lesson 7 homework 3.3 answer key

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• Core Connections Integrated I, 2013
• Core Connections Algebra 1, 2013
• Core Connections Geometry, 2013
• Core Connections Algebra 2, 2013
• Core Connections Integrated I, 2014
• Core Connections Integrated II, 2015
• Core Connections: Course 1
• Core Connections: Course 2
• Core Connections: Course 3
• Core Connections Integrated III, 2015

3.1 Functions and Function Notation

• ⓑ yes (Note: If two players had been tied for, say, 4th place, then the name would not have been a function of rank.)

w = f ( d ) w = f ( d )

g ( 5 ) = 1 g ( 5 ) = 1

m = 8 m = 8

y = f ( x ) = x 3 2 y = f ( x ) = x 3 2

g ( 1 ) = 8 g ( 1 ) = 8

x = 0 x = 0 or x = 2 x = 2

• ⓐ yes, because each bank account has a single balance at any given time;
• ⓑ no, because several bank account numbers may have the same balance;
• ⓒ no, because the same output may correspond to more than one input.
• ⓐ Yes, letter grade is a function of percent grade;
• ⓑ No, it is not one-to-one. There are 100 different percent numbers we could get but only about five possible letter grades, so there cannot be only one percent number that corresponds to each letter grade.

No, because it does not pass the horizontal line test.

3.2 Domain and Range

{ − 5 , 0 , 5 , 10 , 15 } { − 5 , 0 , 5 , 10 , 15 }

( − ∞ , ∞ ) ( − ∞ , ∞ )

( − ∞ , 1 2 ) ∪ ( 1 2 , ∞ ) ( − ∞ , 1 2 ) ∪ ( 1 2 , ∞ )

[ − 5 2 , ∞ ) [ − 5 2 , ∞ )

• ⓐ values that are less than or equal to –2, or values that are greater than or equal to –1 and less than 3
• ⓑ { x | x ≤ − 2 or − 1 ≤ x < 3 } { x | x ≤ − 2 or − 1 ≤ x < 3 }
• ⓒ ( − ∞ , − 2 ] ∪ [ − 1 , 3 ) ( − ∞ , − 2 ] ∪ [ − 1 , 3 )

domain =[1950,2002] range = [47,000,000,89,000,000]

domain: ( − ∞ , 2 ] ; ( − ∞ , 2 ] ; range: ( − ∞ , 0 ] ( − ∞ , 0 ]

3.3 Rates of Change and Behavior of Graphs

$ 2.84 − $ 2.31 5 years = $ 0.53 5 years = $ 0.106 $ 2.84 − $ 2.31 5 years = $ 0.53 5 years = $ 0.106 per year.

a + 7 a + 7

The local maximum appears to occur at ( − 1 , 28 ) , ( − 1 , 28 ) , and the local minimum occurs at ( 5 , − 80 ) . ( 5 , − 80 ) . The function is increasing on ( − ∞ , − 1 ) ∪ ( 5 , ∞ ) ( − ∞ , − 1 ) ∪ ( 5 , ∞ ) and decreasing on ( − 1 , 5 ) . ( − 1 , 5 ) .

3.4 Composition of Functions

( f g ) ( x ) = f ( x ) g ( x ) = ( x − 1 ) ( x 2 − 1 ) = x 3 − x 2 − x + 1 ( f − g ) ( x ) = f ( x ) − g ( x ) = ( x − 1 ) − ( x 2 − 1 ) = x − x 2 ( f g ) ( x ) = f ( x ) g ( x ) = ( x − 1 ) ( x 2 − 1 ) = x 3 − x 2 − x + 1 ( f − g ) ( x ) = f ( x ) − g ( x ) = ( x − 1 ) − ( x 2 − 1 ) = x − x 2

No, the functions are not the same.

A gravitational force is still a force, so a ( G ( r ) ) a ( G ( r ) ) makes sense as the acceleration of a planet at a distance r from the Sun (due to gravity), but G ( a ( F ) ) G ( a ( F ) ) does not make sense.

f ( g ( 1 ) ) = f ( 3 ) = 3 f ( g ( 1 ) ) = f ( 3 ) = 3 and g ( f ( 4 ) ) = g ( 1 ) = 3 g ( f ( 4 ) ) = g ( 1 ) = 3

g ( f ( 2 ) ) = g ( 5 ) = 3 g ( f ( 2 ) ) = g ( 5 ) = 3

[ − 4 , 0 ) ∪ ( 0 , ∞ ) [ − 4 , 0 ) ∪ ( 0 , ∞ )

Possible answer:

g ( x ) = 4 + x 2 h ( x ) = 4 3 − x f = h ∘ g g ( x ) = 4 + x 2 h ( x ) = 4 3 − x f = h ∘ g

3.5 Transformation of Functions

The graphs of f ( x ) f ( x ) and g ( x ) g ( x ) are shown below. The transformation is a horizontal shift. The function is shifted to the left by 2 units.

g ( x ) = 1 x - 1 + 1 g ( x ) = 1 x - 1 + 1

g ( x ) = − f ( x ) g ( x ) = − f ( x )

h ( x ) = f ( − x ) h ( x ) = f ( − x )

Notice: g ( x ) = f ( − x ) g ( x ) = f ( − x ) looks the same as f ( x ) f ( x ) .

g ( x ) = 3 x - 2 g ( x ) = 3 x - 2

g ( x ) = f ( 1 3 x ) g ( x ) = f ( 1 3 x ) so using the square root function we get g ( x ) = 1 3 x g ( x ) = 1 3 x

3.6 Absolute Value Functions

using the variable p p for passing, | p − 80 | ≤ 20 | p − 80 | ≤ 20

f ( x ) = − | x + 2 | + 3 f ( x ) = − | x + 2 | + 3

x = − 1 x = − 1 or x = 2 x = 2

3.7 Inverse Functions

h ( 2 ) = 6 h ( 2 ) = 6

The domain of function f − 1 f − 1 is ( − ∞ , − 2 ) ( − ∞ , − 2 ) and the range of function f − 1 f − 1 is ( 1 , ∞ ) . ( 1 , ∞ ) .

