Question 1

Write the slope-intercept form of the equation of the line through the given point perpendicular to the given line. point: 
$( - 4,7 )$   line: $- 5 x - 15 y = - 6$

Accepted Answer

A)  $y = \frac { 1 } { 5 } x + \frac { 39 } { 5 }$ 
B)  $y = - \frac { 1 } { 3 } x + \frac { 17 } { 3 }$ 
C)  $y = 3 x + 19$ 
D)  $y = - 5 x$ + $27$ 
E)  $y = 3 x - \frac { 5 } { 3 }$ 
A)  $y = \frac { 1 } { 5 } x + \frac { 39 } { 5 }$ 
B)  $y = - \frac { 1 } { 3 } x + \frac { 17 } { 3 }$ 
C)  $y = 3 x + 19$ 
D)  $y = - 5 x$ + $27$ 
E)  $y = 3 x - \frac { 5 } { 3 }$

Question 2

Determine the domain and range of the inverse function $f ^ { - 1 }$ of the following function f .
$f ( x ) = - | x + 6 | + 2$, where $x > - 6$

Accepted Answer

A)  Domain: $[ - 6 , \infty )$; Range: $[ 2 , \infty )$ 
B)  Domain: $( - \infty , 2 ]$; Range: $[ - 6 , \infty )$ 
C)  Domain: $[ - 6,2 ]$; Range: $[ - 6 , \infty )$ 
D)  Domain: $( - \infty , - 6 ]$; Range: $[ - 2 , \infty )$ 
E)  Domain: $( - \infty , \infty )$; Range: $( - \infty , \infty )$ 
A)  Domain: $[ - 6 , \infty )$; Range: $[ 2 , \infty )$ 
B)  Domain: $( - \infty , 2 ]$; Range: $[ - 6 , \infty )$ 
C)  Domain: $[ - 6,2 ]$; Range: $[ - 6 , \infty )$ 
D)  Domain: $( - \infty , - 6 ]$; Range: $[ - 2 , \infty )$ 
E)  Domain: $( - \infty , \infty )$; Range: $( - \infty , \infty )$

Question 3

Determine algebraically whether the following function is one-to-one. 
$| x - 5 | , \quad x \leq 5$
$| a - 5 | = | b - 5 |$

Accepted Answer

A) 
$\begin{aligned}
5 - a & = 5 - b \quad ; \text { one-to-one } \
- a & = - b \quad \
a & = b
\end{aligned}$
$| a - 5 | = | b - 5 |$ 
B) $| a - 5 | = | b - 5 |$
 $\begin{aligned} | a | - 5 & = | b | - 5 \ | a | & = | b | \quad \text { one-to-one } \ a & = b \end{aligned}$ 
C) $| a - 5 | = | b - 5 |$

$\begin{aligned}

a + 5 & = 5 - b \quad ; \text { not one-to-one } \
a & = - b
\end{aligned}$ 
D) $| a - 5 | = | b - 5 |$ 
$\begin{aligned}  \ | - 5 | - a & = | - 5 | + b ; \text { not one-to-one } \ - a & = b \end{aligned}$ 
E)  $| a - 5 | = | b - 5 |$ 
$\begin{aligned}
| - 5 | - a & = | - 5 | - b ; \text { one-to-one } \
- a & = - b \
a & = b
\end{aligned}$ 
A) 
$\begin{aligned}
5 - a & = 5 - b \quad ; \text { one-to-one } \
- a & = - b \quad \
a & = b
\end{aligned}$
$| a - 5 | = | b - 5 |$ 
B) $| a - 5 | = | b - 5 |$
 $\begin{aligned} | a | - 5 & = | b | - 5 \ | a | & = | b | \quad \text { one-to-one } \ a & = b \end{aligned}$ 
C) $| a - 5 | = | b - 5 |$

$\begin{aligned}

a + 5 & = 5 - b \quad ; \text { not one-to-one } \
a & = - b
\end{aligned}$ 
D) $| a - 5 | = | b - 5 |$ 
$\begin{aligned}  \ | - 5 | - a & = | - 5 | + b ; \text { not one-to-one } \ - a & = b \end{aligned}$ 
E)  $| a - 5 | = | b - 5 |$ 
$\begin{aligned}
| - 5 | - a & = | - 5 | - b ; \text { one-to-one } \
- a & = - b \
a & = b
\end{aligned}$

Question 4

Plot the points and find the slope of the line passing through the pair of points.
$( 1,0 ) , ( 5,3 )$

Accepted Answer

A)  slope: $\frac { 4 } { 3 }$ 
B)  slope: $- \frac { 4 } { 3 }$ 
C)  slope: $\frac { 1 } { 2 }$ 
D)  slope: $\frac { 3 } { 4 }$ 
E)  slope: $- \frac { 3 } { 4 }$ 
A)  slope: $\frac { 4 } { 3 }$ 
B)  slope: $- \frac { 4 } { 3 }$ 
C)  slope: $\frac { 1 } { 2 }$ 
D)  slope: $\frac { 3 } { 4 }$ 
E)  slope: $- \frac { 3 } { 4 }$

Question 5

The average lengths L of cellular phone calls in minutes from 1999 to 2004 are shown in the table below.   Use the regression feature of a graphing utility to find a linear model for the data. Let t Represent the year, with t = 9 corresponding to 1999. Use the model to predict the Average lengths of cellular phone calls for the year 2015. Round your answer to two Decimal places.

