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Function definition and input-output rules

Learning goals

  • Determines whether a relation defines a function.
  • Uses function notation correctly.
  • Evaluates a function at given inputs.
  • Interprets inputs and outputs in context.

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Function definition and input-output rules 1
For each table, decide whether \(y\) is a function of \(x\). Briefly justify each answer. a) <table><tr><td>\(x\)</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td></tr><tr><td>\(y\)</td><td>\(7\)</td><td>\(14\)</td><td>\(21\)</td><td>\(28\)</td></tr></table> b) <table><tr><td>\(x\)</td><td>\(-2\)</td><td>\(-1\)</td><td>\(0\)</td><td>\(1\)</td><td>\(2\)</td></tr><tr><td>\(y\)</td><td>\(4\)</td><td>\(1\)</td><td>\(0\)</td><td>\(1\)</td><td>\(4\)</td></tr></table> c) <table><tr><td>\(x\)</td><td>\(5\)</td><td>\(8\)</td><td>\(5\)</td><td>\(10\)</td></tr><tr><td>\(y\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td></tr></table>

Hints

- A function assigns exactly one output to each input. - Repeated outputs are allowed. - Look for an input that appears with two different outputs.

Solution

1. a) Yes. Each input in \(\{1,2,3,4\}\) is paired with exactly one output. 2. b) Yes. Each input has exactly one output. Different inputs may share the same output without violating the definition of a function. 3. c) No. The input \(5\) is paired with both \(2\) and \(4\).

Answer

a) Yes; it is a function. b) Yes; it is a function. c) No; \(x=5\) has two different outputs.
Function definition and input-output rules 2
Determine whether each assignment is a function. Briefly justify each answer. a) Each calendar date in a year \(\rightarrow\) the daily high temperature recorded at one fixed weather station. b) Each vehicle owner \(\rightarrow\) each individual license plate number on a vehicle registered to that owner, with each plate number treated as a separate output. c) Each US state \(\rightarrow\) its state capital.

Hints

- A function must assign exactly one output to each input. - For each assignment, ask whether one input can be paired with two separate outputs. - Treat each individual plate number in part b as its own output.

Solution

1. a) This is a function because each date has one recorded daily-high value at the fixed station. 2. b) This is not necessarily a function because one owner may have several registered vehicles and therefore may be assigned several separate license-plate outputs. 3. c) This is a function because each US state has exactly one state capital.

Answer

a) Function b) Not a function c) Function
Function definition and input-output rules 3
A correspondence is shown with arrows from inputs to outputs: \(1\rightarrow4\) \(2\rightarrow5\) \(2\rightarrow6\) \(3\rightarrow7\) a) Is this correspondence a function from inputs to outputs? Explain. b) Which input causes the function rule to fail? c) Give one arrow that could be removed so the remaining correspondence is a function.

Hints

- Check the arrows leaving each input, one input at a time. - A function may send different inputs to the same output, but one input cannot point to two different outputs. - For part c, change only the arrows connected to the input that violates the rule.

Solution

1. A function must assign exactly one output to each input. 2. Input \(2\) points to both \(5\) and \(6\), so the correspondence is not a function. 3. Removing either \(2\rightarrow5\) or \(2\rightarrow6\) leaves input \(2\) with exactly one output, while inputs \(1\) and \(3\) already have one output each.

Answer

a) No. Input \(2\) has two outputs. b) \(2\) c) Either \(2\rightarrow5\) or \(2\rightarrow6\)
Function definition and input-output rules 4
A rule has these input-output pairs: <table><tr><th>Input</th><td>\(1\)</td><td>\(2\)</td><td>\(2\)</td><td>\(3\)</td></tr><tr><th>Output</th><td>\(5\)</td><td>\(7\)</td><td>\(9\)</td><td>\(11\)</td></tr></table> Is this a function? Explain using one input from the table.

Hints

- Check whether an input is listed more than once. - For a function, what must be true about the output of a particular input?

Solution

1. Input \(2\) is paired with both \(7\) and \(9\). 2. A function cannot assign two different outputs to the same input, so this is not a function.

Answer

No. Input \(2\) has two different outputs, \(7\) and \(9\).
Function definition and input-output rules 5
A vending machine assigns a price to each button: \(A\to\$2\), \(B\to\$2\), \(C\to\$3\). Does the assignment from button to price define a function? Explain.

Hints

- Look at the output of each button separately. - Check whether any single button is assigned more than one price.

Solution

1. Each button is assigned exactly one price. 2. Buttons \(A\) and \(B\) are allowed to share an output. This is a function.

Answer

Yes. Each button gives one price; different buttons may have the same price.
Function definition and input-output rules 6
The dots show a relation. Does the graph represent a function of \(x\)? Explain.
Figure for problem 559940

Hints

- Find points that lie directly above or below one another. - Think about how many outputs can belong to one input in a function.

Solution

1. The graph contains \((2,1)\) and \((2,3)\). 2. Both have input \(2\) but different outputs, so the relation is not a function.

Answer

No. At \(x=2\), the graph has two outputs, \(1\) and \(3\).
Function definition and input-output rules 7
A taxi company charges a base fare of \(\$3.50\) plus \(\$1.80\) per mile. The total cost for a trip of \(x\) miles is \(T(x)=3.50+1.80x\). Find the fare for each distance: a) \(5\) miles b) \(12\) miles c) \(0.5\) mile

Hints

- Substitute each distance for \(x\). - The base fare stays the same for every trip. - Multiply before adding.

Solution

1. For \(x=5\), \(T(5)=3.50+1.80\cdot5=3.50+9.00=12.50\). 2. For \(x=12\), \(T(12)=3.50+1.80\cdot12=3.50+21.60=25.10\). 3. For \(x=0.5\), \(T(0.5)=3.50+1.80\cdot0.5=3.50+0.90=4.40\).

Answer

a) \(\$12.50\) b) \(\$25.10\) c) \(\$4.40\)
Function definition and input-output rules 8
The function is \(T(x)=7x-5\). Complete the table. <table><tr><th>\(x\)</th><td>\(1\)</td><td>\(2\)</td><td>\(5\)</td><td>\(10\)</td><td>\(20\)</td></tr><tr><th>\(T(x)\)</th><td>\(\square\)</td><td>\(9\)</td><td>\(\square\)</td><td>\(65\)</td><td>\(\square\)</td></tr></table>

Hints

- Substitute the missing column's input for \(x\). - Multiply the input by \(7\) before subtracting \(5\). - Use the filled columns to check that your outputs are reasonable.

Solution

1. \(T(1)=7\cdot1-5=2\). 2. \(T(5)=7\cdot5-5=30\). 3. \(T(20)=7\cdot20-5=135\).

Answer

<table><tr><th>\(x\)</th><td>\(1\)</td><td>\(2\)</td><td>\(5\)</td><td>\(10\)</td><td>\(20\)</td></tr><tr><th>\(T(x)\)</th><td>\(2\)</td><td>\(9\)</td><td>\(30\)</td><td>\(65\)</td><td>\(135\)</td></tr></table>
Function definition and input-output rules 9
The functions are \(T_1(x)=4(x+1.5)\) and \(T_2(x)=2-6x\). Evaluate both functions at \(x=-0.75\). Which function has the greater value?

Hints

- Substitute the given input into each function. - Evaluate inside parentheses first. - Be careful when multiplying two negative numbers.

Solution

1. \(T_1(-0.75)=4\cdot(-0.75+1.5)=4\cdot0.75=3\). 2. \(T_2(-0.75)=2-6\cdot(-0.75)=2+4.5=6.5\). 3. Since \(6.5>3\), \(T_2\) has the greater value.

Answer

\(T_1(-0.75)=3\) and \(T_2(-0.75)=6.5\). Therefore, \(T_2\) has the greater value.
Function definition and input-output rules 10
The function is \(T(x)=\frac{2}{3}x-\frac{1}{2}\). Evaluate it for each input. Write each answer as an integer or a fraction in simplest form. a) \(x=3\) b) \(x=\frac{3}{4}\) c) \(x=6\) d) \(x=\frac{1}{4}\)

Hints

- Substitute each input for \(x\). - Multiply before subtracting. - Simplify fractions during the calculation. - Equal quantities have a difference of \(0\).