• ⓐ f ( 60 ) = 50. f ( 60 ) = 50. In 60 minutes, 50 miles are traveled.
• ⓑ f − 1 ( 60 ) = 70. f − 1 ( 60 ) = 70. To travel 60 miles, it will take 70 minutes.

x = 3 y + 5 x = 3 y + 5

f − 1 ( x ) = ( 2 − x ) 2 ; domain of f : [ 0 , ∞ ) ; domain of f − 1 : ( − ∞ , 2 ] f − 1 ( x ) = ( 2 − x ) 2 ; domain of f : [ 0 , ∞ ) ; domain of f − 1 : ( − ∞ , 2 ]

3.1 Section Exercises

A relation is a set of ordered pairs. A function is a special kind of relation in which no two ordered pairs have the same first coordinate.

When a vertical line intersects the graph of a relation more than once, that indicates that for that input there is more than one output. At any particular input value, there can be only one output if the relation is to be a function.

When a horizontal line intersects the graph of a function more than once, that indicates that for that output there is more than one input. A function is one-to-one if each output corresponds to only one input.

not a function

f ( − 3 ) = − 11 ; f ( − 3 ) = − 11 ; f ( 2 ) = − 1 ; f ( 2 ) = − 1 ; f ( − a ) = − 2 a − 5 ; f ( − a ) = − 2 a − 5 ; − f ( a ) = − 2 a + 5 ; − f ( a ) = − 2 a + 5 ; f ( a + h ) = 2 a + 2 h − 5 f ( a + h ) = 2 a + 2 h − 5

f ( − 3 ) = 5 + 5 ; f ( − 3 ) = 5 + 5 ; f ( 2 ) = 5 ; f ( 2 ) = 5 ; f ( − a ) = 2 + a + 5 ; f ( − a ) = 2 + a + 5 ; − f ( a ) = − 2 − a − 5 ; − f ( a ) = − 2 − a − 5 ; f ( a + h ) = 2 − a − h + 5 f ( a + h ) = 2 − a − h + 5

f ( − 3 ) = 2 ; f ( − 3 ) = 2 ; f ( 2 ) = 1 − 3 = − 2 ; f ( 2 ) = 1 − 3 = − 2 ; f ( − a ) = | − a − 1 | − | − a + 1 | ; f ( − a ) = | − a − 1 | − | − a + 1 | ; − f ( a ) = − | a − 1 | + | a + 1 | ; − f ( a ) = − | a − 1 | + | a + 1 | ; f ( a + h ) = | a + h − 1 | − | a + h + 1 | f ( a + h ) = | a + h − 1 | − | a + h + 1 |

g ( x ) − g ( a ) x − a = x + a + 2 , x ≠ a g ( x ) − g ( a ) x − a = x + a + 2 , x ≠ a

a. f ( − 2 ) = 14 ; f ( − 2 ) = 14 ; b. x = 3 x = 3

a. f ( 5 ) = 10 ; f ( 5 ) = 10 ; b. x = − 1 x = − 1 or x = 4 x = 4

• ⓐ f ( t ) = 6 − 2 3 t ; f ( t ) = 6 − 2 3 t ;
• ⓑ f ( − 3 ) = 8 ; f ( − 3 ) = 8 ;
• ⓒ t = 6 t = 6
• ⓐ f ( 0 ) = 1 ; f ( 0 ) = 1 ;
• ⓑ f ( x ) = − 3 , x = − 2 f ( x ) = − 3 , x = − 2 or x = 2 x = 2

not a function so it is also not a one-to-one function

one-to- one function

function, but not one-to-one

f ( x ) = 1 , x = 2 f ( x ) = 1 , x = 2

f ( − 2 ) = 14 ; f ( − 1 ) = 11 ; f ( 0 ) = 8 ; f ( 1 ) = 5 ; f ( 2 ) = 2 f ( − 2 ) = 14 ; f ( − 1 ) = 11 ; f ( 0 ) = 8 ; f ( 1 ) = 5 ; f ( 2 ) = 2

f ( − 2 ) = 4 ;    f ( − 1 ) = 4.414 ; f ( 0 ) = 4.732 ; f ( 1 ) = 5 ; f ( 2 ) = 5.236 f ( − 2 ) = 4 ;    f ( − 1 ) = 4.414 ; f ( 0 ) = 4.732 ; f ( 1 ) = 5 ; f ( 2 ) = 5.236

f ( − 2 ) = 1 9 ; f ( − 1 ) = 1 3 ; f ( 0 ) = 1 ; f ( 1 ) = 3 ; f ( 2 ) = 9 f ( − 2 ) = 1 9 ; f ( − 1 ) = 1 3 ; f ( 0 ) = 1 ; f ( 1 ) = 3 ; f ( 2 ) = 9

[ 0 , 100 ] [ 0 , 100 ]

[ − 0.001 , 0 .001 ] [ − 0.001 , 0 .001 ]

[ − 1 , 000 , 000 , 1,000,000 ] [ − 1 , 000 , 000 , 1,000,000 ]

[ 0 , 10 ] [ 0 , 10 ]

[ −0.1 , 0.1 ] [ −0.1 , 0.1 ]

[ − 100 , 100 ] [ − 100 , 100 ]

• ⓐ g ( 5000 ) = 50 ; g ( 5000 ) = 50 ;
• ⓑ The number of cubic yards of dirt required for a garden of 100 square feet is 1.
• ⓐ The height of a rocket above ground after 1 second is 200 ft.
• ⓑ The height of a rocket above ground after 2 seconds is 350 ft.

3.2 Section Exercises

The domain of a function depends upon what values of the independent variable make the function undefined or imaginary.

There is no restriction on x x for f ( x ) = x 3 f ( x ) = x 3 because you can take the cube root of any real number. So the domain is all real numbers, ( − ∞ , ∞ ) . ( − ∞ , ∞ ) . When dealing with the set of real numbers, you cannot take the square root of negative numbers. So x x -values are restricted for f ( x ) = x f ( x ) = x to nonnegative numbers and the domain is [ 0 , ∞ ) . [ 0 , ∞ ) .

Graph each formula of the piecewise function over its corresponding domain. Use the same scale for the x x -axis and y y -axis for each graph. Indicate inclusive endpoints with a solid circle and exclusive endpoints with an open circle. Use an arrow to indicate − ∞ − ∞ or ∞ . ∞ . Combine the graphs to find the graph of the piecewise function.

