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Exercises 1 A piecewise function is function that is made of two or more equations. An example of a piecewise function can be see below.Each equation is given for a specific domain and the domains cannot overlap. In this case, our function would be written as: f(x)=⎩⎪⎨⎪⎧-21x,-3,-x,-6≤x<00≤x≤33<x≤6 A step function also follows these rules, it is a specific type of piecewise function. A step function's one additional rule is that each segment must be a horizontal line.Each equation in a step function is given for a specific domain and the domains cannot overlap, just as in the piecewise function. In this case, our function would be written as: f(x)=⎩⎪⎨⎪⎧3,-3,-6,-6≤x<00≤x≤33<x≤6 | |

Exercises 2 Let's begin by graphing the function f(x)=∣x∣.We can think of this graph as two lines, one with a slope of -1 and one with a slope of 1, meeting at the point (0,0).These two lines have the equations f(x)=-x and f(x)=x. We cannot have any overlap in the domains though. We need to write the piecewise function as: f(x)={-x,x,x<0x≥0 Note, we could also write our domain as: f(x)={-x,x,x≤0x<0 It doesn't matter which restriction has "or equal to 0" as long as one and only one of them has it. We need our overall domain to have no gaps but we also cannot have any overlaps. | |

Exercises 3 When evaluating a piecewise function, the first thing we need to do is determine into which interval the input falls. This time, we want to find f(-3). This means that we need the interval that contains x=-3. f(x)={5x−1,x+3,if x<-2if x≥-2 The interval x<-2 contains -3, so we need to substitute the value into the first equation of the function. f(-3)=5(-3)−1⇒f(-3)=-16 | |

Exercises 4 When evaluating a piecewise function, the first thing we need to do is determine into which interval the input falls. This time, we want to find f(-2). This means that we need the interval that contains x=-2. f(x)={5x−1,x+3,if x<-2if x≥-2 The interval x≥-2 contains -2, so we need to substitute the value into the second equation of the function. f(-2)=-2+3⇒f(-2)=1 | |

Exercises 5 When evaluating a piecewise function, the first thing we need to do is determine into which interval the input falls. This time, we want to find f(0). This means that we need the interval that contains x=0. f(x)={5x−1,x+3,if x<-2if x≥-2 The interval x≥-2 contains 0, so we need to substitute the value into the second equation of the function. f(0)=0+3⇒f(0)=3 | |

Exercises 6 When evaluating a piecewise function, the first thing we need to do is determine into which interval the input falls. This time, we want to find f(5). This means that we need the interval that contains x=5. f(x)={5x−1,x+3,if x<-2if x≥-2 The interval x≥-2 contains 5, so we need to substitute the value into the second equation of the function. f(5)=5+3⇒f(5)=8 | |

Exercises 7 When evaluating a piecewise function, the first thing we need to do is determine into which interval the input falls. This time, we want to find g(-4). This means that we need the interval that contains x=-4. g(x)=⎩⎪⎨⎪⎧-x+4,3,2x−5,if x≤-1if -1<x<2if x≥2 The interval x≤-1 contains -4, so we need to substitute the value into the first equation of the function. g(-4)=-(-4)+4⇒g(-4)=8 | |

Exercises 8 When evaluating a piecewise function, the first thing we need to do is determine into which interval the input falls. This time, we want to find g(-1). This means that we need the interval that contains x=-1. g(x)=⎩⎪⎨⎪⎧-x+4,3,2x−5,if x≤-1if -1<x<2if x≥2 The interval x≤-1 contains -1, so we need to substitute the value into the first equation of the function. g(-1)=-(-1)+4⇒g(-1)=5 | |

Exercises 9 When evaluating a piecewise function, the first thing we need to do is determine into which interval the input falls. This time, we want to find g(0). This means that we need the interval that contains x=0. g(x)=⎩⎪⎨⎪⎧-x+4,3,2x−5,if x≤-1if -1<x<2if x≥2 The interval -1<x<2 contains 0, so we need to substitute the value into the second equation of the function. g(0)=3 | |

Exercises 10 When evaluating a piecewise function, the first thing we need to do is determine into which interval the input falls. This time, we want to find g(1). This means that we need the interval that contains x=1. g(x)=⎩⎪⎨⎪⎧-x+4,3,2x−5,if x≤-1if -1<x<2if x≥2 The interval -1<x<2 contains 1, so we need to substitute the value into the second equation of the function. g(1)=3 | |

Exercises 11 When evaluating a piecewise function, the first thing we need to do is determine into which interval the input falls. This time, we want to find g(2). This means that we need the interval that contains x=2. g(x)=⎩⎪⎨⎪⎧-x+4,3,2x−5,if x≤-1if -1<x<2if x≥2 The interval x≥2 contains 2, so we need to substitute the value into the third equation of the function. g(2)=2(2)−5⇒g(2)=-1 | |

Exercises 12 When evaluating a piecewise function, the first thing we need to do is determine into which interval the input falls. This time, we want to find g(5). This means that we need the interval that contains x=5. g(x)=⎩⎪⎨⎪⎧-x+4,3,2x−5,if x≤-1if -1<x<2if x≥2 The interval x≥2 contains 5, so we need to substitute the value into the third equation of the function. g(5)=2(5)−5⇒g(5)=5 | |

Exercises 13 It is given that the following piece-wise function can be used to represent d, the total distance traveled in x hours. d(x)={55x,65x−20,0≤x≤22<x≤5 To determine the distance traveled in 4 hours, we can substitute x=4 into d(x)=65x−20. This is because x=4 is a value within the domain of the second piece of the function rule. d(x)=65x−20x=4d(4)=65⋅4−20Multiplyd(4)=260−20Subtract termsd(4)=240 You travel 240 miles in 4 hours. | |

