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Compare the graph of the following function with the graph of f(x) =∣x∣f ( x ) = | x | . y=∣34x∣y = \left| \frac { 3 } { 4 } x \right|


A) vertical shift of 34\frac { 3 } { 4 } units up
B) horizontal stretch of 43\frac { 4 } { 3 } units
C) vertical shrink of 34\frac { 3 } { 4 } units
D) horizontal shrink of 34\frac { 3 } { 4 } units vertical shift of 43\frac { 4 } { 3 } units
E) horizontal shrink of 34\frac { 3 } { 4 } units

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Find the inverse function of ff . f(x) =x5+5f ( x ) = x ^ { 5 } + 5


A) f−1(x) =−x5+5f ^ { - 1 } ( x ) = - \sqrt [ 5 ] { x } + 5
B) f−1(x) =x5+5f ^ { - 1 } ( x ) = \sqrt [ 5 ] { x } + 5
C) f−1(x) =−x+55f ^ { - 1 } ( x ) = - \sqrt [ 5 ] { x + 5 }
D) f−1(x) =x−55f ^ { - 1 } ( x ) = \sqrt [ 5 ] { x - 5 }
E) f−1(x) =x5−5f ^ { - 1 } ( x ) = \sqrt [ 5 ] { x } - 5

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Compare the graph of the following function with the graph of f(x) =x3f ( x ) = x ^ { 3 } . y=[5(x+10) ]3y = [ 5 ( x + 10 ) ] ^ { 3 }


A) vertical shift of 10 units up
B) vertical shift of 10 units up horizontal shrink of 15\frac { 1 } { 5 } units
C) horizontal shift of 10 units to the left horizontal shrink of 1125\frac { 1 } { 125 } units
D) horizontal shift of 10 units to the left horizontal stretch of 15\frac { 1 } { 5 } units
E) horizontal shift of 10 units to the left vertical shift of 5 units up

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Determine an equation that may represented by the graph shown below. Determine an equation that may represented by the graph shown below. A) f ( x ) = | x - 1 | - 1 B) f ( x ) = - | x - 1 | + 1 C) f ( x ) = | x - 1 | + 1 D) f ( x ) = | x + 1 | + 1 E) f ( x ) = | x + 1 | - 1


A) f(x) =∣x−1∣−1f ( x ) = | x - 1 | - 1
B) f(x) =−∣x−1∣+1f ( x ) = - | x - 1 | + 1
C) f(x) =∣x−1∣+1f ( x ) = | x - 1 | + 1
D) f(x) =∣x+1∣+1f ( x ) = | x + 1 | + 1
E) f(x) =∣x+1∣−1f ( x ) = | x + 1 | - 1

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Use the graph of ff to sketch the graph of the function indicated below. Use the graph of f to sketch the graph of the function indicated below. A) B) C) D) E)


A)
Use the graph of f to sketch the graph of the function indicated below. A) B) C) D) E)

B)
Use the graph of f to sketch the graph of the function indicated below. A) B) C) D) E)

C)
Use the graph of f to sketch the graph of the function indicated below. A) B) C) D) E)
D)
Use the graph of f to sketch the graph of the function indicated below. A) B) C) D) E)

E)
Use the graph of f to sketch the graph of the function indicated below. A) B) C) D) E)

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Use the functions f(x) =x+4f ( x ) = x + 4 and g(x) =5x−7g ( x ) = 5 x - 7 to find (g∘f) −1( g \circ f ) ^ { - 1 } .


A) (g∘f) −1=5x+114( g \circ f ) ^ { - 1 } = \frac { 5 x + 11 } { 4 }
B) (g∘f) −1=5x−42( g \circ f ) ^ { - 1 } = 5 x - 42
C) (g∘f) −1=x−135( g \circ f ) ^ { - 1 } = \frac { x - 13 } { 5 }
D) (g∘f) −1=−7x−75( g \circ f ) ^ { - 1 } = \frac { - 7 x - 7 } { 5 }
E) (g∘f) −1=5x+13( g \circ f ) ^ { - 1 } = 5 x + 13

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The average lengths LL of cellular phone calls in minutes from 1999 to 2004 are shown in the table below. 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. A) 4.37 minutes B) 8.74 minutes C) 5.37 minutes D) 3.37 minutes E) 2.19 minutes Use the regression feature of a graphing utility to find a linear model for the data. Let tt represent the year, with t=9t = 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.