( − ∞ , 3 ] ( − ∞ , 3 ]

( − ∞ , − 1 2 ) ∪ ( − 1 2 , ∞ ) ( − ∞ , − 1 2 ) ∪ ( − 1 2 , ∞ )

( − ∞ , − 11 ) ∪ ( − 11 , 2 ) ∪ ( 2 , ∞ ) ( − ∞ , − 11 ) ∪ ( − 11 , 2 ) ∪ ( 2 , ∞ )

( − ∞ , − 3 ) ∪ ( − 3 , 5 ) ∪ ( 5 , ∞ ) ( − ∞ , − 3 ) ∪ ( − 3 , 5 ) ∪ ( 5 , ∞ )

( − ∞ , 5 ) ( − ∞ , 5 )

[ 6 , ∞ ) [ 6 , ∞ )

( − ∞ , − 9 ) ∪ ( − 9 , 9 ) ∪ ( 9 , ∞ ) ( − ∞ , − 9 ) ∪ ( − 9 , 9 ) ∪ ( 9 , ∞ )

domain: ( 2 , 8 ] , ( 2 , 8 ] , range [ 6 , 8 ) [ 6 , 8 )

domain: [ − 4 , 4], [ − 4 , 4], range: [ 0 , 2] [ 0 , 2]

domain: [ − 5 , 3 ) , [ − 5 , 3 ) , range: [ 0 , 2 ] [ 0 , 2 ]

domain: ( − ∞ , 1 ] , ( − ∞ , 1 ] , range: [ 0 , ∞ ) [ 0 , ∞ )

domain: [ − 6 , − 1 6 ] ∪ [ 1 6 , 6 ] ; [ − 6 , − 1 6 ] ∪ [ 1 6 , 6 ] ; range: [ − 6 , − 1 6 ] ∪ [ 1 6 , 6 ] [ − 6 , − 1 6 ] ∪ [ 1 6 , 6 ]

domain: [ − 3 , ∞ ) ; [ − 3 , ∞ ) ; range: [ 0 , ∞ ) [ 0 , ∞ )

domain: ( − ∞ , ∞ ) ( − ∞ , ∞ )

f ( − 3 ) = 1 ; f ( − 2 ) = 0 ; f ( − 1 ) = 0 ; f ( 0 ) = 0 f ( − 3 ) = 1 ; f ( − 2 ) = 0 ; f ( − 1 ) = 0 ; f ( 0 ) = 0

f ( − 1 ) = − 4 ; f ( 0 ) = 6 ; f ( 2 ) = 20 ; f ( 4 ) = 34 f ( − 1 ) = − 4 ; f ( 0 ) = 6 ; f ( 2 ) = 20 ; f ( 4 ) = 34

f ( − 1 ) = − 5 ; f ( 0 ) = 3 ; f ( 2 ) = 3 ; f ( 4 ) = 16 f ( − 1 ) = − 5 ; f ( 0 ) = 3 ; f ( 2 ) = 3 ; f ( 4 ) = 16

domain: ( − ∞ , 1 ) ∪ ( 1 , ∞ ) ( − ∞ , 1 ) ∪ ( 1 , ∞ )

window: [ − 0.5 , − 0.1 ] ; [ − 0.5 , − 0.1 ] ; range: [ 4 , 100 ] [ 4 , 100 ]

window: [ 0.1 , 0.5 ] ; [ 0.1 , 0.5 ] ; range: [ 4 , 100 ] [ 4 , 100 ]

[ 0 , 8 ] [ 0 , 8 ]

Many answers. One function is f ( x ) = 1 x − 2 . f ( x ) = 1 x − 2 .

• ⓐ The fixed cost is $500.
• ⓑ The cost of making 25 items is $750.
• ⓒ The domain is [0, 100] and the range is [500, 1500].

3.3 Section Exercises

Yes, the average rate of change of all linear functions is constant.

The absolute maximum and minimum relate to the entire graph, whereas the local extrema relate only to a specific region around an open interval.

4 ( b + 1 ) 4 ( b + 1 )

4 x + 2 h 4 x + 2 h

− 1 13 ( 13 + h ) − 1 13 ( 13 + h )

3 h 2 + 9 h + 9 3 h 2 + 9 h + 9

4 x + 2 h − 3 4 x + 2 h − 3

increasing on ( − ∞ , − 2.5 ) ∪ ( 1 , ∞ ) , ( − ∞ , − 2.5 ) ∪ ( 1 , ∞ ) , decreasing on ( − 2.5 , 1 ) ( − 2.5 , 1 )

increasing on ( − ∞ , 1 ) ∪ ( 3 , 4 ) , ( − ∞ , 1 ) ∪ ( 3 , 4 ) , decreasing on ( 1 , 3 ) ∪ ( 4 , ∞ ) ( 1 , 3 ) ∪ ( 4 , ∞ )

local maximum: ( − 3 , 60 ) , ( − 3 , 60 ) , local minimum: ( 3 , − 60 ) ( 3 , − 60 )

absolute maximum at approximately ( 7 , 150 ) , ( 7 , 150 ) , absolute minimum at approximately ( −7.5 , −220 ) ( −7.5 , −220 )

Local minimum at ( 3 , − 22 ) , ( 3 , − 22 ) , decreasing on ( − ∞ , 3 ) , ( − ∞ , 3 ) , increasing on ( 3 , ∞ ) ( 3 , ∞ )

Local minimum at ( − 2 , − 2 ) , ( − 2 , − 2 ) , decreasing on ( − 3 , − 2 ) , ( − 3 , − 2 ) , increasing on ( − 2 , ∞ ) ( − 2 , ∞ )

Local maximum at ( − 0.5 , 6 ) , ( − 0.5 , 6 ) , local minima at ( − 3.25 , − 47 ) ( − 3.25 , − 47 ) and ( 2.1 , − 32 ) , ( 2.1 , − 32 ) , decreasing on ( − ∞ , − 3.25 ) ( − ∞ , − 3.25 ) and ( − 0.5 , 2.1 ) , ( − 0.5 , 2.1 ) , increasing on ( − 3.25 , − 0.5 ) ( − 3.25 , − 0.5 ) and ( 2.1 , ∞ ) ( 2.1 , ∞ )

b = 5 b = 5

2.7 gallons per minute

approximately –0.6 milligrams per day

3.4 Section Exercises

Find the numbers that make the function in the denominator g g equal to zero, and check for any other domain restrictions on f f and g , g , such as an even-indexed root or zeros in the denominator.