Exercises 14 We are given the following piecewise function. c(x)=⎩⎨⎧17x+20,15.80x+2014x+20,if 0≤25if 25≤x≤50if x≥50 In order to determine the total cost of ordering 26 shirts, we have to use the middle function, since 26 is greater than 25 and less than 50. c(x)=⎩⎨⎧17x+20,15.80x+2014x+20,if 0≤25if 25≤x≤50if x≥50 Let's find c(26) by substituting 26 into 15.80x+20. c(x)=15.80x+20x=26c(26)=15.80(26)+20Multiplyc(26)=410.8+20Add termsc(26)=430.8 In the context of the word problem this means that the total cost of ordering 26 shirts is equal to $430.8. | |

Exercises 15 To graph the given piecewise function, we should think about the graph of each individual piece of the function. Then we can combine the graphs on one coordinate plane.y=-x First we will graph y=-x for the domain x<2. This function has a slope of -1 and a y-intercept of 0. Since the endpoint is not included, this piece should end with an open circle.Looking at the graph, we can see that all of the possible y-values are greater than -2.y=x−6 Next, we will graph y=x−6 for the domain x≥2. Since the endpoint is included, we will end the piece with a closed circle.From the graph, we can see that all y-values that are greater than or equal to -4 will be produced by this portion.Combining the Pieces Finally, we can combine the pieces onto to one coordinate plane.Looking at the pieces together, we can see that there are no gaps in the possible values of x. We can also see there are no gaps in the possible values of y but that they are all greater than or equal to -4. We can use these facts to write the domain and range of the function. Domain: Range: All real numbers y≥-4 | |

Exercises 16 To graph the given piecewise function, we should think about the graph of each individual piece of the function. Then we can combine the graphs on one coordinate plane.y=2x First we will graph y=2x for the domain x≤-3. This function has a slope of 2 and a y-intercept of 0. Since the endpoint is included, this piece should end with a closed circle.Looking at the graph, we can see that all of the possible y-values are less than or equal to -6.y=-2x Next, we will graph y=-2x for the domain x>-3. Since the endpoint is not included, we will end the piece with an open circle.From the graph, we can see that all y-values that are less than 6 will be produced by this portion.Combining the Pieces Finally, we can combine the pieces onto to one coordinate plane.Looking at the pieces together, we can see that there are no gaps in the possible values of x. We can also see there are no gaps in the possible values of y but that they are all less than 6. We can use these facts to write the domain and range of the function. Domain: Range: All real numbers y<6 | |

Exercises 17 To graph the given piecewise function, we should think about the graph of each individual piece of the function. Then we can combine the graphs on one coordinate plane.y=-3x−2 First we will graph y=-3x−2 for the domain x≤-1. This function has a slope of -3 and a y-intercept of -2. Since the endpoint is included, this piece should end with a closed circle.Looking at the graph, we can see that all of the possible y-values are greater than or equal to 1.y=x+2 Next, we will graph y=x+2 for the domain x>-1. Since the endpoint is not included, we will end the piece with an open circle.From the graph, we can see that all y-values that are greater than 1 will be produced by this portion.Combining the Pieces Finally, we can combine the pieces onto to one coordinate plane.Looking at the pieces together, we can see that there are no gaps in the possible values of x. We can also see there are no gaps in the possible values of y but that they are all greater than or equal to 1. We can use these facts to write the domain and range of the function. Domain: Range: All real numbers y≥1 | |

Exercises 18 To graph the given piecewise function, we should think about the graph of each individual piece of the function. Then we can combine the graphs on one coordinate plane.y=x+8 First we will graph y=x+8 for the domain x<4. This function has a slope of 1 and a y-intercept of 8. Since the endpoint is not included, this piece should end with an open circle.Looking at the graph, we can see that all of the possible y-values are less than 12.y=4x−4 Next, we will graph y=4x−4 for the domain x≥4. Since the endpoint is included, we will end the piece with a closed circle.From the graph, we can see that all y-values that are less than 0 will be produced by this portion.Combining the Pieces Finally, we can combine the pieces onto to one coordinate plane.Looking at the pieces together, we can see that there are no gaps in the possible values of x. We can also see there are no gaps in the possible values of y. We can use these facts to write the domain and range of the function. Domain: Range: All real numbers All real numbers | |

Exercises 19 To graph the given piecewise function, we should think about the graph of each individual piece of the function. Then we can combine the graphs on one coordinate plane.Graphing y=1 First we will graph y=1 for the domain x<-3. This function has a slope of 0, which means that it is constant and its graph is a horizontal line. We also know that its y-intercept is 1. Since the endpoint is not included, this piece should end with an open circle.Looking at the graph, we can see that all of the possible y-values are equal to 1.Graphing y=x−1 Next, we will graph y=x−1 for the domain -3≤x≤3. Since both endpoints are included, we will end the piece with closed circles.From the graph, we can see that all y-values that are greater than or equal to -4 and less than or equal to 2 will be produced by this portion.Graphing y=-2x+4 Now, we will graph y=-2x+4 for the domain x>3. Since the endpoint is not included, this piece should end with an open circle.From the graph, we can see that all y-values that are less than -2 will be produced by this portion.Combining the Pieces Finally, we can combine the pieces onto to one coordinate plane.Looking at the pieces together, we can see that there are no gaps in the possible values of x. We can also see there are no gaps in the possible values of y but that they are all less than or equal to 2. We can use these facts to write the domain and range of the function. Domain: Range: All real numbers y≤2 | |