A) 4.374.37 minutes
B) 8.748.74 minutes
C) 5.375.37 minutes
D) 3.373.37 minutes
E) 2.192.19 minutes

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Determine the domain and range of the inverse function f−1f ^ { - 1 } of the following function ff . f(x) =−∣x+7∣−1f ( x ) = - | x + 7 | - 1 , where x>−7x > - 7


A) Domain: [−7,∞) [ - 7 , \infty ) ; Range: [−1,∞) [ - 1 , \infty )
B) Domain: (−∞,−1]( - \infty , - 1 ] ; Range: [−7,∞) [ - 7 , \infty )
C) Domain: [−7,−1][ - 7 , - 1 ] ; Range: [−7,∞) [ - 7 , \infty )
D) Domain: (−∞,−7]( - \infty , - 7 ] ; Range: [1,∞) [ 1 , \infty )
E) Domain: (−∞,∞) ( - \infty , \infty ) ; Range: (−∞,∞) ( - \infty , \infty )

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Plot the points and find the slope of the line passing through the pair of points. (3,4) ,(−2,4) ( 3,4 ) , ( - 2,4 ) Plot the points and find the slope of the line passing through the pair of points. ( 3,4 ) , ( - 2,4 ) A) slope: 0 B) slope: 1 C) slope: - 5 D) slope: - \frac { 1 } { 5 } E) slope: undefined


A) slope: 0
B) slope: 1
C) slope: −5- 5
D) slope: −15- \frac { 1 } { 5 }
E) slope: undefined

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Find all real values of xx such that f(x) =0f ( x ) = 0 . f(x) =−2x+55f ( x ) = \frac { - 2 x + 5 } { 5 }


A) 12\frac { 1 } { 2 }
B) ±12\pm \frac { 1 } { 2 }
C) ±52\pm \frac { 5 } { 2 }
D) 52\frac { 5 } { 2 }
E) −52- \frac { 5 } { 2 }

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


A)
5a23a2+6=5b23b2+65a23a2+5a26=5b23b2+5b2653+5a26=53+5b2630+5a218=30+5b218; not one-to-one 30+5a2=30+5b25a2=5b2a2=b2±a=±b\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 } \\\frac { 5 } { 3 } + \frac { 5 a ^ { 2 } } { 6 } & = \frac { 5 } { 3 } + \frac { 5 b ^ { 2 } } { 6 } \\\frac { 30 + 5 a ^ { 2 } } { 18 } & = \frac { 30 + 5 b ^ { 2 } } { 18 } ; \text { not one-to-one } \\30 + 5 a ^ { 2 } & = 30 + 5 b ^ { 2 } \\5 a ^ { 2 } & = 5 b ^ { 2 } \\a ^ { 2 } & = b ^ { 2 } \\\pm a & = \quad \pm b\end{aligned}
B)
5a23a2+6=5b23b2+653+6=53+6; one-to-one 56=36a=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)
5a23a2+6=5b23b2+65a23a2=5b23b2; one-to-one 53=53a=b\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)
5a23a2+6=5b23b2+65a29a2=5b29b25a9=5b95a=5b one-to-one a=b\begin{aligned}\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 } } \\\frac { 5 a } { 9 } & = \frac { 5 b } { 9 } \\5 a & = 5 b\text { one-to-one } \\a & = b\end{aligned}
E) 5a23a2+6=5b23b2+6\frac { 5 a ^ { 2 } } { 3 a ^ { 2 } + 6 } = \frac { 5 b ^ { 2 } } { 3 b ^ { 2 } + 6 }
5a23a2+5a26=5b23b2+5b2653+5a26=53+5b2630+5a218=30+5b218; one-to-one 30+5a2=30+5b25a2=5b2a2=b2a=b\begin{aligned}\frac { 5 a ^ { 2 } } { 3 a ^ { 2 } } + \frac { 5 a ^ { 2 } } { 6 } & = \frac { 5 b ^ { 2 } } { 3 b ^ { 2 } } + \frac { 5 b ^ { 2 } } { 6 } \\\frac { 5 } { 3 } + \frac { 5 a ^ { 2 } } { 6 } & = \frac { 5 } { 3 } + \frac { 5 b ^ { 2 } } { 6 } \\\frac { 30 + 5 a ^ { 2 } } { 18 } & = \frac { 30 + 5 b ^ { 2 } } { 18 } ; \text { one-to-one } \\30 + 5 a ^ { 2 } & = 30 + 5 b ^ { 2 } \\5 a ^ { 2 } & = 5 b ^ { 2 } \\a ^ { 2 } & = b ^ { 2 } \\a & = b\end{aligned}

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Evaluate the following function at the specified value of the independent variable and simplify. f(w) =−7w2+20w2;f(0) f ( w ) = \frac { - 7 w ^ { 2 } + 20 } { w ^ { 2 } } ; \quad f ( 0 )