Solution

1. \(T(3)=\frac{2}{3}(3)-\frac{1}{2}=2-\frac{1}{2}=\frac{3}{2}\). 2. \(T\left(\frac{3}{4}\right)=\frac{2}{3}\cdot\frac{3}{4}-\frac{1}{2}=\frac{1}{2}-\frac{1}{2}=0\). 3. \(T(6)=\frac{2}{3}(6)-\frac{1}{2}=4-\frac{1}{2}=\frac{7}{2}\). 4. \(T\left(\frac{1}{4}\right)=\frac{2}{3}\cdot\frac{1}{4}-\frac{1}{2}=\frac{1}{6}-\frac{3}{6}=-\frac{1}{3}\).

Answer

a) \(\frac{3}{2}\) b) \(0\) c) \(\frac{7}{2}\) d) \(-\frac{1}{3}\)
Function definition and input-output rules 11
Evaluate \(T(x)=-15.4+x\) for each input. a) \(x=8.9\) b) \(x=-3.7\) c) \(x=15.4\) d) \(x=\frac{1}{2}\)

Hints

- Substitute the input for \(x\). - Use parentheses when substituting a negative number. - Convert \(\frac{1}{2}\) to a decimal if helpful.

Solution

1. \(T(8.9)=-15.4+8.9=-6.5\). 2. \(T(-3.7)=-15.4+(-3.7)=-19.1\). 3. \(T(15.4)=-15.4+15.4=0\). 4. Since \(\frac{1}{2}=0.5\), \(T\left(\frac{1}{2}\right)=-15.4+0.5=-14.9\).

Answer

a) \(-6.5\) b) \(-19.1\) c) \(0\) d) \(-14.9\)
Function definition and input-output rules 12
The table defines a function \(g\) on the displayed inputs. <table><tr><td>\(x\)</td><td>\(-3\)</td><td>\(-1\)</td><td>\(0\)</td><td>\(2\)</td><td>\(4\)</td></tr><tr><td>\(g(x)\)</td><td>\(5\)</td><td>\(0\)</td><td>\(-4\)</td><td>\(0\)</td><td>\(5\)</td></tr></table> 1) What are the zeros of \(g\)? 2) At what point does the graph intersect the y-axis? 3) Find \(g(4)\).

Hints

- Find the entries where the output is \(0\). - The y-intercept has input \(x=0\). - Match the input \(4\) with its output.

Solution

1. Zeros occur where \(g(x)=0\). The table shows \(g(-1)=0\) and \(g(2)=0\), so the zeros are \(-1\) and \(2\). 2. The y-intercept occurs when \(x=0\). Since \(g(0)=-4\), the y-intercept is \((0, -4)\). 3. The table shows \(g(4)=5\).

Answer

1) \(-1\) and \(2\) 2) \((0, -4)\) 3) \(5\)
Function definition and input-output rules 13
A weather station recorded the outdoor temperature every three hours. <table><thead><tr><th>Time</th><td>\(6{:}00\) a.m.</td><td>\(9{:}00\) a.m.</td><td>\(12{:}00\) p.m.</td><td>\(3{:}00\) p.m.</td><td>\(6{:}00\) p.m.</td><td>\(9{:}00\) p.m.</td></tr></thead><tbody><tr><th>Temperature (\(^\circ\text{F}\))</th><td>\(28\)</td><td>\(34\)</td><td>\(42\)</td><td>\(46\)</td><td>\(39\)</td><td>\(31\)</td></tr></tbody></table> a) Explain why time \(\rightarrow\) temperature is a function for this data set. b) During which three-hour interval did the temperature decrease the most? c) Find the temperature spread, defined here as the highest value minus the lowest value.

Hints

- Apply the definition of a function to the time values. - Compare the differences between consecutive temperatures. - For part c, compare the highest and lowest recorded temperatures.

Solution

1. Each recorded time is paired with exactly one temperature, so the data define a function. 2. From \(3{:}00\) p.m. to \(6{:}00\) p.m., the change is \(39-46=-7\,^\circ\text{F}\). From \(6{:}00\) p.m. to \(9{:}00\) p.m., the change is \(31-39=-8\,^\circ\text{F}\). The greatest decrease occurred from \(6{:}00\) p.m. to \(9{:}00\) p.m. 3. The highest temperature is \(46\,^\circ\text{F}\) and the lowest is \(28\,^\circ\text{F}\). The spread is \(46-28=18\,^\circ\text{F}\).

Answer

a) Each time has exactly one recorded temperature. b) From \(6{:}00\) p.m. to \(9{:}00\) p.m.; the temperature decreased by \(8\,^\circ\text{F}\). c) \(18\,^\circ\text{F}\)
Function definition and input-output rules 14
Consider the set of points \(M=\{P_1(1, 3), P_2(2, 6), P_3(3, 3), P_4(2, 1)\}\). a) Use the definition of a function to explain why \(M\) does not represent \(y=f(x)\). b) Change the x-coordinate of exactly one point so that the new set represents a function. State the coordinates of the changed point. c) For your new set from part b, determine whether the reverse assignment \(y\rightarrow x\) is a function. Explain.

Hints

- In a list of ordered pairs, repeated x-values with different y-values violate the function rule. - Change one of the two points whose x-coordinate is \(2\). - To test the reverse assignment, switch the roles of \(x\) and \(y\).

Solution

1. The input \(x=2\) occurs in \(P_2(2, 6)\) and \(P_4(2, 1)\) with two different outputs. Therefore, \(M\) does not represent a function. 2. One possible change is to replace \(P_4(2, 1)\) with \(P_4(4, 1)\). The new x-values are \(1\), \(2\), \(3\), and \(4\), so each input has exactly one output. 3. In the new set, the output \(y=3\) occurs at both \(x=1\) and \(x=3\). Therefore, the reverse assignment is not a function.

Answer

a) \(x=2\) is paired with both \(y=6\) and \(y=1\). b) One answer is \(P_4(4, 1)\). c) No. The input \(y=3\) in the reverse assignment would have outputs \(x=1\) and \(x=3\).
Function definition and input-output rules 15
The function is \(f(x)=-1.5x+3\). a) Find \(f(-2)\) and \(f(4)\). Does each point \(A(-2,6)\) and \(B(4,-3)\) lie on its graph? b) Find the input \(k\) for which \(f(k)=0\).

Hints

- To test whether a point fits the rule, compare the rule's output with the point's second coordinate. - For part b), look for an input whose output is \(0\). - Can you undo the operations in the rule to find that input?

Solution

1. \(f(-2)=-1.5(-2)+3=6\) and \(f(4)=-1.5(4)+3=-3\). Both listed points lie on the graph. 2. Solve \(-1.5k+3=0\). Then \(k=2\).

Answer

a) \(f(-2)=6\), \(f(4)=-3\); both points lie on the graph. b) \(k=2\)
Function definition and input-output rules 16
Consider the two coordinate axes. a) State the equation of the x-axis. Is the x-axis the graph of a linear function? Explain. b) State the equation of the y-axis. Use the definition of a function to explain why the y-axis cannot be the graph of \(y=f(x)\).

Hints

- Write the coordinate that stays constant on each axis. - A function may assign only one y-value to each x-value. - Apply the vertical line test.

Solution

1. The x-axis has equation \(y=0\). It is the graph of the linear function \(f(x)=0\), because every real input has exactly one output, \(0\). 2. The y-axis has equation \(x=0\). It is not the graph of \(y=f(x)\) because the input \(x=0\) is paired with infinitely many y-values. Equivalently, it fails the vertical line test.

Answer

a) \(y=0\); yes, it is the graph of a linear function. b) \(x=0\); no, it is not the graph of \(y=f(x)\).
Function definition and input-output rules 17
The function \(g\) is defined by \(g(x)=1.2x+0.5\). a) Find \(g(2)\), \(g(-1)\), and \(g(5)\). b) Which of these ordered pairs are input-output pairs of \(g\): \((2,2.9)\), \((-1,-1)\), and \((5,6.5)\)? c) Elīna says that the input \(-1\) has two outputs, \(-0.7\) and \(-1\), because both numbers appear in part b). Explain the error.

Hints

- Calculate the output of the rule for each listed input before judging any ordered pair. - An ordered pair belongs to the function only when its second coordinate matches the function output. - Seeing a number written in a proposed pair does not make it an output of the function.