Exercises 20 To graph the given piecewise function, we should think about the graph of each individual piece of the function. Then we can combine the graphs on one coordinate plane.y=2x+1 First we will graph y=2x+1 for the domain x≤-1. This function has a slope of 2 and a y-intercept of 1. Since the endpoint is included, this piece should end with a closed circle.Looking at the graph, we can see that all of the possible y-values are less than or equal to -1.y=-x+2 Next, we will graph y=-x+2 for the domain -1<x<2. Since both endpoints are not included, we will end the piece with open circles.From the graph, we can see that all y-values that are less than 3 and greater than 0 will be produced by this portion.y=-3 Now, we will graph y=-3 for the domain x≥2. Notice that this function is constant so its graph is a horizontal line. Since the endpoint is included, we will end the piece with a closed circle.Looking at the graph, we can see that all of the possible y-values are equal to -3.Combining the Pieces Finally, we can combine the pieces onto to one coordinate plane.Looking at the pieces together, we can see that there are no gaps in the possible values of x. However, we can see that there is a gap between the first and the second pieces of the function. Our range will include all possible values of y that are not a part of this gap. Domain: Range: All real numbers 0<y<3 or y≤-1 | |

Exercises 21 The function given is f(x)={2x−3x+8x<5x≥5 Notice that each piece of the function rule is defined for different values of x. For values of x that are less than 5, 2x−3 represents the function. For values of x that are greater than or equal to 5 x+8 represents the function. From the work provided, it is apparent that f(5) is being determined. We can see that 2x−3 was used for x=5. This violates the given domain restrictions. As mentioned above, x+8 represents the function when x=5. Thus, this is the piece that should be used. We will solve for f(5) correctly below. f(x)=x+8x=5f(5)=5+8Add termsf(5)=13 | |

Exercises 22 The given function rule is f(x)={x+61x≤-2x>-2s From the given domain restrictions we can see that x=-2 is the point at which the definition of the function changes. Notice that f(x)=x+6 defines f(x) for values of x that are less than or equal to -2. This means the graph for this piece of the function should show a closed dot at x=-2. Similarly, the graph for the second piece of the function should show an open dot at -2, because f(x)=1 defines the function for values of x that are greater than -2. This is where the error lies. The wrong type of dots were used. The correct graph is shown below. | |

Exercises 23 To write the piecewise function for the given graph, we need to find the equation of the line representing each piece and then restrict the domain accordingly.The First Piece Let's begin by finding the slope and y-intercept of the first piece of the function.This line has a slope of 1 and the y-intercept is 2. We can write the equation for this piece in slope-intercept form. y=1x+2⇔y=x+2The Second Piece Now let's take a look at the second piece.Notice that the graph is a part of a horizontal line. We know that a horizontal line has a slope of 0. The y-intercept of our line is 2. We can write the equation for this piece in slope-intercept form. y=0x+2⇔y=2Combining the Pieces We can add the equations of these lines to the piecewise function notation. f(x)={x+22 Finally, we need to determine the domain for each equation. The change occurs at x=0, so this will be where the domains are divided. We cannot have an overlap in our domains so we need to take notice of where the closed and open circles are located.Notice that there is no "jump" in our function. Therefore, when dividing the domains, it doesn't matter where we assign the point of change, as long as it belongs to only one of them. This means that we can correctly write the piecewise function in two different ways. y={x+2,2,if x<0if x≥0ory={x+2,2,if x≤0if x>0 | |

Exercises 24 To write the piecewise function for the given graph, we need to find the equation of the line representing each piece and then restrict the domain accordingly.The First Piece Let's begin by finding the slope and y-intercept of the first piece of the function.Notice that the graph is a part of a horizontal line. We know that horizontal lines have a slope of 0. The y-intercept of our line is -3. We can write the equation for this piece in slope-intercept form. y=0x+(-3)⇔y=-3The Second Piece Now let's take a look at the second piece.This line has a slope of -3 and the y-intercept is 3. We can write the equation for this piece in slope-intercept form. y=-3x+3Combining the Pieces We can add the equations of these lines to the piecewise function notation. f(x)={-3-3x+3 Finally, we need to determine the domain for each equation. The "jump" occurs at x=0, so this will be where the domains are divided. We cannot have an overlap in our domains so we need to take notice of where the closed and open circles are located.The closed circle at the end of the first piece tells us that its domain includes 0. The open circle at the beginning of the second piece tells us that its domain does not include 0. f(x)={-3,-3x+3,if x≤0if x>0 | |

Exercises 25 To write the piecewise function for the given graph, we need to find the equation of the line representing each piece and then restrict the domain accordingly.The First Piece Let's begin by finding the slope and y-intercept of the first piece of the function.This line has a slope of -1 and the y-intercept is 0. We can write the equation for this piece in slope-intercept form. y=-1x+0⇔y=-xThe Second Piece Now let's take a look at the second piece. To help us find the y-intercept we can extend the line to the y-axis.This line has a slope of -1 and the y-intercept is 1. We can write the equation for this piece in slope-intercept form. y=-1x+1⇔y=-x+1Combining the Pieces We can add the equations of these lines to the piecewise function notation. f(x)={-x-x+1 Finally, we need to determine the domain for each equation. The "jump" occurs at x=4, so this will be where the domains are divided. We cannot have an overlap in our domains so we need to take notice of where the closed and open circles are located.The open circle at the end of the first piece tells us that its domain does not include 4. The closed circle at the beginning of the second piece tells us that its domain includes 4. f(x)={-x,-x+1,if x<4if x≥4 | |