Accepted Answer

A)  4.37 minutes 
B)  8.74 minutes 
C)  5.37 minutes 
D)  3.37 minutes 
E)  2.19 minutes 
A)  4.37 minutes 
B)  8.74 minutes 
C)  5.37 minutes 
D)  3.37 minutes 
E)  2.19 minutes

Question 6

Graph the function and determine the interval(s) for which $f ( x ) \geq 0$.
$f ( x ) = - x ^ { 2 } + 4 x$

Accepted Answer

A)  $( - \infty , 0 ] \cup [ 4 , \infty )$ 
B)  $[ 0,4 ]$ 
C)  $( 0,4 )$ 
D)  $( - \infty , 0 ) \cup ( 4 , \infty )$ 
E)  $\{ 4 \}$ 
A)  $( - \infty , 0 ] \cup [ 4 , \infty )$ 
B)  $[ 0,4 ]$ 
C)  $( 0,4 )$ 
D)  $( - \infty , 0 ) \cup ( 4 , \infty )$ 
E)  $\{ 4 \}$

Question 7

Determine algebraically whether the following function is one-to-one. 
$f(x)=\frac{5 x^{2}}{3 x^{2}+6}, \text { where } x>0$

$\begin{aligned}
\frac{5 a^{2}}{3 a^{2}+6} &=\frac{5 b^{2}}{3 b^{2}+6} \
\frac{5 a^{2}}{3 a^{2}}+\frac{5 a^{2}}{6} &=\frac{5 b^{2}}{3 b^{2}}+\frac{5 b^{2}}{6}
\end{aligned}$

Accepted Answer

A)  
$\begin{aligned}
\frac{5}{3}+\frac{5 a^{2}}{6} &=\frac{5}{3}+\frac{5 b^{2}}{6} ;\text { not one-to-one } \
\frac{30+5 a^{2}}{18} &=\frac{30+5 b^{2}}{18} \ 
30+5 a^{2} &=30+5 b^{2} \
5 a^{2} &=5 b^{2} \
a^{2} &=b^{2} \
\pm a &=\pm b
\end{aligned}$ 
B)  
$\begin{aligned}
\frac{5 a^{2}}{3 a^{2}+6} &=\frac{5 b^{2}}{3 b^{2}+6} \
\frac{5}{3+6} &=\frac{5}{3+6} ; \text { one-to-one } \
\frac{5}{6} &=\frac{3}{6} \
a &=b
\end{aligned}$ 
C) 
$\begin{aligned}
\frac{5 a^{2}}{3 a^{2}+6} &=\frac{5 b^{2}}{3 b^{2}+6} \
\frac{5 a^{2}}{3 a^{2}} &=\frac{5 b^{2}}{3 b^{2}} ; \text { one-to-one } \
\frac{5}{3} &=\frac{5}{3} \
a &=b
\end{aligned}$ 
D) 
 $$\frac { 5 a ^ { 2 } } { 3 a ^ { 2 } + 6 } = \frac { 5 b ^ { 2 } } { 3 b ^ { 2 } + 6 }$$
$$\frac { 5 a ^ { 2 } } { 9 a ^ { 2 } } = \frac { 5 b ^ { 2 } } { 9 b ^ { 2 } }$$
$$\begin{aligned}
\frac { 5 a } { 9 } & = \frac { 5 b } { 9 } \quad ; \text { one-to-one } \
5 a & = 5 b \
a & = b
\end{aligned}$$ 
E) 
$\begin{aligned}
\frac { 5 a ^ { 2 } } { 3 a ^ { 2 } + 6 } & = \frac { 5 b ^ { 2 } } { 3 b ^ { 2 } + 6 } \ 
\frac { 5 a ^ { 2 } } { 3 a ^ { 2 } } + \frac { 5 a ^ { 2 } } { 6 } & = \frac { 5 b ^ { 2 } } { 3 b ^ { 2 } } + \frac { 5 b ^ { 2 } } { 6 }
\end{aligned}$ 
$\frac { 5 } { 3 } + \frac { 5 a ^ { 2 } } { 6 } = \frac { 5 } { 3 } + \frac { 5 b ^ { 2 } } { 6 }\ ;\text {  one-to-one }$

