Instructor Checklist for Piecewise Function Projects

Let us represent the information using a Table of Values ... for 0 to 100 meals
Two variables: rental cost and number of meals
The rental cost depends on the number of meals
This implies that:
the rental cost is the dependent variable = y variable
the number of meals is the independent variable = x variable

The Table of Values is:

Number of meals	Rental cost ($\$$)
$x$	$y$
$0$	$500$
$20$	$400$
$40$	$300$
$60$	$200$
$80$	$100$
$100$	$0$

Based on the graphs, Options: F, G, and H satify the Table of Values
However, based on the wording in the question: All 20 meals in a group must be ordered for each reduction in the rental fee:
the correct answer is Option F because the increments must be in 20's only (no in-between values)

The value in the y-axis (the fare) for the closed circle value in the x-axis (the miles)

$ For\;\;a\;\;5-mile\;\;trip: \\[3ex] ABC\;\;Cabs:\;\;fare = \$12 \\[3ex] Tary\;\;Taxicabs:\;\;\; fare = \$9 \\[3ex] Cheaper\;\;fare = \$9 $

This is a Piecewise Function where the values of the output for each domain are contants
It is very evident that Option E is the correct option because it is a Piecewise Function graph that denotes the fact that each output is a constant.
But if you do want me to explain further:
Remember that the closed circle value is the value that actually corresponds to each output
So, $x = 2$ corresponds to $y = \$5.05$
Looking at the graphs in Options A, B, C, and D; when $x = 2$, $y \lt 5.05$
So, those options are eliminated.
When you review the graph in Option E; that graph corresponds to the Table of Values
When $x = 2$, $y = 5.05$
When $x = 3$, $y = 5.70$
When $x = 4$, $y = 6.30$

Let us draw the Table of Values for each function/piece
Then, we shall answer the question by going through each option and eliminating the incorrect options until we find the correct one

First Piece: $x^2 - 2; \enspace x \le 1$

$x$	$-2$	$-1$	$0$	$1$
$y$	$2$	$-1$	$-2$	$-1$
				defined

Second Piece: $x - 7; \enspace 1 \lt x \lt 5$

$x$	$1$	$2$	$3$	$4$	$5$
$y$	$-6$	$-5$	$-4$	$-3$	$-2$
	undefined				undefined

Third Piece: $4 - x; \enspace x \ge 5$

$x$	$5$	$6$	$7$	$8$
$y$	$-1$	$1$	$2$	$3$
	defined

Option F.
First Piece: when $x = 1$, $y = -3$
This is incorrect...as seen from the Table.

Option J.
First Piece: when $x = 1$, $y = 0$
This is incorrect...as seen from the Table.

Option G.
First Piece: when $x = -2$, $y = 0$
This is incorrect...as seen from the Table.

Option K.
This is the correct option because all the Tables of Values corresponds to the graph
This includes where the values are defined (closed circles) and undefined (open circles)
You may STOP here (or if you have time, check the last option)

Option H.
First Piece: when $x = 0$, $y = 0$
This is incorrect...as seen from the Table.

(a.) The Table of Values is:

$y = (x - 2)^2$; $1 \le x \lt 3$
x	y
1 2 3 (undefined)	1 0 1
$y = -3x + 13$; $3 \le x \le 7$
x	y
3 (defined) 4 5 6 7 (defined)	4 1 -2 -5 -8

The graph of the Piecewise Function is:

(b.) The domain of the first piece is: D = [1, 3)
The domain of the second piece is: D = [3, 7]
The domain of the function is: D = [1, 7]

Based on the graph:
(c.) The function has an absolute minimum, f(7) = −8
(d.) The function has an absolute maximum, f(3) = 4

(a.) The domain for the first piece is all real numbers besides zero.
The domain for the second piece is zero.
This implies that the domain for the piecewise function is all real numbers.
D = (−∞, ∞)

(b.) y-intercept
set x = 0 and solve for y
use the domain that includes x = 0
2nd piece: x = 0; y = 3
y-intercept = (0, 3)

x-intercept
set y = 0 and solve for x for each piece
If the value of x is in the domain for that piece, it is the x-intercept; otherwise it is not.

$ \underline{1st\;\;piece} \\[3ex] y = 2x \\[3ex] 2x = y \\[3ex] y = 0 \\[3ex] 2x = 0 \\[3ex] x = \dfrac{0}{2} \\[5ex] x = 0 \\[3ex] $ But 0 is not in that domain

$ \underline{2nd\;\;piece} \\[3ex] y = 3\;\;when\;\;x = 0...this\;\;is\;\;y-intercept \\[3ex] $ There are no x-intercepts.