Exercises 26 To write the piecewise function for the given graph, we need to find the equation of the line representing each piece and then restrict the domain accordingly.The First Piece Let's begin by finding the slope and y-intercept of the first piece of the function. To help us find the y-intercept we can extend the line to the y-axis.This line has a slope of 2 and the y-intercept is 2. We can write the equation for this piece in slope-intercept form. y=2x+2The Second Piece Now let's take a look at the second piece.This line has a slope of 21 and the y-intercept is -1. We can write the equation for this piece in slope-intercept form. y=21x+(-1)⇔y=21x−1Combining the Pieces We can add the equations of these lines to the piecewise function notation. f(x)={2x+221x−1 Finally, we need to determine the domain for each equation. The change occurs at x=-2, so this will be where the domains are divided. We cannot have an overlap in our domains so we need to take notice of where the closed and open circles are located.Notice that there is no "jump" in our function. Therefore, when dividing the domains, it doesn't matter where we assign the point of change, as long as it belongs to only one of them. This means that we can correctly write the piecewise function in two different ways. y={2x+2,21x−1,if x≤-2if x>-2ory={2x+2,21x−1,if x<-2if x≥-2 | |

Exercises 27 To write the piecewise function for the given graph, we need to find the equation of the line representing each piece and then restrict the domain accordingly.The First Piece Let's begin by finding the slope and y-intercept of the first piece of the function. To help us find the y-intercept we can extend the line to the y-axis.Notice that the graph is a part of a horizontal line. We know that a horizontal line has a slope of 0. The y-intercept of our line is 1. We can write the equation for this piece in slope-intercept form. y=0x+1⇔y=1The Second Piece Now let's take a look at the second piece.This line has a slope of 2 and the y-intercept is 0. We can write the equation for this piece in slope-intercept form. y=2x+0⇔y=2xThe Third Piece Now let's take a look at the third piece.This line has a slope of -21 and the y-intercept is 2. We can write the equation for this piece in slope-intercept form. y=-21x+2Combining the Pieces We can add the equations of these lines to the piecewise function notation. f(x)=⎩⎪⎨⎪⎧12x-21x+2 Finally, we need to determine the domain for each equation. The "jumps" occur at x=-2 and at x=0, so this will be where the domains are divided. We cannot have an overlap in our domains so we need to take notice of where the closed and open circles are located.The closed circle at the end of the first piece tells us that its domain includes -2. The open circle at the beginning of the second piece tells us that its domain does not include -2. Similarly, we can conclude that 0 is included in the domain of the second piece. f(x)=⎩⎪⎨⎪⎧1,2x,-21x+2,if x≤-2if -2<x≤0if x>0 | |

Exercises 28 To write the piecewise function for the given graph, we need to find the equation of the line representing each piece and then restrict the domain accordingly.The First Piece Let's begin by finding the slope and y-intercept of the first piece of the function. To help us find the y-intercept we can extend the line to the y-axis.This line has a slope of 1 and the y-intercept is 4. We can write the equation for this piece in slope-intercept form. y=1x+4⇔y=x+4The Second Piece Now let's take a look at the second piece.This line has a slope of -41 and the y-intercept is -41. We can write the equation for this piece in slope-intercept form. y=-41x+(-41)⇔y=-41x−41The Third Piece Now let's take a look at the third piece. Again, we will extend the line to the y-axis.Notice that the graph is a part of a horizontal line. We know that horizontal lines have a slope of 0. The y–intercept of our line is -3. We can now write the equation for this piece in slope-intercept form. y=0x+(-3)⇔y=-3Combining the Pieces We can add the equations of these lines to the piecewise function notation. f(x)=⎩⎪⎨⎪⎧x+4-41x−41-3 Finally, we need to determine the domain for each equation. The "jumps" occur at x=-1 and x=3, so this will be where the domains are divided. We cannot have an overlap in our domains so we need to take notice of where the closed and open circles are located.The closed circle at the end of the first piece tells us that its domain includes -1. Open circles at the beginning and the end of the second piece tell us that its domain does not include -1 or 3. Similarly, we can tell that the domain of the last piece includes 3. f(x)=⎩⎪⎨⎪⎧x+4,-41x−41,-3,if x≤-1if -1<x<3if x≥3 | |

Exercises 29 To write the piecewise function for the given graph, we need to find the equation of the line representing each piece and then restrict the domain accordingly.The First Piece Let's begin by finding the slope and y-intercept of the first piece of the function. To help us find the y-intercept, we can extend the line to the y-axis.Notice that the graph is a part of a horizontal line. We know that horizontal lines have a slope of 0. The y-intercept of our line is -5. We can write the equation for this piece in slope-intercept form. y=0x+(-5)⇔y=-5The Second and Third Piece Now let's take a look at the second and third piece. Again, we can see that both of these pieces are horizontal. Therefore, we find their slope-intercept forms as we did for the first one.Equation in slope-intercept form for second piece. y=0x+(-3)⇔y=-3 And for third piece. y=0x+(-1)⇔y=-1Combining the Pieces We can add the equations of these lines to the piecewise function notation. f(x)=⎩⎪⎨⎪⎧-5-3-1 Finally, we need to determine the domain for each equation. The "jumps" occur at x=-3 and x=-1, so this will be where the domains are divided. We cannot have an overlap in our domains so we need to take notice of where the closed and open circles are located.The closed circle at the beginning of the first piece tells us that its domain includes -5. The open circle at the end of this piece tells us that its domain does not include -3. Similarly, we can find the domains of the remaining pieces. f(x)=⎩⎪⎨⎪⎧-5,-3,-1,if -5≤x<-3if -3≤x<-1if -1≤x<1 | |

