \begin{eqnarray} \hspace{250}& &\\ x\ &=&\ 6y\ -\ 4\\ \vspace*{a \hspace*{-10}}\\ x\ +\ 4\ &=&\ 6y\\ \vspace*{a \hspace*{-10}}\\ \frac{x\ +\ 4}{6}\ &=&\ y\end{eqnarray}\\ \vspace*{9 \hspace*{-10}}\\$
$\text{Finally, replace }y\text{ with }f^{-1}(x).\\ \vspace*{9 \hspace*{-10}}\\ \begin{eqnarray} &\hspace{500}&\\ &f^{-1}(x)\ =\ \frac{x\ +\ 4}{6}&\end{eqnarray}$

12. $\text{To find the inverse of a function, first replace }f(x)\text{ with }y.\\ \vspace*{a \hspace*{-10}}\\ \begin{eqnarray} &\hspace{500}&\\ &y\ =\ -1\ +\ \frac{5}{x}&\end{eqnarray}\\ \vspace*{9 \hspace*{-10}}\\ \text{Next, switch all the }x\text{'s and }y\text{'s, and then solve for }y.\\ \vspace*{a \hspace*{-10}}\\ \begin{eqnarray} \hspace{250}& &\\ x\ &=&\ -1\ +\ \frac{5}{y}\\ \vspace*{a \hspace*{-10}}\\ x\ +\ 1\ &=&\ \frac{5}{y}\\ \vspace*{a \hspace*{-10}}\\ y\ &=&\ \frac{5}{x\ +\ 1}\end{eqnarray}\\ \vspace*{9 \hspace*{-10}}\\$
$\text{Finally, replace }y\text{ with }f^{-1}(x).\\ \vspace*{9 \hspace*{-10}}\\ \begin{eqnarray} &\hspace{500}&\\ &f^{-1}(x)\ =\ \frac{5}{x\ +\ 1}\end{eqnarray}$

13. $\text{To find the inverse of a function, first replace }f(x)\text{ with }y.\\ \vspace*{a \hspace*{-10}}\\ \begin{eqnarray} &\hspace{500}&\\ &y\ =\ (x\ -\ 3)^3\ +\ 8&\end{eqnarray}\\ \vspace*{9 \hspace*{-10}}\\ \text{Next, switch all the }x\text{'s and }y\text{'s, and then solve for }y.\\ \vspace*{a \hspace*{-10}}\\ \begin{eqnarray} \hspace{250}& &\\ x\ &=&\ (y\ -\ 3)^3\ +\ 8\\ \vspace*{a \hspace*{-10}}\\ x\ -\ 8\ &=&\ (y\ -\ 3)^3\\ \vspace*{a \hspace*{-10}}\\ \sqrt[3]{x\ -\ 8}\ &=&\ y\ -\ 3\\ \vspace*{a \hspace*{-10}}\\ \sqrt[3]{x\ -\ 8}\ +\ 3\ &=&\ y\end{eqnarray}\\ \vspace*{9 \hspace*{-10}}\\$
$\text{Finally, replace }y\text{ with }f^{-1}(x).\\ \vspace*{9 \hspace*{-10}}\\ \begin{eqnarray} &\hspace{500}&\\ &f^{-1}(x)\ =\ \sqrt[3]{x\ -\ 8}\ +\ 3&\end{eqnarray}$

14. $\text{To find the inverse of a function, first replace }f(x)\text{ with }y.\\ \vspace*{a \hspace*{-10}}\\ \begin{eqnarray} &\hspace{500}&\\ &y\ =\ \sqrt{x\ -\ 4}&\end{eqnarray}\\ \vspace*{9 \hspace*{-10}}\\ \text{Next, switch all the }x\text{'s and }y\text{'s, and then solve for }y.\\ \vspace*{a \hspace*{-10}}\\ \begin{eqnarray} \hspace{250}& &\\ x\ &=&\ \sqrt{y\ -\ 4}\\ \vspace*{a \hspace*{-10}}\\ x^{2}\ &=&\ y\ -\ 4\\ \vspace*{a \hspace*{-10}}\\ x^{2}\ +\ 4\ &=&\ y\end{eqnarray}\\ \vspace*{9 \hspace*{-10}}\\$
$\text{Finally, replace }y\text{ with }f^{-1}(x).\\ \vspace*{9 \hspace*{-10}}\\ \begin{eqnarray} &\hspace{500}&\\ &f^{-1}(x)\ =\ x^{2}\ +\ 4&\end{eqnarray}$

15. $\text{To find the inverse of a function, first replace }f(x)\text{ with }y.\\ \vspace*{a \hspace*{-10}}\\ \begin{eqnarray} &\hspace{500}&\\ &y\ =\ -x^{3}\ +\ 3&\end{eqnarray}\\ \vspace*{9 \hspace*{-10}}\\ \text{Next, switch all the }x\text{'s and }y\text{'s, and then solve for }y.\\ \vspace*{a \hspace*{-10}}\\ \begin{eqnarray} \hspace{250}& &\\ x\ &=&\ -y^{3}\ +\ 3\\ \vspace*{a \hspace*{-10}}\\ x\ -\ 3\ &=&\ -y^{3}\\ \vspace*{a \hspace*{-10}}\\ -x\ +\ 3\ &=&\ y^{3}\\ \vspace*{a \hspace*{-10}}\\ \sqrt[3]{-x\ +\ 3}\ &=&\ y\end{eqnarray}\\ \vspace*{9 \hspace*{-10}}\\$
$\text{Finally, replace }y\text{ with }f^{-1}(x).\\ \vspace*{9 \hspace*{-10}}\\ \begin{eqnarray} &\hspace{500}&\\ &f^{-1}(x)\ =\ \sqrt[3]{-x\ +\ 3}&\end{eqnarray}$

16. A function must be one-to-one to have an inverse.

An easy way to check for this is by using the horizontal line test. If every horizontal line touches the graph at no more than one point, then the graph passes the horizontal line test and thus represents a function that is one-to-one.

Since the graph of the function f(x) passes the horizontal line test, it is one-to-one and therefore has an inverse.

The function f(x) has an inverse because it is one-to-one.

17. A function must be one-to-one to have an inverse.

An easy way to check for this is by using the horizontal line test. If every horizontal line touches the graph at no more than one point, then the graph passes the horizontal line test and thus represents a function that is one-to-one.

Since the graph of the function f(x) does not pass the horizontal line test, it is not one-to-one and therefore does not have an inverse.

The function f(x) does not have an inverse because it is not one-to-one.

18. A function must be one-to-one to have an inverse.

An easy way to check for this is by using the horizontal line test. If every horizontal line touches the graph at no more than one point, then the graph passes the horizontal line test and thus represents a function that is one-to-one.

Since the graph of the function f(x) passes the horizontal line test, it is one-to-one and therefore has an inverse.

The function f(x) has an inverse because it passes the horizontal line test.