$y = 2x$; $x \ne 0$
x	y
−2 −1 0 (undefined) 1 2	−4 −2 0 2 4
$y = 3$; $x = 0$
x	y
0 (defined)	3

(c.) The graph of the piecewise function is:



(d.) Based on the graph:
The straight line graph covers both the positive and negative values of y but notice that there is a jump discontinuity (open circle) at the origin (0, 0)
So, 0 is not included in the range
Hence, the range is all real numbers besides 0.
R = (−∞ 0) ⋃ (0, ∞)

Beginning from the Left: 1st Piece
Domain
D = [−3, 0]
D = {x | −3 ≤ x ≤ 0}

$ \underline{Function} \\[3ex] Points = (0,0) \;\;and\;\; (-3, 2) \\[3ex] x_1 = 0 \hspace{5em} x_2 = -3 \\[3ex] y_1 = 0 \hspace{5em} y_2 = 2 \\[3ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] = \dfrac{2 - 0}{-3 - 0} \\[5ex] = \dfrac{2}{-3} \\[5ex] = -\dfrac{2}{3} \\[5ex] y - y_1 = m(x - x_1) \\[3ex] y - 0 = -\dfrac{2}{3}(x - 0) \\[5ex] y = -\dfrac{2}{3}x \\[5ex] $ Right: 2nd Piece
Domain
D = (0, 2]
D = {x | 0 < x ≤ 2}

$ \underline{Function} \\[3ex] Points = (0,0) \;\;and\;\; (2, 3) \\[3ex] x_1 = 0 \hspace{5em} x_2 = 2 \\[3ex] y_1 = 0 \hspace{5em} y_2 = 3 \\[3ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] = \dfrac{3 - 0}{2 - 0} \\[5ex] = \dfrac{3}{2} \\[5ex] y - y_1 = m(x - x_1) \\[3ex] y - 0 = \dfrac{3}{2}(x - 0) \\[5ex] y = \dfrac{3}{2}x \\[5ex] $ The piecewise function is:

$ f(x) = \begin{cases} -\dfrac{2}{3}x; \hspace{4em} -3 \le x \le 0 \\[5ex] \dfrac{3}{2}x; \hspace{5em} 0 \lt x \le 2 \end{cases} $

(a.) The domain for the first piece is all real numbers less than 1.
The domain for the second piece is all real numbers greater than or equal to 1.
This implies that the domain for the piecewise function is all real numbers.
D = (−∞, ∞)

(b.) y-intercept
set x = 0 and solve for y
0 lies in the domain for the first piece, so we shall use the function in the first piece.

$ f(x) = -2x + 3 \\[3ex] x = 0 \\[3ex] f(x) = -2(0) + 3 \\[3ex] f(x) = 0 + 3 \\[3ex] f(x) = 3 \\[3ex] $ y-intercept = (0, 3)

x-intercept
set y = 0 and solve for x for each piece
If the value of x is in the domain for that piece, it is the x-intercept; otherwise it is not.

$ \underline{1st\;\;piece} \\[3ex] y = -2x + 3 \\[3ex] 2x = 3 - y \\[3ex] x = \dfrac{3 - y}{2} \\[5ex] y = 0 \\[3ex] x = \dfrac{3 - 0}{2} \\[5ex] x = \dfrac{3}{2} \\[5ex] But:\;\;\dfrac{3}{2} \nless 1 \\[5ex] $ $\dfrac{3}{2}$ is not in that domain.

$ \underline{2nd\;\;piece} \\[3ex] y = 2x - 1 \\[3ex] y + 1 = 2x \\[3ex] 2x = y + 1 \\[3ex] x = \dfrac{y + 1}{2} \\[5ex] y = 0 \\[3ex] x = \dfrac{0 + 1}{2} \\[5ex] x = \dfrac{1}{2} \\[5ex] But:\;\;\dfrac{1}{2} \ngeq 1 \\[5ex] $ $\dfrac{1}{2}$ is not in that domain.
There are no x-intercepts.

$y = -2x + 3$; $x \lt 1$
x	y
−2 −1 0 1 (undefined)	7 5 3 1
$y = 2x - 1$; $x \ge 1$
x	y
1 (defined) 2 3	1 3 5

(c.) The graph of the piecewise function is:



(d.) Based on the graph:
1 is the minimum y-value
(x, y) = (1, 1)
It is closed in the domain for the second piece.
The range are the positive numbers greater than or equal to 1.
R = [1, ∞)

Beginning from the Half Left: 1st Piece
Domain
It is open at 0
It is closed at -5
The domain includes the real numbers from -5 (included) to 0 (excluded)
D = [-5, 0)
D = {x | -5 ≤ x < 0}