$\begin{array} { r l c } 
\frac { 30 + 5 a ^ { 2 } } { 18 } & = \frac { 30 + 5 b ^ { 2 } } { 18 }  \
30 + 5 a ^ { 2 } & = & 30 + 5 b ^ { 2 } \
5 a ^ { 2 } & = & 5 b ^ { 2 } \
a ^ { 2 } & = & b ^ { 2 } \
a & = & b
\end{array}$ 
A)  
$\begin{aligned}
\frac{5}{3}+\frac{5 a^{2}}{6} &=\frac{5}{3}+\frac{5 b^{2}}{6} ;\text { not one-to-one } \
\frac{30+5 a^{2}}{18} &=\frac{30+5 b^{2}}{18} \ 
30+5 a^{2} &=30+5 b^{2} \
5 a^{2} &=5 b^{2} \
a^{2} &=b^{2} \
\pm a &=\pm b
\end{aligned}$ 
B)  
$\begin{aligned}
\frac{5 a^{2}}{3 a^{2}+6} &=\frac{5 b^{2}}{3 b^{2}+6} \
\frac{5}{3+6} &=\frac{5}{3+6} ; \text { one-to-one } \
\frac{5}{6} &=\frac{3}{6} \
a &=b
\end{aligned}$ 
C) 
$\begin{aligned}
\frac{5 a^{2}}{3 a^{2}+6} &=\frac{5 b^{2}}{3 b^{2}+6} \
\frac{5 a^{2}}{3 a^{2}} &=\frac{5 b^{2}}{3 b^{2}} ; \text { one-to-one } \
\frac{5}{3} &=\frac{5}{3} \
a &=b
\end{aligned}$ 
D) 
 $$\frac { 5 a ^ { 2 } } { 3 a ^ { 2 } + 6 } = \frac { 5 b ^ { 2 } } { 3 b ^ { 2 } + 6 }$$
$$\frac { 5 a ^ { 2 } } { 9 a ^ { 2 } } = \frac { 5 b ^ { 2 } } { 9 b ^ { 2 } }$$
$$\begin{aligned}
\frac { 5 a } { 9 } & = \frac { 5 b } { 9 } \quad ; \text { one-to-one } \
5 a & = 5 b \
a & = b
\end{aligned}$$ 
E) 
$\begin{aligned}
\frac { 5 a ^ { 2 } } { 3 a ^ { 2 } + 6 } & = \frac { 5 b ^ { 2 } } { 3 b ^ { 2 } + 6 } \ 
\frac { 5 a ^ { 2 } } { 3 a ^ { 2 } } + \frac { 5 a ^ { 2 } } { 6 } & = \frac { 5 b ^ { 2 } } { 3 b ^ { 2 } } + \frac { 5 b ^ { 2 } } { 6 }
\end{aligned}$ 
$\frac { 5 } { 3 } + \frac { 5 a ^ { 2 } } { 6 } = \frac { 5 } { 3 } + \frac { 5 b ^ { 2 } } { 6 }\ ;\text {  one-to-one }$

$\begin{array} { r l c } 
\frac { 30 + 5 a ^ { 2 } } { 18 } & = \frac { 30 + 5 b ^ { 2 } } { 18 }  \
30 + 5 a ^ { 2 } & = & 30 + 5 b ^ { 2 } \
5 a ^ { 2 } & = & 5 b ^ { 2 } \
a ^ { 2 } & = & b ^ { 2 } \
a & = & b
\end{array}$

Question 8

Use the graphs of $f$ and $g$, shown below, to graph $h ( x ) = ( f + g ) ( x )$.

Accepted Answer

A)   
B)   
C)   
D)   
E)   
A)   
B)   
C)   
D)   
E)

Question 9

Decide whether the two functions shown in the graph below appear to be inverse functions of each other.

Accepted Answer

A)  no 
B)  yes 
C)  not enough information 
A)  no 
B)  yes 
C)  not enough information

Question 10

Determine the domain of $g ( x ) = \frac { 1 } { x ^ { 2 } - 49 }$

Accepted Answer

A)  $[ - 7,7 ]$ 
B)  $( - 7,0 ] \cup [ 0,7 )$ 
C)  $( - \infty , - 7 ) \cup ( - 7,7 ) \cup ( 7 , \infty )$ 
D)  $( - \infty , - 7 ] \cup [ 7 , \infty )$ 
E)  $( - \infty , \infty )$ 
A)  $[ - 7,7 ]$ 
B)  $( - 7,0 ] \cup [ 0,7 )$ 
C)  $( - \infty , - 7 ) \cup ( - 7,7 ) \cup ( 7 , \infty )$ 
D)  $( - \infty , - 7 ] \cup [ 7 , \infty )$ 
E)  $( - \infty , \infty )$

Question 11

Estimate the slope of the line.

Accepted Answer

A)  $- \frac { 1 } { 2 }$ 
B)  2 
C)  $- 2$ 
D)  $\frac { 1 } { 2 }$ 
E)  $- 3$ 
A)  $- \frac { 1 } { 2 }$ 
B)  2 
C)  $- 2$ 
D)  $\frac { 1 } { 2 }$ 
E)  $- 3$

Question 12

Given that $f ( x ) = \sqrt [ 4 ] { x - 4 }$ and $g ( x ) = x ^ { 4 } + 4$ determine the value of the following (if possible).
$( f \circ g ) ( 0 )$

Accepted Answer

A)  0 
B)  2 
C)  4 
D)  $x ^ { 4 } - 16$ 
E)  not possible 
A)  0 
B)  2 
C)  4 
D)  $x ^ { 4 } - 16$ 
E)  not possible

Question 13

Use the graph of the function to find the domain and range of $f$.