Solution

1. \(g(2)=1.2\cdot2+0.5=2.9\). 2. \(g(-1)=1.2(-1)+0.5=-0.7\). 3. \(g(5)=1.2\cdot5+0.5=6.5\). 4. Therefore, \((2,2.9)\) and \((5,6.5)\) are input-output pairs of \(g\), while \((-1,-1)\) is not because \(g(-1)=-0.7\). 5. The number \(-1\) in \((-1,-1)\) is a proposed second coordinate, not another output produced by the rule. For input \(-1\), the rule gives only \(-0.7\).

Answer

a) \(g(2)=2.9\), \(g(-1)=-0.7\), and \(g(5)=6.5\) b) \((2,2.9)\) and \((5,6.5)\) c) The rule assigns the input \(-1\) only the output \(-0.7\); the proposed pair \((-1,-1)\) does not belong to \(g\).
Function definition and input-output rules 18
Evaluate \(T(x)=-4x+\frac{1}{2}\) for each input. a) \(x=3\) b) \(x=-1.5\) c) \(x=\frac{3}{8}\)

Hints

- Substitute each input for \(x\). - Use parentheses around negative inputs. - Work with either fractions or decimals consistently.

Solution

1. \(T(3)=-4\cdot3+\frac{1}{2}=-12+0.5=-11.5\). 2. \(T(-1.5)=-4\cdot(-1.5)+0.5=6.5\). 3. \(T\left(\frac{3}{8}\right)=-4\cdot\frac{3}{8}+\frac{1}{2}=-\frac{3}{2}+\frac{1}{2}=-1\).

Answer

a) \(-11.5\) b) \(6.5\) c) \(-1\)
Function definition and input-output rules 19
The functions are \(T_1(y)=5y-15\) and \(T_2(y)=y^2-3y\). a) Evaluate both functions when \(y=3\). b) Evaluate both functions when \(y=6\). c) For \(y=6\), which function has the greater value, and by how much?

Hints

- Evaluate each function separately at the requested input. - Apply exponents before multiplication and subtraction. - For part c), compare the two outputs at the same input and subtract the smaller from the larger.

Solution

1. For \(y=3\), \(T_1(3)=5\cdot3-15=0\) and \(T_2(3)=3^2-3\cdot3=0\). 2. For \(y=6\), \(T_1(6)=5\cdot6-15=15\) and \(T_2(6)=6^2-3\cdot6=18\). 3. At \(y=6\), \(T_2\) is greater by \(18-15=3\).

Answer

a) \(T_1(3)=0\) and \(T_2(3)=0\) b) \(T_1(6)=15\) and \(T_2(6)=18\) c) \(T_2\) is greater by \(3\).
Function definition and input-output rules 20
A car rental company models the one-day cost with the function \(K(s)=42+0.15s\), where \(s\) is the number of miles driven and \(K(s)\) is the cost in dollars. a) Find and interpret \(K(80)\). b) Find \(K(240)\). c) Find the input \(s\) for which \(K(s)=60\), and interpret that input.

Hints

- In \(K(s)\), the input is mileage and the output is cost. - For a known input, substitute the mileage into the function rule. - For a known output, set the function equal to that cost and solve for the input.

Solution

1. \(K(80)=42+0.15\cdot80=54\). This means an \(80\)-mile day costs \(\$54.00\). 2. \(K(240)=42+0.15\cdot240=78\), so a \(240\)-mile day costs \(\$78.00\). 3. Solve \(60=42+0.15s\). Then \(18=0.15s\), so \(s=120\). A one-day cost of \(\$60.00\) corresponds to driving \(120\) miles.

Answer

a) \(K(80)=54\), meaning \(80\) miles costs \(\$54.00\). b) \(K(240)=78\), so the cost is \(\$78.00\). c) \(s=120\), meaning a \(\$60.00\) day corresponds to \(120\) miles.
Function definition and input-output rules 21
Decide whether each equation defines \(y\) as a function of \(x\). Explain your reasoning. (1) \(y+2x=7\) (2) \(x=y^2\) (3) \(y=|x|+2\)

Hints

- A function gives exactly one output for every allowed input. - Try rewriting the first equation with \(y\) alone. - To show a relation is not a function, one input with two different outputs is enough.

Solution

1. Equation (1) gives \(y=7-2x\). Every input gives exactly one output, so it is a function. 2. In equation (2), \(x=4\) allows both \(y=2\) and \(y=-2\). One input has two outputs, so it is not a function. 3. In equation (3), the absolute value \(|x|\) is one number for each input, so adding \(2\) gives exactly one output. It is a function.

Answer

(1) Function; \(y=7-2x\) gives one output per input. (2) Not a function; when \(x=4\), both \(y=2\) and \(y=-2\) work. (3) Function; each input has one absolute value and hence one output.
Function definition and input-output rules 22
For each relation, decide whether it gives exactly one output \(y\) for each allowed input \(x\). Explain. (1) \(xy=0\) (2) The input-output pairs are \((1,3)\), \((2,4)\), and \((2,5)\). (3) \(y=2x+1\) for \(x\geq0\).

Hints

- Look for an allowed input that can be paired with two different outputs. - For the product equation, think about what happens when one factor is zero. - A restricted domain does not prevent a rule from being a function.

Solution

1. In (1), if \(x=0\), then every value of \(y\) makes \(0y=0\) true. It is not a function. 2. In (2), input \(2\) has both output \(4\) and output \(5\), so it is not a function. 3. In (3), each allowed nonnegative input gives exactly one output when substituted into \(2x+1\), so it is a function.

Answer

(1) Not a function; input \(0\) has many outputs. (2) Not a function; input \(2\) has outputs \(4\) and \(5\). (3) Function; every allowed input gives one output.
Function definition and input-output rules 23
Does each equation define \(y\) as a function of \(x\)? When it does, write the rule with \(y\) alone. a) \(y-x^2=5\) b) \(y^2=9\) c) \(x=-3\) d) \(y=0\) e) \(2y=6x-2\)

Hints

- Pick one input and ask how many different outputs satisfy the equation. - When a condition fixes \(x\) but not \(y\), what happens if you keep that input and change the output? - Rearrange the equations that do give one output so their rules are easier to see.

Solution

1. For a), \(y=x^2+5\), one output for every input. It is a function. 2. For b), every input allows both \(y=3\) and \(y=-3\), so it is not a function. 3. For c), when \(x=-3\), any \(y\) works. It is not a function. 4. For d), all inputs give just the output \(0\), so it is a function. 5. For e), divide by \(2\) to get \(y=3x-1\), one output per input. It is a function.

Answer

a) Function: \(y=x^2+5\) b) Not a function c) Not a function d) Function: \(y=0\) e) Function: \(y=3x-1\)
Function definition and input-output rules 24
Decide whether each assignment is a function. Explain your answer. a) A student \(\rightarrow\) that student's score on one quiz. b) A quiz score \(\rightarrow\) the student who earned it. Mia and Luis both scored \(8\) points. c) A whole number \(n\rightarrow n^2\). d) A square number \(\rightarrow\) a whole number whose square gives it. Here, whole numbers include \(0\) but not negative numbers.

Hints

- Start with one possible input and ask how many outputs it can have. - Reversing an assignment may change whether it is a function. - In part d), pay attention to which numbers are allowed as outputs.

Solution

1. For a), every student has just one score on the specified quiz, so the assignment is a function. 2. For b), score \(8\) is assigned to both Mia and Luis, so it is not a function. 3. For c), each whole number has exactly one square, so it is a function. 4. For d), every square of a whole number has exactly one nonnegative whole-number square root. The assignment is a function.

Answer

a) Function; each student has one quiz score. b) Not a function; score \(8\) belongs to two students. c) Function; each whole number has one square. d) Function; among whole numbers, each square has one square root.
Function definition and input-output rules 25
The graph shows temperature \(T(t)\), in degrees Celsius, as a function of the number of hours \(t\) since midnight during one \(24\)-hour period. a) Find \(T(2)\) and \(T(14)\). b) Find every input \(t\) for which \(T(t)=-8\). c) Explain why the graph represents a function of time. d) Would the reverse relation, with temperature as the input and time as the output, be a function? Use part b) to justify your answer.
Figure for problem 532183

Hints

- To evaluate a graphically defined function, start with the input on the horizontal axis and read its one output. - For part b), look for every place the graph reaches the same horizontal level. - A function may repeat an output for different inputs; the restriction is on one input having more than one output. - Reverse the roles of input and output when considering part d).