$ \underline{Function} \\[3ex] Points = (-5,0) \;\;and\;\; (-2, 3) \\[3ex] x_1 = -5 \hspace{5em} x_2 = -2 \\[3ex] y_1 = 0 \hspace{5em} y_2 = 3 \\[3ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] = \dfrac{3 - 0}{-2 - (-5)} \\[5ex] = \dfrac{3}{-2 + 5} \\[5ex] = \dfrac{3}{3} \\[5ex] = 1 \\[3ex] y - y_1 = m(x - x_1) \\[3ex] y - 0 = 1(x - (-5)) \\[3ex] y = x + 5 \\[3ex] $ Other Half on the Right: 2nd Piece
Domain
0 is included in the domain.
It then extends till positive infinity as seen from the arrow
So, all real numbers from zero to positive infinity are included in the domain.
D = [0, ∞]
D = {x | x ≥ 0}

$ \underline{Function} \\[3ex] Points = (0,0) \;\;and\;\; (3, 3) \\[3ex] x_1 = 0 \hspace{5em} x_2 = 3 \\[3ex] y_1 = 0 \hspace{5em} y_2 = 3 \\[3ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] = \dfrac{3 - 0}{3 - 0} \\[5ex] = \dfrac{3}{3} \\[5ex] = 1 \\[3ex] y - y_1 = m(x - x_1) \\[3ex] y - 0 = 1(x - 0) \\[3ex] y = 1(x) \\[3ex] y = x \\[3ex] $ The piecewise function is:

$ f(x) = \begin{cases} x + 5; \hspace{3em} -5 \le x \lt 0 \\[3ex] x; \hspace{5em} x \ge 0 \end{cases} $

(a.) The domain for the first piece are the real numbers from −2 (included) to 1 (excluded)
The domain for the second piece is 1.
The domain for the third piece are the real numbers greater than 1
This implies that the domain for the piecewise function are the real numbers from −2 (included) to infinity.
D = [−2, ∞)

(b.) y-intercept
set x = 0 and solve for y
0 lies in the domain for the first piece, so we shall use the function in the first piece.

$ f(x) = x + 3 \\[3ex] x = 0 \\[3ex] f(x) = 0 + 3 \\[3ex] f(x) = 3 \\[3ex] $ y-intercept = (0, 3)

x-intercept
set y = 0 and solve for x for each piece
If the value of x is in the domain for that piece, it is the x-intercept; otherwise it is not.

$ \underline{1st\;\;piece} \\[3ex] y = x + 3 \\[3ex] x + 3 = y \\[3ex] x = y - 3 \\[3ex] y = 0 \\[3ex] x = 0 - 3 \\[3ex] x = -3 \\[3ex] But:\;\;-3 \lt -2 \\[3ex] -2 \gt -3 \\[3ex] $ −3 is not in that domain

$ \underline{2nd\;\;piece} \\[3ex] y = 5 \\[3ex] x = 1 \\[3ex] y \ne 0 \\[3ex] $ This piece is not applicable.

$ \underline{3rd\;\;piece} \\[3ex] y = -x + 2 \\[3ex] x = 2 - y \\[3ex] y = 0 \\[3ex] x = 2 - 0 \\[3ex] x = 2 \\[3ex] 2 \gt 1 \\[3ex] $ 2 is in the domain.
x-intercept = (2, 0)

$y = x + 3$; $-2 \le x \lt 1$
x	y
−2 (defined) −1 0 1 (undefined)	1 2 3 4
$y = 5$; $x = 1$
x	y
1 (defined)	5
$y = -x + 2$; $x \gt 1$
x	y
1 (undefined) 2 3	1 0 −1

(c.) The graph of the piecewise function is:



(d.) Based on the graph:
Based on the 2nd table:
5 is the maximum y-value and it is defined.

Based on the 1st table:
1 is defined
4 is undefined
Values between 1 (included) and 4 (excluded) are included.
Also, there is no continuity from 4 to 5

Based on the 3rd table:
1 is not defined (not included)
All other values less than 1 are included.
This implies that the range real numbers from negative infinity up to 4 and the set, 5
R = [−∞, 4) ⋃ {5}

(a.) The domain for the first piece is all real numbers less than 0.
The domain for the second piece is all real numbers greater than or equal to 0.
This implies that the domain for the piecewise function is all real numbers.
D = (−∞, ∞)

(b.) y-intercept
set x = 0 and solve for y
0 lies in the domain for the second piece, so we shall use the function in the first piece.

$ y = x^2 \\[3ex] x = 0 \\[3ex] y = 0^2 \\[3ex] y = 0 \\[3ex] $ y-intercept = (0, 0)
This is also the x-intercept

x-intercept
set y = 0 and solve for x for each piece
If the value of x is in the domain for that piece, it is the x-intercept; otherwise it is not.