Yes. Sample answer: Let f ( x ) = x + 1 and  g ( x ) = x − 1. f ( x ) = x + 1 and  g ( x ) = x − 1. Then f ( g ( x ) ) = f ( x − 1 ) = ( x − 1 ) + 1 = x f ( g ( x ) ) = f ( x − 1 ) = ( x − 1 ) + 1 = x and g ( f ( x ) ) = g ( x + 1 ) = ( x + 1 ) − 1 = x . g ( f ( x ) ) = g ( x + 1 ) = ( x + 1 ) − 1 = x . So f ∘ g = g ∘ f . f ∘ g = g ∘ f .

( f + g ) ( x ) = 2 x + 6 , ( f + g ) ( x ) = 2 x + 6 , domain: ( − ∞ , ∞ ) ( − ∞ , ∞ )

( f − g ) ( x ) = 2 x 2 + 2 x − 6 , ( f − g ) ( x ) = 2 x 2 + 2 x − 6 , domain: ( − ∞ , ∞ ) ( − ∞ , ∞ )

( f g ) ( x ) = − x 4 − 2 x 3 + 6 x 2 + 12 x , ( f g ) ( x ) = − x 4 − 2 x 3 + 6 x 2 + 12 x , domain: ( − ∞ , ∞ ) ( − ∞ , ∞ )

( f g ) ( x ) = x 2 + 2 x 6 − x 2 , ( f g ) ( x ) = x 2 + 2 x 6 − x 2 , domain: ( − ∞ , − 6 ) ∪ ( − 6 , 6 ) ∪ ( 6 , ∞ ) ( − ∞ , − 6 ) ∪ ( − 6 , 6 ) ∪ ( 6 , ∞ )

( f + g ) ( x ) = 4 x 3 + 8 x 2 + 1 2 x , ( f + g ) ( x ) = 4 x 3 + 8 x 2 + 1 2 x , domain: ( − ∞ , 0 ) ∪ ( 0 , ∞ ) ( − ∞ , 0 ) ∪ ( 0 , ∞ )

( f − g ) ( x ) = 4 x 3 + 8 x 2 − 1 2 x , ( f − g ) ( x ) = 4 x 3 + 8 x 2 − 1 2 x , domain: ( − ∞ , 0 ) ∪ ( 0 , ∞ ) ( − ∞ , 0 ) ∪ ( 0 , ∞ )

( f g ) ( x ) = x + 2 , ( f g ) ( x ) = x + 2 , domain: ( − ∞ , 0 ) ∪ ( 0 , ∞ ) ( − ∞ , 0 ) ∪ ( 0 , ∞ )

( f g ) ( x ) = 4 x 3 + 8 x 2 , ( f g ) ( x ) = 4 x 3 + 8 x 2 , domain: ( − ∞ , 0 ) ∪ ( 0 , ∞ ) ( − ∞ , 0 ) ∪ ( 0 , ∞ )

( f + g ) ( x ) = 3 x 2 + x − 5 , ( f + g ) ( x ) = 3 x 2 + x − 5 , domain: [ 5 , ∞ ) [ 5 , ∞ )

( f − g ) ( x ) = 3 x 2 − x − 5 , ( f − g ) ( x ) = 3 x 2 − x − 5 , domain: [ 5 , ∞ ) [ 5 , ∞ )

( f g ) ( x ) = 3 x 2 x − 5 , ( f g ) ( x ) = 3 x 2 x − 5 , domain: [ 5 , ∞ ) [ 5 , ∞ )

( f g ) ( x ) = 3 x 2 x − 5 , ( f g ) ( x ) = 3 x 2 x − 5 , domain: ( 5 , ∞ ) ( 5 , ∞ )

• ⓑ f ( g ( x ) ) = 2 ( 3 x − 5 ) 2 + 1 f ( g ( x ) ) = 2 ( 3 x − 5 ) 2 + 1
• ⓒ f ( g ( x ) ) = 6 x 2 − 2 f ( g ( x ) ) = 6 x 2 − 2
• ⓓ ( g ∘ g ) ( x ) = 3 ( 3 x − 5 ) − 5 = 9 x − 20 ( g ∘ g ) ( x ) = 3 ( 3 x − 5 ) − 5 = 9 x − 20
• ⓔ ( f ∘ f ) ( − 2 ) = 163 ( f ∘ f ) ( − 2 ) = 163

f ( g ( x ) ) = x 2 + 3 + 2 , g ( f ( x ) ) = x + 4 x + 7 f ( g ( x ) ) = x 2 + 3 + 2 , g ( f ( x ) ) = x + 4 x + 7

f ( g ( x ) ) = x + 1 x 3 3 = x + 1 3 x , g ( f ( x ) ) = x 3 + 1 x f ( g ( x ) ) = x + 1 x 3 3 = x + 1 3 x , g ( f ( x ) ) = x 3 + 1 x

( f ∘ g ) ( x ) = 1 2 x + 4 − 4 = x 2 , ( g ∘ f ) ( x ) = 2 x − 4 ( f ∘ g ) ( x ) = 1 2 x + 4 − 4 = x 2 , ( g ∘ f ) ( x ) = 2 x − 4

f ( g ( h ( x ) ) ) = ( 1 x + 3 ) 2 + 1 f ( g ( h ( x ) ) ) = ( 1 x + 3 ) 2 + 1

• ⓐ ( g ∘ f ) ( x ) = − 3 2 − 4 x ( g ∘ f ) ( x ) = − 3 2 − 4 x
• ⓑ ( − ∞ , 1 2 ) ( − ∞ , 1 2 )
• ⓐ ( 0 , 2 ) ∪ ( 2 , ∞ ) ; ( 0 , 2 ) ∪ ( 2 , ∞ ) ;
• ⓑ ( − ∞ , − 2 ) ∪ ( 2 , ∞ ) ; ( − ∞ , − 2 ) ∪ ( 2 , ∞ ) ;
• ⓒ ( 0 , ∞ ) ( 0 , ∞ )