Exercises 30 To write the piecewise function for the given graph, we need to find the equation of the line representing each piece and then restrict the domain accordingly.The First Piece Let's begin by finding the slope and y-intercept of the first piece of the function.Notice that the graph is a part of a horizontal line. We know that horizontal lines have a slope of 0. The y-intercept of our line is 4. We can write the equation for this piece in slope-intercept form. y=0x+4⇔y=4The Other Pieces Now let's take a look at the remaining pieces. Again, we can see that all of these pieces are horizontal. Therefore, we find their slope-intercept form as we did for the first one. To help us find the y-intercepts, we can extend the lines to the y-axis.Let's write the equations for these pieces in slope-intercept form. Second piece:y=0x+3⇔y=3Third piece: y=0x+2⇔y=2Fourth piece:y=0x+1⇔y=1Combining the Pieces We can add the equations of these lines to the piecewise function notation. f(x)=⎩⎪⎪⎪⎨⎪⎪⎪⎧4321 Finally, we need to determine the domain for each equation. The "jumps" occur at x=1, x=2 and x=3. This will be where the domains are divided. We cannot have an overlap in our domains so we need to take notice of where the closed and open circles are located.The open circle at the beginning of the first piece tells us that its domain does not include 0. The closed circle at the end of the first piece tells us that its domain includes 1. Similarly, we can find the domains of the remaining pieces. f(x)=⎩⎪⎪⎪⎨⎪⎪⎪⎧4,3,2,1,if 0<x≤1if 1<x≤2if 2<x≤3if 3<x≤4 | |

Exercises 31 Let's observe the given step function. f(x)=⎩⎪⎪⎪⎨⎪⎪⎪⎧3,4,5,6,if 0≤x<2if 2≤x<4if 4≤x<6if 6≤x<8Graphing the Function To think about how to draw the graph, let's look at the first piece of the function. The restriction on the domain tells us that f(x) equals 3 when x is greater than or equal to 0 and less than 2. f(x)=3 if 0≤x<2 To graph this, we draw a horizontal line at y=3 extending from x=0 to x=2. To indicate that x=0 is contained in the solution set, we place a closed circle at that point. x=2 is not contained in the solution set of this piece, which we indicate with an open circle.Following a similar process, we can graph the other pieces of the function.Domain and Range Now that we've graphed the function, we can describe its domain and range.Domain The domain of a function is the set of x-values for which the function is defined. From the graph (and the function rule), we can see that x can equal any value from 0 to 8, not including 8. 0≤x<8Range The range of a function is the set of y-values for which the function is defined. From the graph (and the function rule), we can see that y can only equal 3,4,5 and 6. {3,4,5,6} | |

Exercises 32 Let's observe the given step function. f(x)=⎩⎪⎪⎪⎨⎪⎪⎪⎧-4,-6,-8,-10,if 1<x≤2if 2<x≤3if 3<x≤4if 4<x≤5Graphing the Function To think about how to draw the graph, let's look at the first piece of the function. The restriction on the domain tells us that f(x) equals -4 when x is greater than 1 and less than or equal to 2. f(x)=-4 if 1<x≤2 To graph this, we draw a horizontal line at y=-4 extending from x=1 to x=2. To indicate that x=1 is not contained in the solution set, we place an open circle at that point. x=2 is contained in the solution set of this piece, which we indicate with a closed circle.Following a similar process, we can graph the other pieces of the function.Domain and Range Now that we've graphed the function, we can describe its domain and range.Domain The domain of a function is the set of x-values for which the function is defined. From the graph (and the function rule), we can see that x can equal any value from 1 to 5, not including 1. 1<x≤5Range The range of a function is the set of y-values for which the function is defined. From the graph (and the function rule), we can see that y can only equal -4,-6,-8 and -10. {-10,-8,-6,-4} | |

Exercises 33 Let's observe the given step function. f(x)=⎩⎪⎪⎪⎨⎪⎪⎪⎧9,6,5,1,if 1<x≤2if 2<x≤4if 4<x≤9if 9<x≤12Graphing the Function To think about how to draw the graph, let's look at the first piece of the function. The restriction on the domain tells us that f(x) equals 9 when x is greater than 1 and less than or equal to 2. f(x)=9 if 1<x≤2 To graph this, we draw a horizontal line at y=9 extending from x=1 to x=2. To indicate that x=1 is not contained in the solution set, we place an open circle at that point. x=2 is contained in the solution set of this piece, which we indicate with a closed circle.Following a similar process, we can graph the other pieces of the function.Domain and Range Now that we've graphed the function, we can describe its domain and range.Domain The domain of a function is the set of x-values for which the function is defined. From the graph (and the function rule), we can see that x can equal any value from 1 to 12, not including 1. 1<x≤12Range The range of a function is the set of y-values for which the function is defined. From the graph (and the function rule), we can see that y can only equal 9,6,5 and 1. {1,5,6,9} | |

Exercises 34 Let's observe the given step function. f(x)=⎩⎪⎪⎪⎨⎪⎪⎪⎧-2,-1,0,1,if -6≤x<-5if -5≤x<-3if -3≤x<-2if -2≤x<0Graphing the Function To think about how to draw the graph, let's look at the first piece of the function. The restriction on the domain tells us that f(x) equals -2 when x is greater than or equal to -6 and less than -5. f(x)=-2 if -6≤x<-5 To graph this, we draw a horizontal line at y=-2 extending from x=-6 to x=-5. To indicate that x=-6 is contained in the solution set, we place a closed circle at that point. x=-5 is not contained in the solution set of this piece, which we indicate with an open circle.Following a similar process, we can graph the other pieces of the function.Domain and Range Now that we've graphed the function, we can describe its domain and range.Domain The domain of a function is the set of x-values for which the function is defined. From the graph (and the function rule), we can see that x can equal any value from -6 to 0, not including 0. -6≤x<0Range The range of a function is the set of y-values for which the function is defined. From the graph (and the function rule), we can see that y can only equal -2,-1,0, and 1. {-2,-1,0,1} | |