Accepted Answer

A)  domain : all real numbers
range : $( - \infty , - 2 ) \cup ( - 1 , \infty )$ 
B)  domain : all real numbers
range : all real numbers 
C)  domain: $( - \infty , - 2 ) \cup ( - 2 , \infty )$
range: $( - \infty , - 2 ) \cup ( - 1 , \infty )$ 
D)  domain: $( - \infty , - 2 ) \cup ( - 1 , \infty )$
range : $( - \infty , - 2 ) \cup ( - 2 , \infty )$ 
E)  Domain: all real numbers
Range: $( - \infty , - 2 ] \cup [ - 1 , \infty )$ 
A)  domain : all real numbers
range : $( - \infty , - 2 ) \cup ( - 1 , \infty )$ 
B)  domain : all real numbers
range : all real numbers 
C)  domain: $( - \infty , - 2 ) \cup ( - 2 , \infty )$
range: $( - \infty , - 2 ) \cup ( - 1 , \infty )$ 
D)  domain: $( - \infty , - 2 ) \cup ( - 1 , \infty )$
range : $( - \infty , - 2 ) \cup ( - 2 , \infty )$ 
E)  Domain: all real numbers
Range: $( - \infty , - 2 ] \cup [ - 1 , \infty )$

Question 14

Use the graph of $f ( x ) = x ^ { 2 }$
to write an equation for the function whose graph is shown.

Accepted Answer

A)  $f ( x ) = ( x + 1 ) ^ { 2 } - 2$ 
B)  $f ( x ) = ( x - 1 ) ^ { 2 } - 2$ 
C)  $f ( x ) = ( x + 1 ) ^ { 2 } + 2$ 
D)  $f ( x ) = \frac { 1 } { 2 } ( x - 1 ) ^ { 2 } - 2$ 
E)  $f ( x ) = \frac { 1 } { 2 } ( x + 1 ) ^ { 2 } - 2$ 
A)  $f ( x ) = ( x + 1 ) ^ { 2 } - 2$ 
B)  $f ( x ) = ( x - 1 ) ^ { 2 } - 2$ 
C)  $f ( x ) = ( x + 1 ) ^ { 2 } + 2$ 
D)  $f ( x ) = \frac { 1 } { 2 } ( x - 1 ) ^ { 2 } - 2$ 
E)  $f ( x ) = \frac { 1 } { 2 } ( x + 1 ) ^ { 2 } - 2$

Question 15

Show algebraically that the functions $f$ and $g$ shown below are inverse functions.
$f ( x ) = \frac { 2 } { 2 + x } , x \geq 0 , \quad g ( x ) = \frac { 2 - 2 x } { x } , 0 < x \leq 1$

Accepted Answer

A) $\begin{aligned}
f(g(x)) &=\frac{2}{2+\left(\frac{2-2 x}{x}\right)} \
&=\frac{2}{2+\left(\frac{1}{x}\right)} \
&=\frac{1}{\left(\frac{1}{x}\right)} \
&=1 \cdot \frac{x}{1} \
&=x
\end{aligned}$

$\begin{aligned}
g(f(x)) &=\frac{2-2\left(\frac{2}{2+x}\right)}{\left(\frac{2}{2+x}\right)} \\
&=\frac{0-\left(\frac{2}{2+x}\right)}{\left(\frac{2}{2+x}\right)} \\
&=\frac{\frac{-2}{2+x}}{\left(\frac{2}{2+x}\right)} \\
&=\left(\frac{-2}{2+x}\right)\left(\frac{2+x}{2}\right) \\
&=\frac{2 x+2}{2+x}\
&=x
\end{aligned}$ 
B) $\begin{aligned}
f(g(x)) &=\frac{2}{2+\left(\frac{2-2 x}{x}\right)} \
&=\frac{1}{1+\frac{2-2 x}{x}} \
&=\frac{1}{\left(\frac{0}{x}\right)} \
&=x
\end{aligned}$

$\begin{array}{l}
g(f(x))=\frac{2-2\left(\frac{2}{2+x}\right)}{\left(\frac{2}{2+x}\right)}\\\
=\frac{2-\left(\frac{4}{2+x}\right)}{\left(\frac{2}{2+x}\right)}\\
=\frac{\frac{4+2 x-4}{2+x}}{\left(\frac{2}{2+x}\right)}\\
=\frac{\frac{2 x}{2+x}}{\left(\frac{x}{2+x}\right)}\\
=\frac{2 x}{x}\\
=x
\end{array}$ 
C) $\begin{aligned}
f(g(x)) &=\frac{2}{2+\left(\frac{2-2 x}{x}\right)} \
&=\frac{4}{\frac{2-2 x}{x}} \
&=\frac{2}{\left(\frac{2 x}{x}\right)} \
&=2 \cdot \frac{x}{2} \
&=x
\end{aligned}$
$\begin{aligned}
g(f(x)) &=\frac{2-2\left(\frac{2}{2+x}\right)}{\left(\frac{2}{2+x}\right)} \\
&=\frac{\left(\frac{2}{2+2 x}\right)}{\left(\frac{2}{2+x}\right)} \\
&=\left(\frac{2}{2+2 x}\right)\left(\frac{2+x}{2}\right) \\
&=\frac{2+x}{2+2 x} \\
&=\frac{x}{2 x} \\
&=x
\end{aligned}$ 
D) $\begin{aligned}
f(g(x)) &=\frac{2}{2+\left(\frac{2-2 x}{x}\right)} \
&=\frac{2}{\frac{2 x+2-2 x}{x}} \
&=\frac{2-2 x}{\left(\frac{2}{x}\right)} \
&=\frac{1-x}{1} \
&=x
\end{aligned}$
$\begin{aligned}
g(f(x)) &=\frac{2-2\left(\frac{2}{2+x}\right)}{\left(\frac{2}{2+x}\right)} \
&=\frac{\left(\frac{4}{2+x}\right)}{\left(\frac{2}{2+x}\right)} \
&=\left(\frac{4}{2+x}\right)\left(\frac{2+x}{2}\right) \
&=\frac{2(2+x)}{2+x} \
&=\frac{2 x}{2} \
&=x
\end{aligned}$ 
E) $\begin{aligned}
f(g(x)) &=\frac{2}{2+\left(\frac{2-2 x}{x}\right)} \
&=\frac{2}{\frac{2 x+2-2 x}{x}} \
&=\frac{2}{\left(\frac{2}{x}\right)} \
&=2 \cdot \frac{x}{2} \
&=x
\end{aligned}$