Solution

1. Reading the graph gives \(T(2)=-12\) and \(T(14)=-4\). 2. The horizontal level \(-8\) meets the graph at \(t=8\) and \(t=20\). 3. Every time input on the displayed domain corresponds to exactly one temperature output, so \(T\) is a function of time. 4. The reverse relation is not a function because the input temperature \(-8\) would have two outputs, \(8\) and \(20\).

Answer

a) \(T(2)=-12\) and \(T(14)=-4\) b) \(t=8\) and \(t=20\) c) Each time input has exactly one temperature output. d) No. In the reverse relation, the input \(-8\) would have two outputs, \(8\) and \(20\).
Function definition and input-output rules 26
A graph shows the water depth at a harbor during the first \(12\) hours after midnight. a) Read the water depth at \(t = 0, 2, 4, 6, 8, 10,\) and \(12\) hours, and record the values in a table. Is the relation from time to water depth a function? Explain. b) At what times is the water depth exactly \(4\,\text{m}\)? Is the reverse relation from water depth to time a function? Explain.
Figure for problem 532188

Hints

- Recall that a function assigns exactly one output to each input. - Read the graph at \(t = 0, 2, 4, 6, 8, 10,\) and \(12\). - To find when the depth is \(4\,\text{m}\), trace horizontally from \(h = 4\) to the graph. - For the reverse relation, ask whether one depth can occur at two different times.

Solution

1. Read the plotted values: <table> <tr><th>Time \(t\) (h)</th><td>\(0\)</td><td>\(2\)</td><td>\(4\)</td><td>\(6\)</td><td>\(8\)</td><td>\(10\)</td><td>\(12\)</td></tr> <tr><th>Depth \(h\) (m)</th><td>\(1\)</td><td>\(2\)</td><td>\(4\)</td><td>\(5\)</td><td>\(4\)</td><td>\(2\)</td><td>\(1\)</td></tr> </table> 2. The relation from time to depth is a function because each time is paired with exactly one depth. 3. The graph has depth \(4\,\text{m}\) at \(t = 4\,\text{h}\) and \(t = 8\,\text{h}\). 4. The reverse relation is not a function because the input depth \(4\,\text{m}\) would be paired with two different times.

Answer

a) <table> <tr><th>Time \(t\) (h)</th><td>\(0\)</td><td>\(2\)</td><td>\(4\)</td><td>\(6\)</td><td>\(8\)</td><td>\(10\)</td><td>\(12\)</td></tr> <tr><th>Depth \(h\) (m)</th><td>\(1\)</td><td>\(2\)</td><td>\(4\)</td><td>\(5\)</td><td>\(4\)</td><td>\(2\)</td><td>\(1\)</td></tr> </table> Yes. Time to depth is a function because each time has exactly one depth. b) The depth is \(4\,\text{m}\) at \(t = 4\,\text{h}\) and \(t = 8\,\text{h}\). The reverse relation is not a function because one depth can correspond to more than one time.
Function definition and input-output rules 27
Graphs \(g\) and \(h\) show two lines. a) For each line, write an equation in the form \(ax+by=c\). b) Decide whether each line is the graph of a function \(y=f(x)\). Justify each answer.
Figure for problem 532195

Hints

- Identify which coordinate stays constant on each line. - A vertical line has a constant x-coordinate; a horizontal line has a constant y-coordinate. - Apply the vertical line test.

Solution

1. Line \(g\) is vertical at \(x=3\), so one equation is \(x+0y=3\). 2. Line \(g\) is not the graph of a function \(y=f(x)\) because the input \(x=3\) corresponds to infinitely many y-values. It fails the vertical line test. 3. Line \(h\) is horizontal at \(y=-2\), so one equation is \(0x+y=-2\). 4. Line \(h\) is the graph of the constant function \(f(x)=-2\), because each input has exactly one output.

Answer

a) \(g:\ x+0y=3\); \(h:\ 0x+y=-2\) b) \(g\) is not a function graph; \(h\) is a function graph.
Function definition and input-output rules 28
a) At \(x=2\), list every output shown by each graph. b) Which graph represents a function of \(x\)? Explain why the other one does not.
Figure for problem 533220

Hints

- Follow the vertical grid line at the stated input in each panel. - After looking at that input, think about the one-output rule for functions.

Solution

1. In graph a), \(x=2\) gives the single output \(y=2\). 2. In graph b), \(x=2\) gives both \(y=2\) and \(y=-2\). 3. Graph a) is a function because each input has one output. Graph b) is not because the input \(2\) already has two outputs.

Answer

a) Graph a): \(2\); Graph b): \(-2\) and \(2\). b) Graph a) is a function; Graph b) is not, since one input has two outputs.
Function definition and input-output rules 29
Determine whether the graph represents a function. Justify your answer.
Figure for problem 533565

Hints

- Compare an upper semicircle with a full circle. - Apply the vertical line test.

Solution

1. The graph is the upper semicircle \(y=\sqrt{4-x^2}\) for \(-2\le x\le2\). 2. Every input in \([-2, 2]\) corresponds to exactly one output on the upper semicircle. Equivalently, every vertical line meets the graph at most once. 3. Therefore, the graph represents a function.

Answer

Yes. The graph represents a function because each \(x\in[-2, 2]\) has exactly one y-value.
Function definition and input-output rules 30
Determine whether the curve is the graph of a function \(y=f(x)\). Justify your answer.
Figure for problem 533566

Hints

- Look for a vertical line that intersects the graph more than once. - Check the graph at \(x=2\).

Solution

1. A graph represents a function only if every input \(x\) has at most one output \(y\). 2. The curve is a sideways V. For example, the vertical line \(x=2\) meets the curve at \((2, 3)\) and \((2, 1)\). 3. Therefore, the curve fails the vertical line test and is not the graph of a function.

Answer

No. At \(x=2\), the graph has two y-values, \(3\) and \(1\).
Function definition and input-output rules 31
Wind turbines produce power \(w(t)\), and solar panels produce power \(s(t)\). Their total power is \(g(t)=w(t)+s(t)\). a) Read \(w(12)\) and \(s(12)\). Find \(g(12)\). b) During which hours is \(g(t)=w(t)\)? Explain. c) If \(w\) and \(s\) are functions, why is \(g\) also a function?
Figure for problem 533585

Hints

- At one hour, find the heights of both curves before adding them. - What must the solar contribution be if adding it does not change the wind amount? - For one fixed time, how many possible values can the sum have?

Solution

1. At hour \(12\), the wind power is \(2\,\text{MW}\) and the solar power is \(4\,\text{MW}\). Their sum is \(g(12)=2+4=6\,\text{MW}\). 2. The total equals wind power when solar power is zero. This happens from midnight to \(6{:}00\) a.m. and from \(6{:}00\) p.m. to midnight, including the endpoints. 3. At each time, \(w\) and \(s\) each give one output. Adding them gives exactly one total, so \(g\) is a function.

Answer

a) \(w(12)=2\,\text{MW}\), \(s(12)=4\,\text{MW}\), and \(g(12)=6\,\text{MW}\) b) Midnight to \(6{:}00\) a.m. and \(6{:}00\) p.m. to midnight; solar power is zero then. c) Each time gives one wind value and one solar value, so their sum has one value.
Function definition and input-output rules 32
Which graphs represent functions \(y=f(x)\)? Justify each decision with the vertical line test.
Figure for problem 534934

Hints

- Move an imaginary vertical line across each graph. - A function graph may be intersected at most once by any vertical line.

Solution

1. Graph a) is a function because every vertical line intersects the parabola at most once. 2. Graph b) is not a function because vertical lines through \(-2<x<2\) intersect the circle twice. 3. Graph c) is a function because every vertical line intersects the line at most once. 4. Graph d) is not a function because vertical lines through \(0<x\le4\) intersect the sideways V twice.

Answer

Graphs a) and c) are functions. Graphs b) and d) are not functions.
Function definition and input-output rules 33
a) Find the input values that give an output of \(0\). b) Two different inputs have that same output. Does this mean \(f\) is not a function? Explain.
Figure for problem 535027

Hints

- Where would an output of \(0\) appear on the graph? - Check what happens at a single input, rather than comparing different inputs.

Solution

1. The graph crosses the x-axis at \(x=-2\) and \(x=2\). These are the inputs that give output \(0\). 2. The graph is still a function. Having two inputs share an output is allowed; no single input has two outputs.