Exercises 35 Following the steps in the given example, let's organize the information in the table.Number of playersTotal cost 0≤p≤5$180 6$210 7$240 8$270 9$300 Now we can build a piecewise function on the basis of the table. Note that we change the number of players for intervals, so it will be easier to graph it later. C(p)=⎩⎪⎪⎪⎪⎨⎪⎪⎪⎪⎧180,210240270300if 0<p≤5if 5<p≤6if 6<p≤7if 7<p≤8if 8<p≤9 Finally, we graph the function interval by interval paying extra attention that each of the intervals is half-open. | |

Exercises 36 We can organize the information in the table first, so we better understand the charging structure.Number of hoursTotal cost 1$4 2$8 3$12 4≤h$15 Now we build a piecewise function on the basis of the table. Note that we change the number o hours into intervals, so it will be easier to graph it later. C(p)=⎩⎪⎪⎨⎪⎪⎧4,8,12,15,if 0<p≤1if 1<p≤2if 2<p≤3if 4<p Finally, we graph the function interval by interval paying extra attention that each of the intervals is half-open. | |

Exercises 37 We are asked to write the absolute value function ∣x∣+1 as a piecewise function. Let's start by looking at the graph of our given function.As we can see, the behavior of the absolute value function changes at the vertex. Therefore, this is a good point to separate the graph into two pieces, both being straight lines. Our next step will then be to find the equation for both lines. For this, let's recall the slope-intercept form of an equation. y=mx+b In this form, m is the slope of the line and b is the y-intercept. Now, we can start with the line to the left of the vertex first. We can identify the y-intercept and slope from the graph shown below.For this line, the y-intercept is b=1 and the slope is m=-1. Therefore, the equation for it is y=-x+1. We will now do the same with the line to the right of the vertex.Then, for this second line, the y-intercept is b=1 and the slope is m=1. Therefore, the equation for it is y=x+1. Knowing the equations for both lines, we can now write the absolute value function ∣x∣+1 as the combination of two functions. y={-x+1-x+1 The final step will be deciding the domain of each piece. In this case, the pieces meet and change directions at x=0, so this is where our domain needs to change. The point at x=0 can belong to either piece, as long as it belongs to only one of them. This means that we can correctly write the piecewise function in two different ways. y={-x+1-x+1ififx≤0x>0ory={-x+1-x+1ififx<0x≥0 | |

Exercises 38 We are asked to write the absolute value function ∣x∣−3 as a piecewise function. Let's start by looking at the graph of our given function.As we can see, the behavior of the absolute value function changes at the vertex. Therefore, this is a good point to separate the graph into two pieces, both being straight lines. Our next step will then be to find the equation for both lines. For this, let's recall the slope-intercept form of an equation. y=mx+b In this form, m is the slope of the line and b is the y-intercept. Now, we can start with the line to the left of the vertex first. We can identify the y-intercept and slope from the graph shown below.For this line, the y-intercept is b=-3 and the slope is m=-1. Therefore, the equation for it is y=-x−3. We will now do the same with the line to the right of the vertex.Then, for this second line, the y-intercept is b=-3 and the slope is m=1. Therefore, the equation for it is y=x−3. Knowing the equations for both lines, we can now write the absolute value function ∣x∣−3 as the combination of two functions. y={-x−3-x−3 The final step will be deciding the domain of each piece. In this case, the pieces meet and change directions at x=0, so this is where our domain needs to change. The point at x=0 can belong to either piece, as long as it belongs to only one of them. This means that we can correctly write the piecewise function in two different ways. y={-x−3-x−3ififx≤0x>0ory={-x−3-x−3ififx<0x≥0 | |

Exercises 39 We are asked to write the absolute value function x−2 as a piecewise function. Let's start by looking at the graph of our given function.As we can see, the behavior of the absolute value function changes at the vertex. Therefore, this is a good point to separate the graph into two pieces, both being straight lines. Our next step will then be to find the equation for both lines. For this, let's recall the slope-intercept form of an equation. y=mx+b In this form, m is the slope of the line and b is the y-intercept. Now, we can start with the line to the left of the vertex first. We can identify the y-intercept and slope from the graph shown below.For this line, the y-intercept is b=2 and the slope is m=-1. Therefore, the equation for it is y=-x+2. We will now do the same with the line to the right of the vertex.Then, for this second line, the y-intercept is b=-2 and the slope is m=1. Therefore, the equation for it is y=x−2. Knowing the equations for both lines, we can now write the absolute value function ∣x−2∣ as the combination of two functions. y={-x+2-x−2 The final step will be deciding the domain of each piece. In this case, the pieces meet and change directions at x=2, so this is where our domain needs to change. The point at x=2 can belong to either piece, as long as it belongs to only one of them. This means that we can correctly write the piecewise function in two different ways. y={-x+2,-x−2,if x≤2if x>2ory={-x+2,-x−2,if x<2if x≥2 | |

Exercises 40 We are asked to write the absolute value function ∣x+5∣ as a piecewise function. Let's start by looking at the graph of our given function.As we can see, the behavior of the absolute value function changes at the vertex. Therefore, this is a good point to separate the graph into two pieces, both being straight lines. Our next step will then be to find the equation for both lines. For this, let's recall the slope-intercept form of an equation. y=mx+b In this form, m is the slope of the line and b is the y-intercept. Now, we can start with the line to the left of the vertex first. We can identify the y-intercept and slope from the graph shown below.For this line, the y-intercept is b=-5 and the slope is m=-1. Therefore, the equation for it is y=-x−5. We will now do the same with the line to the right of the vertex.Then, for this second line, the y-intercept is b=5 and the slope is m=1. Therefore, the equation for it is y=x+5. Knowing the equations for both lines, we can now write the absolute value function ∣x+5∣ as the combination of two functions. y={-x−5-x+5 The final step will be deciding the domain of each piece. In this case, the pieces meet and change directions at x=-5, so this is where our domain needs to change. The point at x=-5 can belong to either piece, as long as it belongs to only one of them. This means that we can correctly write the piecewise function in two different ways. y={-x−5,-x+5,if x≤-5if x>-5ory={-x−5,-x+5,if x<-5if x≥-5 | |