$\begin{array}{l}
g(f(x))=\frac{2-2\left(\frac{2}{2+x}\right)}{\left(\frac{2}{2+x}\right)}\\
=\frac{2-\left(\frac{4}{2+x}\right)}{\left(\frac{2}{2+x}\right)}\\
=\frac{\frac{4+2 x-4}{2+x}}{\left(\frac{2}{2+x}\right)}\\
=\frac{\frac{2 x}{2+x}}{\left(\frac{2}{2+x}\right)}\\
=\frac{2 x}{2}\\
=x
\end{array}$ 
A) $\begin{aligned}
f(g(x)) &=\frac{2}{2+\left(\frac{2-2 x}{x}\right)} \
&=\frac{2}{2+\left(\frac{1}{x}\right)} \
&=\frac{1}{\left(\frac{1}{x}\right)} \
&=1 \cdot \frac{x}{1} \
&=x
\end{aligned}$

$\begin{aligned}
g(f(x)) &=\frac{2-2\left(\frac{2}{2+x}\right)}{\left(\frac{2}{2+x}\right)} \\
&=\frac{0-\left(\frac{2}{2+x}\right)}{\left(\frac{2}{2+x}\right)} \\
&=\frac{\frac{-2}{2+x}}{\left(\frac{2}{2+x}\right)} \\
&=\left(\frac{-2}{2+x}\right)\left(\frac{2+x}{2}\right) \\
&=\frac{2 x+2}{2+x}\
&=x
\end{aligned}$ 
B) $\begin{aligned}
f(g(x)) &=\frac{2}{2+\left(\frac{2-2 x}{x}\right)} \
&=\frac{1}{1+\frac{2-2 x}{x}} \
&=\frac{1}{\left(\frac{0}{x}\right)} \
&=x
\end{aligned}$

$\begin{array}{l}
g(f(x))=\frac{2-2\left(\frac{2}{2+x}\right)}{\left(\frac{2}{2+x}\right)}\\\
=\frac{2-\left(\frac{4}{2+x}\right)}{\left(\frac{2}{2+x}\right)}\\
=\frac{\frac{4+2 x-4}{2+x}}{\left(\frac{2}{2+x}\right)}\\
=\frac{\frac{2 x}{2+x}}{\left(\frac{x}{2+x}\right)}\\
=\frac{2 x}{x}\\
=x
\end{array}$ 
C) $\begin{aligned}
f(g(x)) &=\frac{2}{2+\left(\frac{2-2 x}{x}\right)} \
&=\frac{4}{\frac{2-2 x}{x}} \
&=\frac{2}{\left(\frac{2 x}{x}\right)} \
&=2 \cdot \frac{x}{2} \
&=x
\end{aligned}$
$\begin{aligned}
g(f(x)) &=\frac{2-2\left(\frac{2}{2+x}\right)}{\left(\frac{2}{2+x}\right)} \\
&=\frac{\left(\frac{2}{2+2 x}\right)}{\left(\frac{2}{2+x}\right)} \\
&=\left(\frac{2}{2+2 x}\right)\left(\frac{2+x}{2}\right) \\
&=\frac{2+x}{2+2 x} \\
&=\frac{x}{2 x} \\
&=x
\end{aligned}$ 
D) $\begin{aligned}
f(g(x)) &=\frac{2}{2+\left(\frac{2-2 x}{x}\right)} \
&=\frac{2}{\frac{2 x+2-2 x}{x}} \
&=\frac{2-2 x}{\left(\frac{2}{x}\right)} \
&=\frac{1-x}{1} \
&=x
\end{aligned}$
$\begin{aligned}
g(f(x)) &=\frac{2-2\left(\frac{2}{2+x}\right)}{\left(\frac{2}{2+x}\right)} \
&=\frac{\left(\frac{4}{2+x}\right)}{\left(\frac{2}{2+x}\right)} \
&=\left(\frac{4}{2+x}\right)\left(\frac{2+x}{2}\right) \
&=\frac{2(2+x)}{2+x} \
&=\frac{2 x}{2} \
&=x
\end{aligned}$ 
E) $\begin{aligned}
f(g(x)) &=\frac{2}{2+\left(\frac{2-2 x}{x}\right)} \
&=\frac{2}{\frac{2 x+2-2 x}{x}} \
&=\frac{2}{\left(\frac{2}{x}\right)} \
&=2 \cdot \frac{x}{2} \
&=x
\end{aligned}$