( 1 , ∞ ) ( 1 , ∞ )

sample: f ( x ) = x 3 g ( x ) = x − 5 f ( x ) = x 3 g ( x ) = x − 5

sample: f ( x ) = 4 x g ( x ) = ( x + 2 ) 2 f ( x ) = 4 x g ( x ) = ( x + 2 ) 2

sample: f ( x ) = x 3 g ( x ) = 1 2 x − 3 f ( x ) = x 3 g ( x ) = 1 2 x − 3

sample: f ( x ) = x 4 g ( x ) = 3 x − 2 x + 5 f ( x ) = x 4 g ( x ) = 3 x − 2 x + 5

sample: f ( x ) = x g ( x ) = 2 x + 6 f ( x ) = x g ( x ) = 2 x + 6

sample: f ( x ) = x 3 g ( x ) = ( x − 1 ) f ( x ) = x 3 g ( x ) = ( x − 1 )

sample: f ( x ) = x 3 g ( x ) = 1 x − 2 f ( x ) = x 3 g ( x ) = 1 x − 2

sample: f ( x ) = x g ( x ) = 2 x − 1 3 x + 4 f ( x ) = x g ( x ) = 2 x − 1 3 x + 4

f ( g ( 0 ) ) = 27 , g ( f ( 0 ) ) = − 94 f ( g ( 0 ) ) = 27 , g ( f ( 0 ) ) = − 94

f ( g ( 0 ) ) = 1 5 , g ( f ( 0 ) ) = 5 f ( g ( 0 ) ) = 1 5 , g ( f ( 0 ) ) = 5

18 x 2 + 60 x + 51 18 x 2 + 60 x + 51

g ∘ g ( x ) = 9 x + 20 g ∘ g ( x ) = 9 x + 20

( f ∘ g ) ( 6 ) = 6 ( f ∘ g ) ( 6 ) = 6 ; ( g ∘ f ) ( 6 ) = 6 ( g ∘ f ) ( 6 ) = 6

( f ∘ g ) ( 11 ) = 11 , ( g ∘ f ) ( 11 ) = 11 ( f ∘ g ) ( 11 ) = 11 , ( g ∘ f ) ( 11 ) = 11

A ( t ) = π ( 25 t + 2 ) 2 A ( t ) = π ( 25 t + 2 ) 2 and A ( 2 ) = π ( 25 4 ) 2 = 2500 π A ( 2 ) = π ( 25 4 ) 2 = 2500 π square inches

A ( 5 ) = π ( 2 ( 5 ) + 1 ) 2 = 121 π A ( 5 ) = π ( 2 ( 5 ) + 1 ) 2 = 121 π square units

• ⓐ N ( T ( t ) ) = 23 ( 5 t + 1.5 ) 2 − 56 ( 5 t + 1.5 ) + 1 N ( T ( t ) ) = 23 ( 5 t + 1.5 ) 2 − 56 ( 5 t + 1.5 ) + 1
• ⓑ 3.38 hours

3.5 Section Exercises

A horizontal shift results when a constant is added to or subtracted from the input. A vertical shifts results when a constant is added to or subtracted from the output.

A horizontal compression results when a constant greater than 1 is multiplied by the input. A vertical compression results when a constant between 0 and 1 is multiplied by the output.

For a function f , f , substitute ( − x ) ( − x ) for ( x ) ( x ) in f ( x ) . f ( x ) . Simplify. If the resulting function is the same as the original function, f ( − x ) = f ( x ) , f ( − x ) = f ( x ) , then the function is even. If the resulting function is the opposite of the original function, f ( − x ) = − f ( x ) , f ( − x ) = − f ( x ) , then the original function is odd. If the function is not the same or the opposite, then the function is neither odd nor even.

g ( x ) = | x - 1 | − 3 g ( x ) = | x - 1 | − 3

g ( x ) = 1 ( x + 4 ) 2 + 2 g ( x ) = 1 ( x + 4 ) 2 + 2

The graph of f ( x + 43 ) f ( x + 43 ) is a horizontal shift to the left 43 units of the graph of f . f .

The graph of f ( x - 4 ) f ( x - 4 ) is a horizontal shift to the right 4 units of the graph of f . f .

The graph of f ( x ) + 8 f ( x ) + 8 is a vertical shift up 8 units of the graph of f . f .

The graph of f ( x ) − 7 f ( x ) − 7 is a vertical shift down 7 units of the graph of f . f .

The graph of f ( x + 4 ) − 1 f ( x + 4 ) − 1 is a horizontal shift to the left 4 units and a vertical shift down 1 unit of the graph of f . f .

decreasing on ( − ∞ , − 3 ) ( − ∞ , − 3 ) and increasing on ( − 3 , ∞ ) ( − 3 , ∞ )

decreasing on ( 0 , ∞ ) ( 0 , ∞ )

g ( x ) = f ( x - 1 ) , h ( x ) = f ( x ) + 1 g ( x ) = f ( x - 1 ) , h ( x ) = f ( x ) + 1

f ( x ) = | x - 3 | − 2 f ( x ) = | x - 3 | − 2

f ( x ) = x + 3 − 1 f ( x ) = x + 3 − 1

f ( x ) = ( x - 2 ) 2 f ( x ) = ( x - 2 ) 2

f ( x ) = | x + 3 | − 2 f ( x ) = | x + 3 | − 2

f ( x ) = − x f ( x ) = − x

f ( x ) = − ( x + 1 ) 2 + 2 f ( x ) = − ( x + 1 ) 2 + 2

f ( x ) = − x + 1 f ( x ) = − x + 1

The graph of g g is a vertical reflection (across the x x -axis) of the graph of f . f .

The graph of g g is a vertical stretch by a factor of 4 of the graph of f . f .

The graph of g g is a horizontal compression by a factor of 1 5 1 5 of the graph of f . f .

The graph of g g is a horizontal stretch by a factor of 3 of the graph of f . f .