$ \underline{1st\;\;piece} \\[3ex] y = 2 + 2x \\[3ex] y - 2 = 2x \\[3ex] 2x = y - 2 \\[3ex] x = \dfrac{y - 2}{2} \\[5ex] y = 0 \\[3ex] x = \dfrac{0 - 2}{2} \\[5ex] x = -\dfrac{2}{2} \\[5ex] x = -1 \\[3ex] $ −1 is in the domain.
x-intercept = (-1, 0)

$ \underline{2nd\;\;piece} \\[3ex] y = x^2 \\[3ex] x^2 = y \\[3ex] x = \sqrt{y} \\[3ex] y = 0 \\[3ex] x = \sqrt{0} \\[3ex] x = 0 \\[3ex] $ 0 is in the domain.
x-intercept = (0, 0)
This is also the y-intercept

$y = 2 + 2x$; $x \lt 0$
x	y
−2 −1 0 (undefined)	−2 0 2
$y = x^2$; $x \ge 0$
x	y
0 (defined) 1 2	0 1 4

(c.) The graph of the piecewise function is:



(d.) Based on the graph:
On the left:
Real numbers from negative infinity up to 2 are included.
But 2 is excluded.

On the right:
Real numbers from 0 to infinity are included.
0 is included.
2 is also included.

Therefore, the range are all the real numbers.
R = (−∞, ∞)

Beginning from the Half Left: 1st Piece
Domain
0 is included in the domain.
It then extends till negative infinity as seen from the arrow
So, all real numbers from negative infinity to zero are included in the domain.
D = (−∞, 0]
D = {x | x ≤ 0}

$ \underline{Function} \\[3ex] Points = (0,0) \;\;and\;\; (-2, 2) \\[3ex] x_1 = 0 \hspace{5em} x_2 = -2 \\[3ex] y_1 = 0 \hspace{5em} y_2 = 2 \\[3ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] = \dfrac{2 - 0}{-2 - 0} \\[5ex] = \dfrac{2}{-2} \\[5ex] = -1 \\[3ex] y - y_1 = m(x - x_1) \\[3ex] y - 0 = -1(x - 0) \\[3ex] y = -x \\[3ex] $ Other Half on the Right: 2nd Piece
Domain
It is open at 0
It is closed at 3
The domain includes the real numbers from 0 (excluded) to 3 (included)
D = (0, 3]
D = {x | 0 < x ≤ 3}

$ \underline{Function} \\[3ex] Points = (3,0) \;\;and\;\; (2, 1) \\[3ex] x_1 = 3 \hspace{5em} x_2 = 2 \\[3ex] y_1 = 0 \hspace{5em} y_2 = 1 \\[3ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] = \dfrac{1 - 0}{2 - 3} \\[5ex] = \dfrac{1}{-1} \\[5ex] = -1 \\[3ex] y - y_1 = m(x - x_1) \\[3ex] y - 0 = -1(x - 3) \\[3ex] y = -x + 3 \\[3ex] y = 3 - x \\[3ex] $ The piecewise function is:

$ f(x) = \begin{cases} -x; \hspace{4em} x \le 0 \\[3ex] 3 - x; \hspace{3em} 0 \lt x \le 3 \end{cases} $

(a.) The domain for the first piece are the real numbers from −2 (included) to 1 (excluded)
The domain for the second piece is all real numbers greater than or equal to 0.
This implies that the domain for the piecewise function are the real numbers from −2 (included) to infinity.
D = [−2, ∞)

(b.) y-intercept
set x = 0 and solve for y
0 lies in the domain for the second piece, so we shall use the function in the first piece.

$ y = x^3 \\[3ex] x = 0 \\[3ex] y = 0^3 \\[3ex] y = 0 \\[3ex] $ y-intercept = (0, 0)
This is also the x-intercept

x-intercept
set y = 0 and solve for x for each piece
If the value of x is in the domain for that piece, it is the x-intercept; otherwise it is not.

$ \underline{1st\;\;piece} \\[3ex] y = |x| \\[3ex] |x| = y \\[3ex] |x| = 0 \\[3ex] x = 0 \;\;OR\;\; -x = 0 \\[3ex] x = 0 \\[3ex] $ But: 0 is not in the domain.

$ \underline{2nd\;\;piece} \\[3ex] y = x^3 \\[3ex] x^3 = y \\[3ex] x = \sqrt[3]{y} \\[3ex] y = 0 \\[3ex] x = \sqrt[3]{0} \\[3ex] x = 0 \\[3ex] $ 0 is in the domain.
x-intercept = (0, 0)
This is also the y-intercept

$y = |x|$; $-2 \le x \lt 0$
x	y
−2 −1 0 (undefined)	2 1 0
$y = x^3$; $x \ge 0$
x	y
0 (defined) 1 2	0 1 8

(c.) The graph of the piecewise function is:



(d.) Based on the graph:
The range does not have any negative number.
0 is defined for the second piece. So, 0 is included. All positive numbers are included.
Therefore, the range are all non-negative numbers.
R = [0, ∞)