Exercises 41 We are asked to write the absolute value function 2∣x+3∣ as a piecewise function. Let's start by looking at the graph of our given function.As we can see, the behavior of the absolute value function changes at the vertex. Therefore, this is a good point to separate the graph into two pieces, both being straight lines. Our next step will then be to find the equation for both lines. For this, let's recall the slope-intercept form of an equation. y=mx+b In this form, m is the slope of the line and b is the y-intercept. Now, we can start with the line to the left of the vertex first. We can identify the y-intercept and slope from the graph shown below.For this line, the y-intercept is b=-6 and the slope is m=-2. Therefore, the equation for it is y=-2x−6. We will now do the same with the line to the right of the vertex.Then, for this second line, the y-intercept is b=6 and the slope is m=2. Therefore, the equation for it is y=2x+6. Knowing the equations for both lines, we can now write the absolute value function 2∣x+3∣ as the combination of two functions. y={-2x−6-2x+6 The final step will be deciding the domain of each piece. In this case, the pieces meet and change directions at x=-3, so this is where our domain needs to change. The point at x=-3 can belong to either piece, as long as it belongs to only one of them. This means that we can correctly write the piecewise function in two different ways. y={-2x−6,-2x+6,if x≤-3if x>-3ory={-2x−6,-2x+6,if x<-3if x≥-3 | |

Exercises 42 We are asked to write the absolute value function 4∣x−1∣ as a piecewise function. Let's start by looking a the graph of our given function.As we can see, the behavior of the absolute value function changes at the vertex. Therefore, this is a good point to separate the graph into two pieces, both being straight lines. Our next step will then be to find the equation for both lines. For this, let's recall the slope-intercept form of an equation. y=mx+b In this form, m is the slope of the line and b is the y-intercept. Now, we can start with the line to the left of the vertex first. We can identify the y-intercept and slope from the graph shown below.For this line, the y-intercept is b=4 and the slope is m=-4. Therefore, the equation for it is y=-4x+4. We will now do the same with the line to the right of the vertex.Then, for this second line, the y-intercept is b=-4 and the slope is m=4. Therefore, the equation for it is y=4x−4. Knowing the equations for both lines, we can now write the absolute value function 4∣x−1∣ as the combination of two functions. y={-4x+4-4x−4 The final step will be deciding the domain of each piece. In this case, the pieces meet and change directions at x=1, so this is where our domain needs to change. The point at x=1 can belong to either piece, as long as it belongs to only one of them. This means that we can correctly write the piecewise function in two different ways. y={-4x+4,-4x−4,if x≤1if x>1ory={-4x+4,-4x−4,if x<1if x≥1 | |

Exercises 43 We are asked to write the absolute value function -5∣x−8∣ as a piecewise function. Let's start by looking at the graph of our given function.As we can see, the behavior of the absolute value function changes at the vertex. Therefore, this is a good point to separate the graph into two pieces, both being straight lines. Our next step will then be to find the equation for both lines. For this, let's recall the slope-intercept form of an equation. y=mx+b In this form, m is the slope of the line and b is the y-intercept. Now, we can start with the line to the left of the vertex first. We can identify the y-intercept and slope from the graph shown below.For this line, the y-intercept is b=-40 and the slope is m=5. Therefore, the equation for it is y=5x−40. We will now do the same with the line to the right of the vertex.Then, for this second line, the y-intercept is b=40 and the slope is m=-5. Therefore, the equation for it is y=-5x+40. Knowing the equations for both lines, we can now write the absolute value function -5∣x−8∣ as the combination of two functions. y={-5x−40-5x+40 The final step will be deciding the domain of each piece. In this case, the pieces meet and change directions at x=8, so this is where our domain needs to change. The point at x=8 can belong to either piece, as long as it belongs to only one of them. This means that we can correctly write the piecewise function in two different ways. y={-5x−40,-5x+40,if x≤8if x>8ory={-5x−40,-5x+40,if x<8if x≥8 | |

Exercises 44 We are asked to write the absolute value function -3∣x+6∣ as a piecewise function. Let's start by looking at the graph of our given function.As we can see, the behavior of the absolute value function changes at the vertex. Therefore, this is a good point to separate the graph into two pieces, both being straight lines. Our next step will then be to find the equation for both lines. For this, let's recall the slope-intercept form of an equation. y=mx+b In this form, m is the slope of the line and b is the y-intercept. Now, we can start with the line to the left of the vertex first. We can identify the y-intercept and slope from the graph shown below.For this line, the y-intercept is b=18 and the slope is m=3. Therefore, the equation for it is y=3x+18. We will now do the same with the line to the right of the vertex.Then, for this second line, the y-intercept is b=-18 and the slope is m=-3. Therefore, the equation for it is y=-3x−18. Knowing the equations for both lines, we can now write the absolute value function -3∣x+6∣ as the combination of two functions. y={-3x+18-3x−18 The final step will be deciding the domain of each piece. In this case, the pieces meet and change directions at x=-6, so this is where our domain needs to change. The point at x=-6 can belong to either piece, as long as it belongs to only one of them. This means that we can correctly write the piecewise function in two different ways. y={-3x+18,-3x−18,if x≤-6if x>-6ory={-3x+18,-3x−18,if x<-6if x≥-6 | |