The graph of g g is a horizontal reflection across the y y -axis and a vertical stretch by a factor of 3 of the graph of f . f .

g ( x ) = | − 4 x | g ( x ) = | − 4 x |

g ( x ) = 1 3 ( x + 2 ) 2 − 3 g ( x ) = 1 3 ( x + 2 ) 2 − 3

g ( x ) = 1 2 ( x - 5 ) 2 + 1 g ( x ) = 1 2 ( x - 5 ) 2 + 1

The graph of the function f ( x ) = x 2 f ( x ) = x 2 is shifted to the left 1 unit, stretched vertically by a factor of 4, and shifted down 5 units.

The graph of f ( x ) = | x | f ( x ) = | x | is stretched vertically by a factor of 2, shifted horizontally 4 units to the right, reflected across the horizontal axis, and then shifted vertically 3 units up.

The graph of the function f ( x ) = x 3 f ( x ) = x 3 is compressed vertically by a factor of 1 2 . 1 2 .

The graph of the function is stretched horizontally by a factor of 3 and then shifted vertically downward by 3 units.

The graph of f ( x ) = x f ( x ) = x is shifted right 4 units and then reflected across the vertical line x = 4. x = 4.

3.6 Section Exercises

Isolate the absolute value term so that the equation is of the form | A | = B . | A | = B . Form one equation by setting the expression inside the absolute value symbol, A , A , equal to the expression on the other side of the equation, B . B . Form a second equation by setting A A equal to the opposite of the expression on the other side of the equation, − B . − B . Solve each equation for the variable.

The graph of the absolute value function does not cross the x x -axis, so the graph is either completely above or completely below the x x -axis.

The distance from x to 8 can be represented using the absolute value statement: ∣ x − 8 ∣ = 4.

∣ x − 10 ∣ ≥ 15

There are no x-intercepts.

(−4, 0) and (2, 0)

( 0 , − 4 ) , ( 4 , 0 ) , ( − 2 , 0 ) ( 0 , − 4 ) , ( 4 , 0 ) , ( − 2 , 0 )

( 0 , 7 ) , ( 25 , 0 ) , ( − 7 , 0 ) ( 0 , 7 ) , ( 25 , 0 ) , ( − 7 , 0 )

range: [ – 400 , 100 ] [ – 400 , 100 ]

There is no solution for a a that will keep the function from having a y y -intercept. The absolute value function always crosses the y y -intercept when x = 0. x = 0.

| p − 0.08 | ≤ 0.015 | p − 0.08 | ≤ 0.015

| x − 5.0 | ≤ 0.01 | x − 5.0 | ≤ 0.01

3.7 Section Exercises

Each output of a function must have exactly one output for the function to be one-to-one. If any horizontal line crosses the graph of a function more than once, that means that y y -values repeat and the function is not one-to-one. If no horizontal line crosses the graph of the function more than once, then no y y -values repeat and the function is one-to-one.

Yes. For example, f ( x ) = 1 x f ( x ) = 1 x is its own inverse.

Given a function y = f ( x ) , y = f ( x ) , solve for x x in terms of y . y . Interchange the x x and y . y . Solve the new equation for y . y . The expression for y y is the inverse, y = f − 1 ( x ) . y = f − 1 ( x ) .

f − 1 ( x ) = x − 3 f − 1 ( x ) = x − 3

f − 1 ( x ) = 2 − x f − 1 ( x ) = 2 − x

f − 1 ( x ) = − 2 x x − 1 f − 1 ( x ) = − 2 x x − 1

domain of f ( x ) : [ − 7 , ∞ ) ; f − 1 ( x ) = x − 7 f ( x ) : [ − 7 , ∞ ) ; f − 1 ( x ) = x − 7

domain of f ( x ) : [ 0 , ∞ ) ; f − 1 ( x ) = x + 5 f ( x ) : [ 0 , ∞ ) ; f − 1 ( x ) = x + 5

a. f ( g ( x ) ) = x f ( g ( x ) ) = x and g ( f ( x ) ) = x . g ( f ( x ) ) = x . b. This tells us that f f and g g are inverse functions

  f ( g ( x ) ) = x , g ( f ( x ) ) = x   f ( g ( x ) ) = x , g ( f ( x ) ) = x

not one-to-one

[ 2 , 10 ] [ 2 , 10 ]

f − 1 ( x ) = ( 1 + x ) 1 / 3 f − 1 ( x ) = ( 1 + x ) 1 / 3

f − 1 ( x ) = 5 9 ( x − 32 ) . f − 1 ( x ) = 5 9 ( x − 32 ) . Given the Fahrenheit temperature, x , x , this formula allows you to calculate the Celsius temperature.

t ( d ) = d 50 , t ( d ) = d 50 , t ( 180 ) = 180 50 . t ( 180 ) = 180 50 . The time for the car to travel 180 miles is 3.6 hours.

Review Exercises

f ( − 3 ) = − 27 ; f ( − 3 ) = − 27 ; f ( 2 ) = − 2 ; f ( 2 ) = − 2 ; f ( − a ) = − 2 a 2 − 3 a ; f ( − a ) = − 2 a 2 − 3 a ; − f ( a ) = 2 a 2 − 3 a ; − f ( a ) = 2 a 2 − 3 a ; f ( a + h ) = − 2 a 2 + 3 a − 4 a h + 3 h − 2 h 2 f ( a + h ) = − 2 a 2 + 3 a − 4 a h + 3 h − 2 h 2

x = − 1.8 x = − 1.8 or or  x = 1.8 or  x = 1.8

− 64 + 80 a − 16 a 2 − 1 + a = − 16 a + 64 − 64 + 80 a − 16 a 2 − 1 + a = − 16 a + 64

( − ∞ , − 2 ) ∪ ( − 2 , 6 ) ∪ ( 6 , ∞ ) ( − ∞ , − 2 ) ∪ ( − 2 , 6 ) ∪ ( 6 , ∞ )

increasing ( 2 , ∞ ) ; ( 2 , ∞ ) ; decreasing ( − ∞ , 2 ) ( − ∞ , 2 )

increasing ( − 3 , 1 ) ; ( − 3 , 1 ) ; constant ( − ∞ , − 3 ) ∪ ( 1 , ∞ ) ( − ∞ , − 3 ) ∪ ( 1 , ∞ )

local minimum ( − 2 , − 3 ) ; ( − 2 , − 3 ) ; local maximum ( 1 , 3 ) ( 1 , 3 )