$\begin{array}{l}
g(f(x))=\frac{2-2\left(\frac{2}{2+x}\right)}{\left(\frac{2}{2+x}\right)}\\
=\frac{2-\left(\frac{4}{2+x}\right)}{\left(\frac{2}{2+x}\right)}\\
=\frac{\frac{4+2 x-4}{2+x}}{\left(\frac{2}{2+x}\right)}\\
=\frac{\frac{2 x}{2+x}}{\left(\frac{2}{2+x}\right)}\\
=\frac{2 x}{2}\\
=x
\end{array}$

Question 16

Use function notation to write $g$ in terms of $f ( x ) = x ^ { 3 }$. $g ( x ) = - \frac { 1 } { 2 } ( x + 9 ) ^ { 3 }$

Accepted Answer

A)  $g ( x ) = - \frac { 1 } { 2 } [ f ( x ) ] ^ { 3 } + 9$ 
B)  $g ( x ) = - \frac { 1 } { 2 } [ f ( x ) + 9 ]$ 
C)  $g ( x ) = - [ f ( x ) ] ^ { 3 } + \frac { 729 } { 2 }$ 
D)  $g ( x ) = - \frac { 1 } { 2 } [ f ( x ) ] ^ { 3 } + 729$ 
E)  $g ( x ) = - \frac { 1 } { 2 } [ f ( x + 9 ) ]$ 
A)  $g ( x ) = - \frac { 1 } { 2 } [ f ( x ) ] ^ { 3 } + 9$ 
B)  $g ( x ) = - \frac { 1 } { 2 } [ f ( x ) + 9 ]$ 
C)  $g ( x ) = - [ f ( x ) ] ^ { 3 } + \frac { 729 } { 2 }$ 
D)  $g ( x ) = - \frac { 1 } { 2 } [ f ( x ) ] ^ { 3 } + 729$ 
E)  $g ( x ) = - \frac { 1 } { 2 } [ f ( x + 9 ) ]$

Question 17

Find $( f + g ) ( x )$.
$$\begin{array} { l } 
f ( x ) = - 8 x ^ { 2 } + 5 x - 2 \
g ( x ) = 4 x ^ { 2 } + 7 x + 4
\end{array}$$

Accepted Answer

A)  $( f + g ) ( x ) = - 12 x ^ { 4 } - 2 x ^ { 2 } - 6$ 
B)  $( f + g ) ( x ) = - 4 x ^ { 4 } + 12 x ^ { 2 } + 2$ 
C)  $( f + g ) ( x ) = - 12 x ^ { 2 } - 2 x - 6$ 
D)  $( f + g ) ( x ) = - 4 x ^ { 2 } + 12 x + 2$ 
E)  $( f + g ) ( x ) = 4 x ^ { 2 } - 12 x - 2$ 
A)  $( f + g ) ( x ) = - 12 x ^ { 4 } - 2 x ^ { 2 } - 6$ 
B)  $( f + g ) ( x ) = - 4 x ^ { 4 } + 12 x ^ { 2 } + 2$ 
C)  $( f + g ) ( x ) = - 12 x ^ { 2 } - 2 x - 6$ 
D)  $( f + g ) ( x ) = - 4 x ^ { 2 } + 12 x + 2$ 
E)  $( f + g ) ( x ) = 4 x ^ { 2 } - 12 x - 2$

Question 18

Find the difference quotient and simplify your answer. 
$f ( w ) = - 9 w ^ { 2 } + 2 w$,
$\frac { f ( 4 + h ) - f ( 4 ) } { h } , h \neq 0$

Accepted Answer

A)  $10 + h$ 
B)  $- 70 - 9 w + \frac { 16 } { w }$ 
C)  $2 - 9 w + \frac { 16 } { w }$ 
D)  $2 - 9 h$ 
E)  $- 70 - 9 h$ 
A)  $10 + h$ 
B)  $- 70 - 9 w + \frac { 16 } { w }$ 
C)  $2 - 9 w + \frac { 16 } { w }$ 
D)  $2 - 9 h$ 
E)  $- 70 - 9 h$

Question 19

Find the domain of the function.
$$g ( p ) = \sqrt { 4 - p ^ { 2 } }$$

Accepted Answer

A)  $- 2 \leq p \leq 2$ 
B)  $p \leq - 2$ or $p \geq 2$ 
C)  $p \geq 0$ 
D)  $p \leq 2$ 
E)  all real numbers 
A)  $- 2 \leq p \leq 2$ 
B)  $p \leq - 2$ or $p \geq 2$ 
C)  $p \geq 0$ 
D)  $p \leq 2$ 
E)  all real numbers

Question 20

Compare the graph of the following function with the graph of $f ( x ) = | x |$.
$y = \left| \frac { 7 } { 9 } x \right|$

Accepted Answer

A)  vertical shift of $\frac { 7 } { 9 }$ units up 
B)  horizontal stretch of $\frac { 9 } { 7 }$ units 
C)  vertical shrink of $\frac { 7 } { 9 }$ units horizontal shrink of $\frac { 7 } { 9 }$ units 
D)  vertical shift of $\frac { 9 } { 7 }$ units 
E)  horizontal shrink of $\frac { 7 } { 9 }$ units 
A)  vertical shift of $\frac { 7 } { 9 }$ units up 
B)  horizontal stretch of $\frac { 9 } { 7 }$ units 
C)  vertical shrink of $\frac { 7 } { 9 }$ units horizontal shrink of $\frac { 7 } { 9 }$ units 
D)  vertical shift of $\frac { 9 } { 7 }$ units 
E)  horizontal shrink of $\frac { 7 } { 9 }$ units