Exercises 45 We are asked to write the absolute value function -∣x−3∣+2 as a piecewise function. Let's start by looking at the graph of our given function.As we can see, the behavior of the absolute value function changes at the vertex. Therefore, this is a good point to separate the graph into two pieces, both being straight lines. Our next step will then be to find the equation for both lines. For this, let's recall the slope-intercept form of an equation. y=mx+b In this form, m is the slope of the line and b is the y-intercept. Now, we can start with the line to the left of the vertex first. We can identify the y-intercept and slope from the graph shown below.For this line, the y-intercept is b=-1 and the slope is m=1. Therefore, the equation for it is y=x−1. We will now do the same with the line to the right of the vertex.Then, for this second line, the y-intercept is b=5 and the slope is m=-1. Therefore, the equation for it is y=-x+5. Knowing the equations for both lines, we can now write the absolute value function -∣x−3∣+2 as the combination of two functions. y={x−1-x+5 The final step will be deciding the domain of each piece. In this case, the pieces meet and change directions at x=3, so this is where our domain needs to change. The point at x=3 can belong to either piece, as long as it belongs to only one of them. This means that we can correctly write the piecewise function in two different ways. y={x−1,-x+5,if x≤3if x>3ory={x−1,-x+5,if x<3if x≥3 | |

Exercises 46 We are asked to write the absolute value function 7∣x+1∣−5 as a piecewise function. Let's start by looking at the graph of our given function.As we can see, the behavior of the absolute value function changes at the vertex. Therefore, this is a good point to separate the graph into two pieces, both being straight lines. Our next step will then be to find the equation for both lines. For this, let's recall the slope-intercept form of an equation. y=mx+b In this form, m is the slope of the line and b is the y-intercept. Let's start with the line to the left of the vertex. We can identify the y-intercept and slope from the graph shown below.For this line, the y-intercept is b=-11 and the slope is m=-6.5. Therefore, the equation is y=-6.5x−11. We will now do the same with the line to the right of the vertex.For the second line, the y-intercept is b=2 and the slope is m=6.5. Therefore, the equation is y=6.5x+2. Knowing the equations for both lines, we can now write the absolute value function 7∣x+1∣−5 as the combination of two functions. y={-6.5x−116.5x+2 The final step will be deciding the domain of each piece. In this case, the pieces meet and change directions at x=-1, so this is where our domain needs to change. The point at x=-1 can belong to either piece, as long as it belongs to only one of them. This means that we can correctly write the piecewise function in two different ways. y={-6.5x−116.5x+2ififx≤-1x>-1ory={-6.5x−116.5x+2ififx<-1x≥-1 | |

Exercises 47 aSince the vertex of the reflected sunlight is (3,0), the function has the form f(x)=a∣x−3∣.bf(x)={-2x+6,-2x−6,if x<3if x≥3 | |

Exercises 48 | |

Exercises 49 aBefore we can evaluate f(-10), the functions value when x=-10, we need to determine into which interval it falls and then find the equation of that line. Because the "jump" occurs at x=3, we need to know if our value is less than, greater than, or equal to 3. Since -10<3, we know that we are evaluating using the equation for the left "piece." Now, we need to write an equation for this piece using the graph.We can see that the slope is 21 and the y-intercept is 1. We can write this in slope-intercept form as y=21x+1, which we can use to solve for f(-10) by substituting -10 for x. y=21x+1x=-10y=21(-10)+1 Simplify b1⋅a=bay=2-10+1Calculate quotienty=-5+1Add terms y=-4 Therefore, f(-10)=-4.bBefore we can evaluate f(8), we need to determine into which interval it falls and then find the equation of that line. Because the "jump" occurs at x=3, we need to know if our value is less than, greater than, or equal to 3. Since 8>3, we know that we are evaluating using the equation for the right "piece." Now, we need to write an equation for this piece using the graph.We can see that the slope is 1 but we cannot see what the y-intercept would be if the line was extended. We need to use a point on the line, such as (4,2), to solve for b. y=x+bx=4, y=22=4+bLHS−4=RHS−4-2=bRearrange equationb=-2 So we can write the equation for the right "piece" of the function as: y=x−2, which we can use to solve for f(8) by substituting 8 for x. y=x−2x=8y=8−2Subtract termy=6 Therefore, f(8)=6. | |

Exercises 50 aLet's begin by graphing the piecewise function as it has been given to us.The domain of this function is all real numbers. Thanks to the placement of the closed and open dots, there are no gaps in the possible values of x. The range of the function starts at negative infinity, as there are no lower endpoints, and goes up to, but not including, 4 because the open dot is on the highest point. Domain: Range: (-∞,∞)(-∞,4) Now we can graph the same function but with the signs changed so < becomes ≤ and ≥ becomes >.The domain of the function is still all real numbers, there are still no gaps in the possible values of x. The range of the function still starts at negative infinity, as there are still no lower endpoints, but it now goes up to, and includes, 4. Domain: Range: (-∞,∞)(-∞,4]bLet's begin by graphing the piecewise function as it has been given to us.Due to the placement of the open and closed dots, the domain of this function is all values less than, but not including, 1 and all values greater than or equal to 2. The range of the function starts at negative infinity, as there are no lower endpoints, and goes up to, but not including, 2. Domain: Range: (-∞,1)∪[2,∞)(-∞,2) Now we can graph the same function but with the signs changed so < becomes ≤ and ≥ becomes >.The domain and range will both change to reflect the switching of the open and closed dots. Domain: Range: (-∞,1]∪(2,∞)(-∞,2] | |

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##### Other subchapters in Writing Linear Functions

- Maintaining Mathematical Proficiency
- Mathematical Practices
- Writing Equations in Slope-Intercept Form
- Writing Equations in Point-Slope Form
- Writing Equations of Parallel and Perpendicular Lines
- Quiz
- Scatter Plots and Lines of Fit
- Analyzing Lines of Fit
- Arithmetic Sequences
- Chapter Review
- Chapter Test
- Cumulative Assessment