Answer

a) \(x=-2\) and \(x=2\) b) No. Each input still has just one output.
Function definition and input-output rules 34
The table defines a function from \(x\) to \(y\). <table><thead><tr><th>\(x\)</th><th>\(y\)</th></tr></thead><tbody><tr><td>\(-2\)</td><td>\(4\)</td></tr><tr><td>\(0\)</td><td>\(1\)</td></tr><tr><td>\(3\)</td><td>\(4\)</td></tr><tr><td>\(5\)</td><td>\(9\)</td></tr></tbody></table> a) State the domain. b) State the range. c) Which inputs have output \(4\)? d) If every input-output pair were reversed, would the reversed relation be a function? Explain.

Hints

- The domain consists of the values used as inputs. - The range consists of distinct output values; repeated outputs are not repeated when writing a set. - For the reversed relation, imagine swapping the two entries in every row and then apply the one-output-per-input rule again.

Solution

1. The domain is the set of input values: \(\{-2,0,3,5\}\). 2. The range is the set of output values that occur: \(\{1,4,9\}\). The repeated output \(4\) is listed only once in the set. 3. The inputs \(-2\) and \(3\) both have output \(4\). 4. After reversing the pairs, input \(4\) would be paired with outputs \(-2\) and \(3\). Because one reversed input would have two outputs, the reversed relation would not be a function.

Answer

a) \(\{-2,0,3,5\}\) b) \(\{1,4,9\}\) c) \(-2\) and \(3\) d) No. In the reversed relation, input \(4\) would have two outputs, \(-2\) and \(3\).
Function definition and input-output rules 35
A function has the following values: <table><tr><th>Input \(x\)</th><td>\(-2\)</td><td>\(0\)</td><td>\(3\)</td><td>\(6\)</td></tr><tr><th>Output \(f(x)\)</th><td>\(4\)</td><td>\(1\)</td><td>\(4\)</td><td>\(7\)</td></tr></table> a) List every input shown that has output \(4\). b) If you reverse the pairs in this table, will the new relation be a function? Explain.

Hints

- Start by reading along the output row for the specified value. - Reversing a pair swaps the roles of its input and output. - Check whether any reversed input receives two outputs.

Solution

1. Inputs \(-2\) and \(3\) both give output \(4\). 2. After reversing the pairs, input \(4\) would have outputs \(-2\) and \(3\), so the reversed relation is not a function.

Answer

a) \(-2\) and \(3\) b) No. Reversed input \(4\) would have two outputs.
Function definition and input-output rules 36
The water level in a pool is \(h(t)=50-2t\) inches, where \(t\) is hours since draining began. The pool is drained for the first \(6\) hours only. a) Which inputs are allowed: \(t=-1\), \(t=2.5\), or \(t=8\)? Explain. b) Find the water level at the allowed input.

Hints

- The input is elapsed time during an observation period. - Ask whether time can be negative, fractional, or beyond the last observed hour. - Substitute only the permitted input into the rule.

Solution

1. Time since draining began during the first six hours satisfies \(0\le t\le6\). Only \(2.5\) is allowed. 2. \(h(2.5)=50-2(2.5)=45\) inches.

Answer

a) Only \(t=2.5\) hours; the allowed inputs run from \(0\) through \(6\) hours. b) \(45\) inches
Function definition and input-output rules 37
Consider the functions \(A(x)=2x+10\) and \(B(x)=4x\). a) Complete a table for both functions using \(x=3,4,5,6,\) and \(7\). b) Use the table to determine the value of \(x\) for which the functions have the same output.

Hints

- Evaluate both functions for each input. - Organize the outputs in matching rows. - Look for a column with equal outputs.

Solution

1. The outputs for \(A(x)\) are \(16,18,20,22,\) and \(24\). 2. The outputs for \(B(x)\) are \(12,16,20,24,\) and \(28\). 3. Both functions have output \(20\) when \(x=5\).

Answer

a) <table> <tr><th>\(x\)</th><th>\(3\)</th><th>\(4\)</th><th>\(5\)</th><th>\(6\)</th><th>\(7\)</th></tr> <tr><th>\(A(x)\)</th><td>\(16\)</td><td>\(18\)</td><td>\(20\)</td><td>\(22\)</td><td>\(24\)</td></tr> <tr><th>\(B(x)\)</th><td>\(12\)</td><td>\(16\)</td><td>\(20\)</td><td>\(24\)</td><td>\(28\)</td></tr> </table> b) \(x=5\); both outputs are \(20\).
Function definition and input-output rules 38
An ice cream shop sells single scoops and discounted three-scoop cups. The table shows the lowest price for each number of scoops. <table><tr><th>Scoops</th><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td><td>\(6\)</td><td>\(7\)</td><td>\(8\)</td></tr><tr><th>Lowest price</th><td>\(\$1.50\)</td><td>\(\$3.00\)</td><td>\(\$4.00\)</td><td>\(\$5.50\)</td><td>\(\$7.00\)</td><td>\(\$8.00\)</td><td>\(\$9.50\)</td><td>\(\$11.00\)</td></tr></table> a) Does the number of scoops determine exactly one lowest price? Explain whether this is a function. b) A customer can pay \(\$4.00\) for a three-scoop cup or \(\$4.50\) for three single scoops. Is the assignment from number of scoops to price actually paid a function? Explain. c) What is the lowest price for \(7\) scoops? How many scoops have a lowest price of \(\$8.00\)? d) Should a graph of the lowest prices use separate dots or a connected line? Explain.

Hints

- Check whether each input has only one possible output in each assignment. - Does buying the same number of scoops always lead to the same bill? - Decide which scoop counts can occur before choosing how to graph them.

Solution

1. Every listed scoop count has exactly one lowest price, so scoop count to lowest price is a function. 2. The price actually paid is not a function of scoop count: buying \(3\) scoops can cost either \(\$4.00\) or \(\$4.50\). 3. The table gives \(\$9.50\) for \(7\) scoops and \(6\) scoops for \(\$8.00\). 4. Scoop counts are whole numbers, so the graph uses separate dots.

Answer

a) Yes. Each scoop count has exactly one lowest price. b) No. The same \(3\) scoops can cost \(\$4.00\) or \(\$4.50\). c) \(\$9.50\) for \(7\) scoops; \(6\) scoops cost \(\$8.00\). d) Separate dots, because the number of scoops must be a whole number.
Function definition and input-output rules 39
A textbook weighs \(1.5\,\text{lb}\), and a thin workbook weighs \(0.25\,\text{lb}\). Putri packs exactly \(6\) items total, using any combination of textbooks and workbooks. a) Find the minimum and maximum possible total weights. b) Complete the table. <table><tr><td>Number of textbooks</td><td>\(0\)</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td><td>\(6\)</td></tr><tr><td>Total weight (lb)</td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr></table> c) In this context, should the plotted points be connected by a line? Explain.

Hints

- Compare the lightest and heaviest possible all-one-type combinations. - Look for the constant change in total weight when one workbook is replaced by one textbook. - Decide whether a fractional number of textbooks is meaningful in this context.

Solution

1. The minimum occurs with \(6\) workbooks: \(6(0.25)=1.5\,\text{lb}\). The maximum occurs with \(6\) textbooks: \(6(1.5)=9\,\text{lb}\). 2. Replacing one workbook with one textbook increases the total weight by \(1.5-0.25=1.25\,\text{lb}\). The completed table is: <table><tr><td>Number of textbooks</td><td>\(0\)</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td><td>\(6\)</td></tr><tr><td>Total weight (lb)</td><td>\(1.5\)</td><td>\(2.75\)</td><td>\(4\)</td><td>\(5.25\)</td><td>\(6.5\)</td><td>\(7.75\)</td><td>\(9\)</td></tr></table> 3. The number of textbooks must be a whole number. Inputs such as \(1.5\) textbooks have no meaning, so the graph is discrete and the points should not be connected.

Answer

a) Minimum: \(1.5\,\text{lb}\); maximum: \(9\,\text{lb}\) b) <table><tr><td>Number of textbooks</td><td>\(0\)</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td><td>\(6\)</td></tr><tr><td>Total weight (lb)</td><td>\(1.5\)</td><td>\(2.75\)</td><td>\(4\)</td><td>\(5.25\)</td><td>\(6.5\)</td><td>\(7.75\)</td><td>\(9\)</td></tr></table> c) No. Only whole-number inputs are meaningful.
Function definition and input-output rules 40
The function is \(T(x)=4x-6\). a) Find \(T(-1.5)\). b) Find \(T\left(\frac{3}{4}\right)\). c) For what value of \(x\) is \(T(x)=0\)? d) Find \(T(2)\) and \(T(3)\). How much does the output change when \(x\) increases by \(1\)?