Write the slope-intercept form of the equation of the line through the given point perpendicular to the given line. point: $( - 4,7 )$ line: $- 5 x - 15 y = - 6$

Determine the domain and range of the inverse function $f ^ { - 1 }$ of the following function f . $f ( x ) = - | x + 6 | + 2$ , where $x > - 6$

Determine algebraically whether the following function is one-to-one. $| x - 5 | , \quad x \leq 5$ $| a - 5 | = | b - 5 |$

Plot the points and find the slope of the line passing through the pair of points. $( 1,0 ) , ( 5,3 )$

Graph the function and determine the interval(s) for which $f ( x ) \geq 0$ . $f ( x ) = - x ^ { 2 } + 4 x$ $Graph the function and determine the interval(s) for which f ( x ) \geq 0 . f ( x ) = - x ^ { 2 } + 4 x$

Determine algebraically whether the following function is one-to-one. $f(x)=\frac{5 x^{2}}{3 x^{2}+6}, \text { where } x>0$ = + =+

Use the graphs of $f$ and $g$ , shown below, to graph $h ( x ) = ( f + g ) ( x )$ .

Decide whether the two functions shown in the graph below appear to be inverse functions of each other.

Determine the domain of $g ( x ) = \frac { 1 } { x ^ { 2 } - 49 }$

Estimate the slope of the line.

Given that $f ( x ) = \sqrt [ 4 ] { x - 4 }$ and $g ( x ) = x ^ { 4 } + 4$ determine the value of the following (if possible). $( f \circ g ) ( 0 )$

Use the graph of the function to find the domain and range of $f$ .

Use the graph of $f ( x ) = x ^ { 2 }$ to write an equation for the function whose graph is shown. $Use the graph of f ( x ) = x ^ { 2 } to write an equation for the function whose graph is shown.$

Show algebraically that the functions $f$ and $g$ shown below are inverse functions. $f ( x ) = \frac { 2 } { 2 + x } , x \geq 0 , \quad g ( x ) = \frac { 2 - 2 x } { x } , 0 < x \leq 1$

Use function notation to write $g$ in terms of $f ( x ) = x ^ { 3 }$ . $g ( x ) = - \frac { 1 } { 2 } ( x + 9 ) ^ { 3 }$

Find $( f + g ) ( x )$ . f(x)=-8+5x-2 g(x)=4+7x+4

Find the difference quotient and simplify your answer. $f ( w ) = - 9 w ^ { 2 } + 2 w$ , $\frac { f ( 4 + h ) - f ( 4 ) } { h } , h \neq 0$

Find the domain of the function. $g ( p ) = \sqrt { 4 - p ^ { 2 } }$

Compare the graph of the following function with the graph of $f ( x ) = | x |$ . $y = \left| \frac { 7 } { 9 } x \right|$

Polynomial and Rational Functions

Exponential and Logarithmic Functions

Trigonometric Functions

Analytic Trigonometry

Additional Topics in Trigonometry

Linear Systems and Matrices

Sequences, Series, and Probability

Topics in Analytic Geometry

Analytic Geometry in Three Dimensions

Limits and an Introduction to Calculus

Review of Graphs, Equations, and Inequalities

Filters

Exam 1: Functions and Their Graphs

Write the slope-intercept form of the equation of the line through the given point perpendicular to the given line. point: (−4,7)( - 4,7 )(−4,7) line: −5x−15y=−6- 5 x - 15 y = - 6−5x−15y=−6

Determine the domain and range of the inverse function f−1f ^ { - 1 }f−1 of the following function f . f(x)=−∣x+6∣+2f ( x ) = - | x + 6 | + 2f(x)=−∣x+6∣+2 , where x>−6x > - 6x>−6

Determine algebraically whether the following function is one-to-one. ∣x−5∣,x≤5| x - 5 | , \quad x \leq 5∣x−5∣,x≤5 ∣a−5∣=∣b−5∣| a - 5 | = | b - 5 |∣a−5∣=∣b−5∣

Plot the points and find the slope of the line passing through the pair of points. (1,0),(5,3)( 1,0 ) , ( 5,3 )(1,0),(5,3)

Graph the function and determine the interval(s) for which f(x)≥0f ( x ) \geq 0f(x)≥0 . f(x)=−x2+4xf ( x ) = - x ^ { 2 } + 4 xf(x)=−x2+4x

Determine algebraically whether the following function is one-to-one. f(x)=5x23x2+6, where x>0f(x)=\frac{5 x^{2}}{3 x^{2}+6}, \text { where } x>0f(x)=3x2+65x2​, where x>0 = + =+

Use the graphs of fff and ggg , shown below, to graph h(x)=(f+g)(x)h ( x ) = ( f + g ) ( x )h(x)=(f+g)(x) .

Decide whether the two functions shown in the graph below appear to be inverse functions of each other.