( − 1.8 , 10 ) ( − 1.8 , 10 )

( f ∘ g ) ( x ) = 17 − 18 x ; ( g ∘ f ) ( x ) = − 7 − 18 x ( f ∘ g ) ( x ) = 17 − 18 x ; ( g ∘ f ) ( x ) = − 7 − 18 x

( f ∘ g ) ( x ) = 1 x + 2 ; ( g ∘ f ) ( x ) = 1 x + 2 ( f ∘ g ) ( x ) = 1 x + 2 ; ( g ∘ f ) ( x ) = 1 x + 2

( f ∘ g ) ( x ) = 1 + x 1 + 4 x ,   x ≠ 0 ,   x ≠ − 1 4 ( f ∘ g ) ( x ) = 1 + x 1 + 4 x ,   x ≠ 0 ,   x ≠ − 1 4

( f ∘ g ) ( x ) = 1 x , x > 0 ( f ∘ g ) ( x ) = 1 x , x > 0

sample: g ( x ) = 2 x − 1 3 x + 4 ; f ( x ) = x g ( x ) = 2 x − 1 3 x + 4 ; f ( x ) = x

f ( x ) = | x − 3 | f ( x ) = | x − 3 |

f ( x ) = 1 2 | x + 2 | + 1 f ( x ) = 1 2 | x + 2 | + 1

f ( x ) = − 3 | x − 3 | + 3 f ( x ) = − 3 | x − 3 | + 3

f − 1 ( x ) = x - 9 10 f − 1 ( x ) = x - 9 10

f − 1 ( x ) = x - 1 f − 1 ( x ) = x - 1

The function is one-to-one.

Practice Test

The relation is a function.

The graph is a parabola and the graph fails the horizontal line test.

2 a 2 − a 2 a 2 − a

− 2 ( a + b ) + 1 − 2 ( a + b ) + 1

f − 1 ( x ) = x + 5 3 f − 1 ( x ) = x + 5 3

( − ∞ , − 1.1 ) and  ( 1.1 , ∞ ) ( − ∞ , − 1.1 ) and  ( 1.1 , ∞ )

( 1.1 , − 0.9 ) ( 1.1 , − 0.9 )

f ( 2 ) = 2 f ( 2 ) = 2

f ( x ) = { | x | if x ≤ 2 3 if x > 2 f ( x ) = { | x | if x ≤ 2 3 if x > 2

x = 2 x = 2

f − 1 ( x ) = − x − 11 2 f − 1 ( x ) = − x − 11 2

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Eureka Math Grade 3 Module 3 Lesson 17 Answer Key

Students of Grade 3 can get a strong foundation by referring to the Eureka Math Book. It was developed by highly professional mathematics educators and the solutions prepared by them are in a concise manner for easy grasping. To achieve high scores in Grade 3, students need to solve all questions and exercises included in Eureka Grade 3 Book.

Engage NY Eureka Math 3rd Grade Module 3 Lesson 17 Answer Key

Teachers and students can find this Eureka Book Answer Key for Grade 3 more helpful in raising students’ scores and supporting teachers to educate the students. Learning activities are the best option to educate elementary school kids and make them understand the basic mathematical concepts like addition, subtraction, multiplication, division, etc. Grade 3 elementary school students can find these fun-learning exercises for all math concepts through the Eureka Book.

Eureka Math Grade 3 Module 3 Lesson 17 Problem Set Answer Key

a. Color all the squares with even products orange. Can an even product ever have an odd factor?

Answer: 1 x 1 = 1, 1 x 2 = 2, 1 x 3 = 3, 1 x 4 = 4, 1 x 5 = 5, 1 x 6 = 6, 1 x 7 = 8, 1 x 8 = 8, 2 x 1 = 2, 2 x 2 = 4, 2 x 3 = 6, 2 x 4 = 8, 2 x 5 = 10, 2 x 6 = 12, 2 x 7 = 14, 2 x 8 = 16, 3 x 1 = 3, 3 x 2 = 6, 3 x 3 = 9, 3 x 4 = 12, 3 x 5 = 15, 3 x 6 = 18, 3 x 7 = 21, 3 x 8 = 24, 4 x 1 = 4, 4 x 2 = 8, 4 x 3 = 12, 4 x 4 = 16, 4 x 5 = 20, 4 x 6 = 24, 4 x 7 = 28, 4 x 8 = 32, 5 x 1 = 5, 5 x 2 = 10, 5 x 3 = 15, 5 x 4 = 20, 5 x 5 = 25, 5 x 6 = 30, 5 x 7 = 35, 5 x 8 = 40, 6 x 1 = 6, 6 x 2 = 12, 6 x 3 = 18, 6 x 4 = 24, 6 x 5 = 30, 6 x 6 = 36, 6 x 7 = 42, 6 x 8 = 48, 7 x 1 = 7, 7 x 2 = 14, 7 x 3 = 21, 7 x 4 = 28, 7 x 5 = 35, 7 x 6 = 42, 7 x 7 = 49, 7 x 8 = 56, 8 x 1 = 8, 8 x 2 = 16, 8 x 3 = 24, 8 x 4 = 32, 8 x 5 = 40, 8 x 6 = 48, 8 x 7 = 56, 8 x 8 = 64.

b. Can an odd product ever have an even factor?

Answer: No, an odd product ever has an even factor.