Hints

- Substitute each input into the function. - For part c), set the output equal to \(0\) and solve. - Compare the outputs for inputs that differ by \(1\). - Relate the change in output to the coefficient of \(x\).

Solution

1. \(T(-1.5)=4\cdot(-1.5)-6=-6-6=-12\). 2. \(T\left(\frac{3}{4}\right)=4\cdot\frac{3}{4}-6=3-6=-3\). 3. Set \(4x-6=0\). Then \(4x=6\), so \(x=1.5\). 4. \(T(2)=4\cdot2-6=2\) and \(T(3)=4\cdot3-6=6\). The output increases by \(6-2=4\) when \(x\) increases by \(1\).

Answer

a) \(-12\) b) \(-3\) c) \(x=1.5\) d) \(T(2)=2\), \(T(3)=6\); the output increases by \(4\).
Function definition and input-output rules 41
The function is \(T(x)=\frac{2}{3}x-4\left(x+\frac{1}{2}\right)\). Find \(T\left(\frac{3}{4}\right)\).

Hints

- Substitute the given input for \(x\). - Evaluate the expression inside parentheses before multiplying. - Simplify fractions before multiplying when possible. - Check the sign of the final result.

Solution

1. Substitute \(x=\frac{3}{4}\): \(T\left(\frac{3}{4}\right)=\frac{2}{3}\cdot\frac{3}{4}-4\cdot\left(\frac{3}{4}+\frac{1}{2}\right)\). 2. The first term is \(\frac{2}{3}\cdot\frac{3}{4}=\frac{1}{2}\). 3. Inside the parentheses, \(\frac{3}{4}+\frac{1}{2}=\frac{5}{4}\). 4. Then \(4\cdot\frac{5}{4}=5\). 5. Therefore, \(T\left(\frac{3}{4}\right)=\frac{1}{2}-5=-\frac{9}{2}=-4.5\).

Answer

\(-\frac{9}{2}\), or \(-4.5\)
Function definition and input-output rules 42
The functions are \(T_1(k)=2k+10\) and \(T_2(k)=40-3k\). a) Evaluate both functions for \(k=2\) and \(k=10\). b) As \(k\) increases, which function increases and which decreases? c) Test integer values of \(k\) to find when the functions have the same output.

Hints

- Substitute each given input into both functions. - Compare the outputs at the smaller and larger inputs. - Test integer values between \(2\) and \(10\).

Solution

1. For \(k=2\), \(T_1(2)=2\cdot2+10=14\) and \(T_2(2)=40-3\cdot2=34\). 2. For \(k=10\), \(T_1(10)=2\cdot10+10=30\) and \(T_2(10)=40-3\cdot10=10\). 3. \(T_1\) increases as \(k\) increases, while \(T_2\) decreases. 4. Testing \(k=6\) gives \(T_1(6)=22\) and \(T_2(6)=22\), so the outputs are equal.

Answer

a) For \(k=2\): \(T_1=14\), \(T_2=34\). For \(k=10\): \(T_1=30\), \(T_2=10\). b) \(T_1\) increases and \(T_2\) decreases. c) \(k=6\); both outputs are \(22\).
Function definition and input-output rules 43
Two phone plans have different monthly costs. Plan A charges a \(\$5.00\) monthly fee plus \(\$0.12\) per minute: \(K_A(m)=5+0.12m\). Plan B has no monthly fee and charges \(\$0.20\) per minute: \(K_B(m)=0.20m\). a) Find the cost of each plan for \(50\) minutes. b) Which plan costs less for \(100\) minutes? Show both costs. c) Plan A costs \(\$14.00\) in one month. How many minutes were used?

Hints

- Substitute each number of minutes into both cost rules. - Compare the total charges, not only the per-minute prices. - For part c), use Plan A's rule and the given total cost to find its input.

Solution

1. At \(m=50\), Plan A costs \(5+0.12\cdot50=\$11.00\), while Plan B costs \(0.20\cdot50=\$10.00\). 2. At \(m=100\), Plan A costs \(5+0.12\cdot100=\$17.00\), while Plan B costs \(0.20\cdot100=\$20.00\). Plan A is cheaper. 3. Solve \(5+0.12m=14\). Then \(0.12m=9\), so \(m=75\) minutes.

Answer

a) Plan A: \(\$11.00\); Plan B: \(\$10.00\) b) Plan A; \(\$17.00\) compared with Plan B's \(\$20.00\). c) \(75\) minutes
Function definition and input-output rules 44
The functions are \(T_1(x)=24-4x\) and \(T_2(x)=2x+6\). a) Evaluate both functions at \(x=3\). b) Nikos claims, “The functions have the same value when \(x=5\).” Check the claim. c) Find \(T_1(10)\).

Hints

- Substitute each requested input into the corresponding function rule. - Equal function values must have identical outputs at the same input. - Follow the order of operations when evaluating.

Solution

1. \(T_1(3)=24-4\cdot3=12\) and \(T_2(3)=2\cdot3+6=12\). 2. \(T_1(5)=24-4\cdot5=4\) and \(T_2(5)=2\cdot5+6=16\). Since \(4\ne 16\), the claim is false. 3. \(T_1(10)=24-4\cdot10=-16\).

Answer

a) \(T_1(3)=12\) and \(T_2(3)=12\) b) The claim is false because \(4\ne 16\). c) \(T_1(10)=-16\)
Function definition and input-output rules 45
Determine whether each statement about functions and their graphs is true or false. Justify each answer with an explanation or counterexample. a) Every function has at least one zero. b) A function graph can intersect the y-axis at more than one point. c) If a function graph contains \((4, 0)\), then \(4\) is a zero of the function.

Hints

- Think of a graph that never meets the x-axis. - Apply the function rule at \(x=0\). - Interpret what the coordinate \(y=0\) means.

Solution

1. Statement a is false. For example, the constant function \(f(x)=2\) never equals \(0\). 2. Statement b is false. A y-axis intersection has input \(x=0\), and a function can assign at most one output to that input. 3. Statement c is true. If \((4, 0)\) lies on the graph, then \(f(4)=0\), which means \(4\) is a zero.

Answer

a) False; for example, \(f(x)=2\) has no zero. b) False; a function can have at most one output at \(x=0\). c) True; \((4, 0)\) means \(f(4)=0\).
Function definition and input-output rules 46
A smartphone's battery level is recorded during the day. <table><thead><tr><th>Time</th><td>\(7{:}00\) a.m.</td><td>\(10{:}00\) a.m.</td><td>\(12{:}00\) p.m.</td><td>\(2{:}00\) p.m.</td><td>\(4{:}00\) p.m.</td><td>\(6{:}00\) p.m.</td></tr></thead><tbody><tr><th>Battery level</th><td>\(100\%\)</td><td>\(76\%\)</td><td>\(60\%\)</td><td>\(60\%\)</td><td>\(44\%\)</td><td>\(88\%\)</td></tr></tbody></table> a) During which interval was the phone probably connected to a charger? Explain. b) Find the average battery use, in percentage points per hour, from \(7{:}00\) a.m. to \(12{:}00\) p.m. c) Consider the reverse assignment battery level \(\rightarrow\) time. Does this reverse assignment define a function for the data in the table? Explain.

Hints

- Look for an interval in which the battery level increases. - Divide the total decrease by the elapsed time. - For the reverse assignment, check whether one battery percentage appears at more than one time.

Solution

1. The battery level rises from \(44\%\) at \(4{:}00\) p.m. to \(88\%\) at \(6{:}00\) p.m., so the phone was probably charging during that interval. 2. From \(7{:}00\) a.m. to \(12{:}00\) p.m. is \(5\) hours. The battery decreases by \(100-60=40\) percentage points, so the average use is \(40\div5=8\) percentage points per hour. 3. The reverse assignment is not a function because the battery level \(60\%\) corresponds to both \(12{:}00\) p.m. and \(2{:}00\) p.m.