Determine the domain of g(x)=1x2−49g ( x ) = \frac { 1 } { x ^ { 2 } - 49 }g(x)=x2−491​

Estimate the slope of the line.

Given that f(x)=x−44f ( x ) = \sqrt [ 4 ] { x - 4 }f(x)=4x−4​ and g(x)=x4+4g ( x ) = x ^ { 4 } + 4g(x)=x4+4 determine the value of the following (if possible). (f∘g)(0)( f \circ g ) ( 0 )(f∘g)(0)

Use the graph of the function to find the domain and range of fff .

Use the graph of f(x)=x2f ( x ) = x ^ { 2 }f(x)=x2 to write an equation for the function whose graph is shown.

Show algebraically that the functions fff and ggg shown below are inverse functions. f(x)=22+x,x≥0,g(x)=2−2xx,0<x≤1f ( x ) = \frac { 2 } { 2 + x } , x \geq 0 , \quad g ( x ) = \frac { 2 - 2 x } { x } , 0 < x \leq 1f(x)=2+x2​,x≥0,g(x)=x2−2x​,0<x≤1

Use function notation to write ggg in terms of f(x)=x3f ( x ) = x ^ { 3 }f(x)=x3 . g(x)=−12(x+9)3g ( x ) = - \frac { 1 } { 2 } ( x + 9 ) ^ { 3 }g(x)=−21​(x+9)3

Find (f+g)(x)( f + g ) ( x )(f+g)(x) . f(x)=-8+5x-2 g(x)=4+7x+4

Find the difference quotient and simplify your answer. f(w)=−9w2+2wf ( w ) = - 9 w ^ { 2 } + 2 wf(w)=−9w2+2w , f(4+h)−f(4)h,h≠0\frac { f ( 4 + h ) - f ( 4 ) } { h } , h \neq 0hf(4+h)−f(4)​,h=0

Find the domain of the function. g(p)=4−p2g ( p ) = \sqrt { 4 - p ^ { 2 } }g(p)=4−p2​

Compare the graph of the following function with the graph of f(x)=∣x∣f ( x ) = | x |f(x)=∣x∣ . y=∣79x∣y = \left| \frac { 7 } { 9 } x \right|y=​97​x​

Polynomial and Rational Functions

Exponential and Logarithmic Functions

Trigonometric Functions

Analytic Trigonometry

Additional Topics in Trigonometry

Linear Systems and Matrices

Sequences, Series, and Probability

Topics in Analytic Geometry

Analytic Geometry in Three Dimensions

Limits and an Introduction to Calculus

Review of Graphs, Equations, and Inequalities

Filters

Write the slope-intercept form of the equation of the line through the given point perpendicular to the given line. point: $( - 4,7 )$ line: $- 5 x - 15 y = - 6$

Determine the domain and range of the inverse function $f ^ { - 1 }$ of the following function f . $f ( x ) = - | x + 6 | + 2$ , where $x > - 6$

Determine algebraically whether the following function is one-to-one. $| x - 5 | , \quad x \leq 5$ $| a - 5 | = | b - 5 |$

Plot the points and find the slope of the line passing through the pair of points. $( 1,0 ) , ( 5,3 )$

Graph the function and determine the interval(s) for which $f ( x ) \geq 0$ . $f ( x ) = - x ^ { 2 } + 4 x$ $Graph the function and determine the interval(s) for which f ( x ) \geq 0 . f ( x ) = - x ^ { 2 } + 4 x$

Determine algebraically whether the following function is one-to-one. $f(x)=\frac{5 x^{2}}{3 x^{2}+6}, \text { where } x>0$ = + =+

Use the graphs of $f$ and $g$ , shown below, to graph $h ( x ) = ( f + g ) ( x )$ .

Determine the domain of $g ( x ) = \frac { 1 } { x ^ { 2 } - 49 }$

Given that $f ( x ) = \sqrt [ 4 ] { x - 4 }$ and $g ( x ) = x ^ { 4 } + 4$ determine the value of the following (if possible). $( f \circ g ) ( 0 )$

Use the graph of the function to find the domain and range of $f$ .

Use the graph of $f ( x ) = x ^ { 2 }$ to write an equation for the function whose graph is shown. $Use the graph of f ( x ) = x ^ { 2 } to write an equation for the function whose graph is shown.$

Show algebraically that the functions $f$ and $g$ shown below are inverse functions. $f ( x ) = \frac { 2 } { 2 + x } , x \geq 0 , \quad g ( x ) = \frac { 2 - 2 x } { x } , 0 < x \leq 1$

Use function notation to write $g$ in terms of $f ( x ) = x ^ { 3 }$ . $g ( x ) = - \frac { 1 } { 2 } ( x + 9 ) ^ { 3 }$

Find $( f + g ) ( x )$ . f(x)=-8+5x-2 g(x)=4+7x+4

Find the difference quotient and simplify your answer. $f ( w ) = - 9 w ^ { 2 } + 2 w$ , $\frac { f ( 4 + h ) - f ( 4 ) } { h } , h \neq 0$

Find the domain of the function. $g ( p ) = \sqrt { 4 - p ^ { 2 } }$

Compare the graph of the following function with the graph of $f ( x ) = | x |$ . $y = \left| \frac { 7 } { 9 } x \right|$