A) 20
B) 0
C) −7- 7
D) 13
E) undefined

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Find (fg) (x) ( f g ) ( x ) . f(x) =−5xg(x) =−8x+6f ( x ) = \sqrt { - 5 x } \quad g ( x ) = \sqrt { - 8 x + 6 }


A) (fg) (x) =2x10−30x( f g ) ( x ) = 2 x \sqrt { 10 } - \sqrt { 30 x }
B) (fg) (x) =2x10−30x( f g ) ( x ) = 2 x \sqrt { 10 - 30 x }
C) (fg) (x) =−13x+6( f g ) ( x ) = \sqrt { - 13 x + 6 }
D) (fg) (x) =40x2+6( f g ) ( x ) = \sqrt { 40 x ^ { 2 } + 6 }
E) (fg) (x) =40x2−30x( f g ) ( x ) = \sqrt { 40 x ^ { 2 } - 30 x }

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Use the graphs of ff and gg to evaluate the function. (f∘g) (1) ( f \circ g ) ( 1 ) Use the graphs of f and g to evaluate the function. ( f \circ g ) ( 1 ) A) 9 B) - 1 C) 0 D) - 4 E) - 2


A) 9
B) −1- 1
C) 0
D) −4- 4
E) −2- 2

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Find the slope-intercept form of the line passing through the points. (−4,−2) ,(−1,7) ( - 4 , - 2 ) , ( - 1,7 )


A) y=3x+2y = 3 x + 2
B) y=3x+10y = 3 x + 10
C) y=13x−23y = \frac { 1 } { 3 } x - \frac { 2 } { 3 }
D) y=−13x−103y = - \frac { 1 } { 3 } x - \frac { 10 } { 3 }
E) y=−3x−14y = - 3 x - 14

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Find all real values of xx such that f(x) =0f ( x ) = 0 . f(x) =49x2−64f ( x ) = 49 x ^ { 2 } - 64


A) ±78\pm \frac { 7 } { 8 }
B) ±87\pm \frac { 8 } { 7 }
C) ±6449\pm \frac { 64 } { 49 }
D) −6449- \frac { 64 } { 49 }
E) 87\frac { 8 } { 7 }

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Find f∘gf \circ g . f(x) =x+2g(x) =5x2−4f ( x ) = x + 2 \quad g ( x ) = \frac { 5 } { x ^ { 2 } - 4 }


A) (f∘g) (x) =5x2( f \circ g ) ( x ) = \frac { 5 } { x ^ { 2 } }
B) (f∘g) (x) =5x2+4x( f \circ g ) ( x ) = \frac { 5 } { x ^ { 2 } + 4 x }
C) (f∘g) (x) =2x2+3x2−4( f \circ g ) ( x ) = \frac { 2 x ^ { 2 } + 3 } { x ^ { 2 } - 4 }
D) (f∘g) (x) =7x2−4( f \circ g ) ( x ) = \frac { 7 } { x ^ { 2 } - 4 }
E) (f∘g) (x) =2x2−3x2−4( f \circ g ) ( x ) = \frac { 2 x ^ { 2 } - 3 } { x ^ { 2 } - 4 }

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Use the graphs of ff and gg , shown below, to graph h(x) =(f+g) (x) h ( x ) = ( f + g ) ( x ) . Use the graphs of f and g , shown below, to graph h ( x ) = ( f + g ) ( x ) . A) B) C) D) E)


A)
Use the graphs of f and g , shown below, to graph h ( x ) = ( f + g ) ( x ) . A) B) C) D) E)
B)
Use the graphs of f and g , shown below, to graph h ( x ) = ( f + g ) ( x ) . A) B) C) D) E)
C)
Use the graphs of f and g , shown below, to graph h ( x ) = ( f + g ) ( x ) . A) B) C) D) E)
D)
Use the graphs of f and g , shown below, to graph h ( x ) = ( f + g ) ( x ) . A) B) C) D) E)
E)
Use the graphs of f and g , shown below, to graph h ( x ) = ( f + g ) ( x ) . A) B) C) D) E)

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Decide whether the two functions shown in the graph below appear to be inverse functions of each other. Decide whether the two functions shown in the graph below appear to be inverse functions of each other. A) yes B) no C) not enough information


A) yes
B) no
C) not enough information

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Find the domain of the function. g(w) =4ww+9g ( w ) = \frac { 4 w } { w + 9 }


A) all real numbers w≠−9w \neq - 9
B) all real numbers w≠−9,w≠0w \neq - 9 , w \neq 0
C) all real numbers
D) w=−9,w=0w = - 9 , w = 0
E) w=−9w = - 9

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