Answer

a) From \(4{:}00\) p.m. to \(6{:}00\) p.m. b) \(8\) percentage points per hour c) No. The input \(60\%\) would have two outputs: \(12{:}00\) p.m. and \(2{:}00\) p.m.
Function definition and input-output rules 47
The linear function \(f\) is defined by \(f(x) = 1.5x - 2\). Point \(P(4, y_P)\) lies on the graph of \(f\). 1. Find the missing coordinate \(y_P\). 2. Point \(Q\) has the same x-coordinate as \(P\) and is exactly \(5\) units above the graph of \(f\). Find the coordinates of \(Q\). 3. Determine whether \(Q\) lies on the graph of \(g(x) = 1.5x + 3\).

Hints

- Substitute the given x-value into the function to find its output. - Moving vertically changes the y-coordinate but not the x-coordinate. - To test a point, evaluate the function at the point's x-coordinate and compare the result with its y-coordinate.

Solution

1. \(f(4) = 1.5 \cdot 4 - 2 = 4\), so \(P(4, 4)\). 2. Moving \(5\) units vertically upward keeps the x-coordinate the same and adds \(5\) to the y-coordinate. Thus \(Q(4, 9)\). 3. \(g(4) = 1.5 \cdot 4 + 3 = 9\). This matches the y-coordinate of \(Q\), so \(Q\) lies on the graph of \(g\).

Answer

1) \(y_P = 4\) 2) \(Q(4, 9)\) 3) Yes, because \(g(4) = 9\).
Function definition and input-output rules 48
Let \(D=\{1,2,3,4,5,6,7,8,9,10\}\). Consider this rule: “Assign each number \(n\in D\) each of its distinct prime factors as a separate output.” a) Explain why this rule does not define a function. Give a specific example. b) Rewrite the rule so that it defines a function without changing the domain. c) Does the rule “assign each rational number \(x\) its absolute value \(|x|\)” define a function? Explain.

Hints

- Find a number in \(D\) with more than one distinct prime factor. - The revised rule must produce exactly one output even for \(n=1\). - Ask how many absolute values one rational number has.

Solution

1. The input \(6\) has the distinct prime factors \(2\) and \(3\), so it would receive two different outputs. Therefore, the rule does not define a function. 2. One valid revision is: “Assign each \(n\) the number of its distinct prime factors.” This rule assigns exactly one output to every input, including assigning \(0\) to \(1\). 3. Every rational number has exactly one absolute value, so \(x\mapsto|x|\) is a function.

Answer

a) No. For example, \(6\) would have outputs \(2\) and \(3\). b) Example: Assign each \(n\) the number of its distinct prime factors, with \(1\mapsto0\). c) Yes. Every rational number has exactly one absolute value.
Function definition and input-output rules 49
A mail service uses these fictional rates for letters with \(0<m\le16\), where \(m\) is the weight in ounces: - Up to \(1\,\text{oz}\): \(\$0.85\) - More than \(1\,\text{oz}\) and up to \(2\,\text{oz}\): \(\$1.00\) - More than \(2\,\text{oz}\) and up to \(16\,\text{oz}\): \(\$1.60\) a) Make a table for weight \(m\) and postage \(P\) at \(m=0.5\), \(1\), \(1.5\), \(2\), and \(5\) ounces. b) Explain why weight \(\rightarrow\) postage is a function on the stated domain. c) Is the reverse assignment postage \(\rightarrow\) weight a function? Explain.

Hints

- Pay close attention to which endpoints are included in each interval. - A function assigns exactly one output to every input in its domain. - In the reverse direction, ask whether a postage value determines one exact weight.

Solution

1. Apply the rate interval that contains each weight. The postage values are \(\$0.85\), \(\$0.85\), \(\$1.00\), \(\$1.00\), and \(\$1.60\). 2. Each weight in \(0<m\le16\) belongs to exactly one rate interval, so it has exactly one postage value. Therefore, weight \(\rightarrow\) postage is a function. 3. The reverse assignment is not a function. For example, the postage value \(\$1.00\) corresponds to every weight in \(1<m\le2\), not to one unique weight.

Answer

a) <table><tr><td>Weight (oz)</td><td>\(0.5\)</td><td>\(1\)</td><td>\(1.5\)</td><td>\(2\)</td><td>\(5\)</td></tr><tr><td>Postage</td><td>\(\$0.85\)</td><td>\(\$0.85\)</td><td>\(\$1.00\)</td><td>\(\$1.00\)</td><td>\(\$1.60\)</td></tr></table> b) Yes. Each allowed weight has exactly one postage value. c) No. One postage value corresponds to many possible weights.
Function definition and input-output rules 50
The table gives four ordered pairs. <table><tr><td>\(x\)</td><td>\(4\)</td><td>\(4\)</td><td>\(4\)</td><td>\(4\)</td></tr><tr><td>\(y\)</td><td>\(-2\)</td><td>\(0\)</td><td>\(2\)</td><td>\(5\)</td></tr></table> a) Describe the location of the four points. On what line do they lie? b) Explain why the assignment \(x\rightarrow y\) is not a function. c) Switch the input and output values. Using \(u\) as the new input variable, write the resulting function and state its domain.

Hints

- Compare the x-coordinates of the four points. - One input cannot have several outputs in a function. - After switching the coordinates, identify the new input set and the common output.

Solution

1. The points \((4, -2)\), \((4, 0)\), \((4, 2)\), and \((4, 5)\) all have x-coordinate \(4\), so they lie on the vertical line \(x=4\). 2. The input \(x=4\) is assigned four different outputs, so \(x\rightarrow y\) is not a function. 3. After switching the coordinates, the inputs are \(-2\), \(0\), \(2\), and \(5\), and every output is \(4\). Thus, \(f(u)=4\) for \(u\in\{-2, 0, 2, 5\}\).

Answer

a) The points lie on \(x=4\). b) It is not a function because \(x=4\) has multiple outputs. c) \(f(u)=4\) for \(u\in\{-2, 0, 2, 5\}\).
Function definition and input-output rules 51
The functions are \(T_1(y)=0.5y-3\) and \(T_2(y)=y^2-4\). Complete a table for \(y\in\{-4,-2,0,2,4\}\). For which listed input is \(T_1(y)>T_2(y)\)?

Hints

- A negative number squared is positive. - Evaluate both functions for each listed input. - Compare the two outputs in each row.

Solution

1. For \(y=-4\), the outputs are \(-5\) and \(12\). 2. For \(y=-2\), the outputs are \(-4\) and \(0\). 3. For \(y=0\), the outputs are \(-3\) and \(-4\). 4. For \(y=2\), the outputs are \(-2\) and \(0\). 5. For \(y=4\), the outputs are \(-1\) and \(12\). 6. Only at \(y=0\) is \(T_1(y)>T_2(y)\).

Answer

<table> <tr><th>\(y\)</th><th>\(T_1(y)\)</th><th>\(T_2(y)\)</th></tr> <tr><td>\(-4\)</td><td>\(-5\)</td><td>\(12\)</td></tr> <tr><td>\(-2\)</td><td>\(-4\)</td><td>\(0\)</td></tr> <tr><td>\(0\)</td><td>\(-3\)</td><td>\(-4\)</td></tr> <tr><td>\(2\)</td><td>\(-2\)</td><td>\(0\)</td></tr> <tr><td>\(4\)</td><td>\(-1\)</td><td>\(12\)</td></tr> </table> \(T_1(y)>T_2(y)\) only when \(y=0\).
Function definition and input-output rules 52
The function is \(T(x)=5x-7\). 1) Complete the table. <table> <tr> <td>\(x\)</td> <td>\(-3\)</td> <td>\(0\)</td> <td>\(2.5\)</td> </tr> <tr> <td>\(T(x)\)</td> <td>\(\square\)</td> <td>\(\square\)</td> <td>\(\square\)</td> </tr> </table> 2) Find the value of \(x\) for which \(T(x)=13\). 3) For what value of \(x\) do \(T(x)=5x-7\) and \(Q(x)=2x+8\) have the same output?

Hints

- Substitute each input into the function rule. - Represent a required output with an equation. - Set two function rules equal when their outputs must match. - Use equivalent operations to isolate the input.

Solution

1. Evaluate \(T(x)=5x-7\) at each input: \(T(-3)=5\cdot(-3)-7=-22\), \(T(0)=5\cdot0-7=-7\), and \(T(2.5)=5\cdot2.5-7=5.5\). 2. Solve \(5x-7=13\). Add \(7\): \(5x=20\). Divide by \(5\): \(x=4\). 3. Set the outputs equal: \(5x-7=2x+8\). Subtract \(2x\) and add \(7\): \(3x=15\). Divide by \(3\): \(x=5\).