Explanation: In the above-given question, given that, 1 x 1 = 1, 1 x 2 = 2, 1 x 3 = 3, 1 x 4 = 4, 1 x 5 = 5, 1 x 6 = 6, 1 x 7 = 8, 1 x 8 = 8, 2 x 1 = 2, 2 x 2 = 4, 2 x 3 = 6, 2 x 4 = 8, 2 x 5 = 10, 2 x 6 = 12, 2 x 7 = 14, 2 x 8 = 16, 3 x 1 = 3, 3 x 2 = 6, 3 x 3 = 9, 3 x 4 = 12, 3 x 5 = 15, 3 x 6 = 18, 3 x 7 = 21, 3 x 8 = 24, 4 x 1 = 4, 4 x 2 = 8, 4 x 3 = 12, 4 x 4 = 16, 4 x 5 = 20, 4 x 6 = 24, 4 x 7 = 28, 4 x 8 = 32, 5 x 1 = 5, 5 x 2 = 10, 5 x 3 = 15, 5 x 4 = 20, 5 x 5 = 25, 5 x 6 = 30, 5 x 7 = 35, 5 x 8 = 40, 6 x 1 = 6, 6 x 2 = 12, 6 x 3 = 18, 6 x 4 = 24, 6 x 5 = 30, 6 x 6 = 36, 6 x 7 = 42, 6 x 8 = 48, 7 x 1 = 7, 7 x 2 = 14, 7 x 3 = 21, 7 x 4 = 28, 7 x 5 = 35, 7 x 6 = 42, 7 x 7 = 49, 7 x 8 = 56, 8 x 1 = 8, 8 x 2 = 16, 8 x 3 = 24, 8 x 4 = 32, 8 x 5 = 40, 8 x 6 = 48, 8 x 7 = 56, 8 x 8 = 64. No, an odd product ever has an even factor.

c. Everyone knows that 7 × 4 = (5 × 4) + (2 × 4). Explain how this is shown in the table.

Answer: 7 x 4 = 28. 5 x 4 = 20. 2 x 4 = 8. 20 + 8 = 28.

Explanation: In the above-given question, given that, 7 x 4 = 28. 5 x 4 = 20. 2 x 4 = 8. 20 + 8 = 28.

d. Use what you know to find the product of 7 × 16 or 8 sevens + 8 sevens.

Answer: 7 x 16 = 112.

Explanation: In the above-given question, given that, 7 x 6 = 112. 8 sevens + 8 sevens. 8 x 7 + 8 x 7. 56 + 56. 112.

Answer: 1 x 2 = 2, 1 x 3 = 3, 1 x 4 = 4, 1 x 5 = 5, 1 x 6 = 6, 2 x 1 = 2, 2 x 3 = 6, 2 x 4 = 8, 2 x 5 = 10, 2 x 6 = 12, 3 x 1 = 3, 3 x 2 = 6, 3 x 4 = 12, 3 x 5 = 15, 3 x 6 = 18, 4 x 1 = 4, 4 x 2 = 8, 4 x 3 = 12, 4 x 5 = 20, 4 x 6 = 24, 5 x 1 = 5, 5 x 2 = 1, 5 x 3 = 15, 5 x 4 = 20, 5 x 6 = 30, 6 x 1 = 6, 6 x 2 = 12, 6 x 3 = 18, 6 x 4 = 24, 6 x 5 = 30.

b. Draw an array to match each expression in the table below. Then, label the number of squares you added to make each new array. The first two arrays have been done for you.

Answer: 1 + 2 = 3. 3 + 6 = 9. 6 + 10 = 16. 10 + 15 = 25. 15 + 21 = 36.

c. What pattern do you notice in the number of squares that are added to each new array?

Answer: 1 + 3 = 4. 4 + 6 = 9. 6 + 10 = 16. 10 + 15 = 25. 15 + 21 = 36.

Explanation: In the above-given question, given that, 1 + 3 = 4. 4 + 6 = 9. 6 + 10 = 16. 10 + 15 = 25. 15 + 21 = 36.

d. Use the pattern you discovered in Part (b) to prove this: 9 × 9 is the sum of the first 9 odd numbers.

Answer: 9 x 9 = 81.

Explanation: In the above-given question, given that, 9 x 9 = 81. the sum of the first 9 odd numbers is 81. 9 x 9 = 81.

Eureka Math Grade 3 Module 3 Lesson 17 Exit Ticket Answer Key

Question 1. Use what you know to find the product of 8 × 12 or 6 eights + 6 eights.

Answer: 8 x 12 = 96.

Explanation: In the above-given question, given that, 8 x 12 = 96. 6 eights + 6 eights. 6 x 8 + 6 x 8. 48 + 48. 96.

Question 2. Luis says 3 × 233 = 626. Use what you learned about odd times odd to explain why Luis is wrong.

Answer: 3 x 233 = 626.

Explanation: In the above-given question, given that, 3 x 233 = 626. odd x odd = odd. but in the given question, Luis is wrong. 3 x 233 = 626. 625 + 1.

Eureka Math Grade 3 Module 3 Lesson 17 Homework Answer Key

b. Color the rows and columns with even factors yellow.

Answer: The even factors are color with yellow.

c. What do you notice about the factors and products that are left unshaded

Answer: The factors that are left unshaded is odd numbers.

Explanation: In the above-given question, given that, the factors that are left unshaded is odd. the factors that are in diagonals is even.

d. Complete the chart by filling in each blank and writing an example for each rule.

Answer: odd times odd equals = odd. even times even equals = even. even times odd equals = even.

Explanation: In the above-given question, given that, odd times odd equals = odd. even times even equals = even. even times odd equals = even. 9 x 9 = 81. 2 x 2 = 4. 4 x 3 = 12.

e. Explain how 7 × 6 = (5 × 6) + (2 × 6) is shown in the table

Answer: 7 x 6 = 42.

Explanation: In the above-given question, given that, 7 x 6 = 42. 5 x 6 = 30. 2 x 6 = 12. ( 5 x 6 ) + (2 x 6 ). 30 + 12 = 42.

f. Use what you know to find the product of 4 × 16 or 8 fours + 8 fours.

Answer: 4 x 16 = 64.

Explanation: In the above-given question, given that, 4 x 16 = 64. 8 fours + 8 fours. 8 x 4 + 8 x 4. 32 + 32. 64.

Question 2. Today in class, we found that n × n is the sum of the first n odd numbers. Use this pattern to find the value of n for each equation below. The first is done for you. a. 1 + 3 + 5 = n × n

b. 1 + 3 + 5 + 7 = n × n

Answer: 16 = 4 x 4.

Explanation: In the above-given question, given that, 16 = 4 x 4. 1 + 3 + 5 + 7 = n x n. n = 4. 4 x 4 = 16. 1 + 3 + 5 + 7 = 16.

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