Answer

1) The missing outputs are \(-22\), \(-7\), and \(5.5\). 2) \(x=4\) 3) \(x=5\)
Function definition and input-output rules 53
Two mobile phone plans have monthly costs in dollars, where \(m\) is the number of minutes used. Plan A: \(K=10+0.05m\) Plan B: \(K=5+0.10m\) a) Complete the table. <table> <tr><td>Minutes \((m)\)</td><td>\(0\)</td><td>\(50\)</td><td>\(100\)</td><td>\(150\)</td></tr> <tr><td>Plan A</td><td></td><td></td><td></td><td></td></tr> <tr><td>Plan B</td><td></td><td></td><td></td><td></td></tr> </table> b) For which listed number of minutes is Plan A less expensive than Plan B? Explain using the table.

Hints

- Substitute each value of \(m\) into both formulas. - The constant term is the cost at \(0\) minutes. - Compare the two entries in each column.

Solution

1. Plan A costs \(\$10.00\), \(\$12.50\), \(\$15.00\), and \(\$17.50\) for the listed inputs. 2. Plan B costs \(\$5.00\), \(\$10.00\), \(\$15.00\), and \(\$20.00\). 3. The plans cost the same at \(100\) minutes. At \(150\) minutes, Plan A is less expensive.

Answer

a) <table> <tr><td>Minutes \((m)\)</td><td>\(0\)</td><td>\(50\)</td><td>\(100\)</td><td>\(150\)</td></tr> <tr><td>Plan A</td><td>\(\$10.00\)</td><td>\(\$12.50\)</td><td>\(\$15.00\)</td><td>\(\$17.50\)</td></tr> <tr><td>Plan B</td><td>\(\$5.00\)</td><td>\(\$10.00\)</td><td>\(\$15.00\)</td><td>\(\$20.00\)</td></tr> </table> b) At \(150\) minutes, Plan A is less expensive: \(\$17.50<\$20.00\).
Function definition and input-output rules 54
Three graphs show relations between inputs \(x\) and outputs \(y\). a) Which graph is not a function of \(x\)? Give one input and two outputs that show why. b) Describe one part of that graph you could erase so that the remaining graph is a function. Explain why your change works.
Figure for problem 532193

Hints

- Pick one input and look for multiple points directly above or below it. - If several outputs share an input, consider which plotted points could be left out.

Solution

1. Graph 2 is not a function: at input \(x=1\), it has both outputs \(y=1\) and \(y=-1\). 2. Erase the branch below the x-axis, leaving the upper branch and its endpoint at \((0,0)\). Each remaining input has exactly one output. Erasing the upper branch instead would also work.

Answer

a) Graph 2; input \(1\) has outputs \(-1\) and \(1\). b) Erase either the upper or lower branch, keeping the point at the origin. Each input would then have one output.
Function definition and input-output rules 55
Neema walks from home to school. Graphs A and B show distance from home as a function of time. a) Explain which graph could represent a realistic walk to school and why the other graph is impossible in this context. b) Determine whether Graph A and Graph B are function graphs when time in minutes is the input and distance in feet is the output. Justify each decision.
Figure for problem 532207

Hints

- Interpret increasing, horizontal, and vertical segments in the context. - Time cannot reverse direction in a realistic time graph. - Use the vertical line test to check whether each input has only one output.

Solution

1. Graph A is realistic. Increasing segments represent walking away from home, and horizontal segments represent stopping, such as at a crosswalk. 2. Graph B is not realistic. At \(t=2\) minutes, it contains a vertical segment, which would place Neema at many distances at the same instant. It also moves backward in time from \(t=6\) to \(t=5\), which cannot represent a time-ordered trip. 3. Graph A passes the vertical line test, so it is a function graph. 4. Graph B fails the vertical line test. For example, \(t=2\) has many distances, and some inputs between \(5\) and \(6\) minutes have multiple distances.

Answer

a) Graph A is realistic. Graph B is impossible because it includes an instantaneous change in distance and a segment that moves backward in time. b) Graph A is a function graph. Graph B is not a function graph.
Function definition and input-output rules 56
A computer animation shows a ball rising and falling. The graph gives its height \(h\), in feet, as a function of time \(t\), in seconds. a) Find the ball's maximum height and when it occurs. b) How high is the ball after \(1\) second and after \(5\) seconds? c) If you know only the ball's height, can you always tell exactly when it was at that height? Explain.
Figure for problem 533196

Hints

- Find the graph's highest point. - Read the vertical value at each stated time. - For part c), test the claim at a specific height on the graph.

Solution

1. The highest point is \((3,18)\), so the maximum is \(18\,\text{ft}\) at \(t=3\,\text{s}\). 2. The graph gives \(h(1)=10\,\text{ft}\) and \(h(5)=10\,\text{ft}\). 3. No. A height of \(10\,\text{ft}\) occurs at both \(1\) and \(5\) seconds, so height does not determine exactly one time.

Answer

a) \(18\,\text{ft}\) after \(3\) seconds b) \(10\,\text{ft}\) at both times c) No. The same height may occur at two different times.
Function definition and input-output rules 57
A circular lake is shown on a coordinate map. a) Explain mathematically why the entire shoreline cannot be represented by one function \(y=f(x)\). b) Describe how to divide the shoreline into two parts so that each part is the graph of a function.
Figure for problem 533219

Hints

- Apply the vertical line test to an x-value near the center of the circle. - For part b), look for a way to divide the shoreline so that each resulting part has at most one point for every x-value.

Solution

1. For most x-values between the leftmost and rightmost points of the circle, a vertical line meets the shoreline twice: once on the upper half and once on the lower half. Therefore, one input would have two outputs, so the entire circle is not the graph of \(y=f(x)\). 2. Divide the circle horizontally into an upper semicircle and a lower semicircle. Each semicircle passes the vertical line test and can be represented as a separate function.

Answer

a) The circle is not a function graph because many x-values correspond to two y-values. b) Split it into the upper semicircle and the lower semicircle.
Function definition and input-output rules 58
A bicycle courier completes a delivery. The graph shows the courier's distance from the distribution center over time. a) How long does the entire trip last, from departure until return? b) How far from the center is the courier after \(20\) minutes? What is the courier probably doing from minute \(15\) to minute \(25\)? c) How far from the center is the courier after \(45\) minutes? d) Explain why the relation \(\text{time}\mapsto\text{distance}\) is a function.
Figure for problem 533249

Hints

- The trip ends when the graph returns to distance \(0\). - A horizontal segment means the distance from the center is unchanged. - Use the constant rate on the final segment to find the value at \(45\) minutes. - Apply the definition of a function or the vertical line test.

Solution

1. The graph begins at \(t=0\) and returns to distance \(0\) at \(t=60\) minutes, so the trip lasts \(60\) minutes. 2. At \(t=20\), the graph is on a horizontal segment at \(4\) miles. The courier is probably making a delivery or taking a short break. 3. From minute \(40\) to minute \(60\), the distance decreases linearly from \(6\) miles to \(0\). In \(5\) minutes, it decreases by \(\frac{5}{20}\cdot6=1.5\) miles, so after \(45\) minutes the distance is \(6-1.5=4.5\) miles. 4. Every input time is paired with exactly one distance. Equivalently, every vertical line intersects the graph at most once, so the relation is a function.

Answer

a) \(60\) minutes b) \(4\) miles; probably making a delivery or taking a break c) \(4.5\) miles d) Each time corresponds to exactly one distance.
Function definition and input-output rules 59
The graph shows the internet data rate in a shared apartment over one day. a) At what time is the data rate greatest, and approximately what is the maximum rate? b) During what approximate time intervals is the data rate greater than \(15\,\text{Mbps}\)? c) Someone claims, “The data rate never falls below \(5\,\text{Mbps}\).” Evaluate the claim using the graph. d) Is the reverse relation \(\text{data rate}\mapsto\text{time of day}\) a function? Explain.
Figure for problem 533250

Hints

- Locate the graph's highest point. - Compare the graph with a horizontal line at \(15\). - Check the early-morning portion of the graph for part c). - For the reverse relation, ask whether one data-rate value can occur at more than one time.

Solution

1. The graph reaches its maximum at about \(8{:}00\) p.m., with a data rate of about \(42\,\text{Mbps}\). 2. The graph is above \(15\,\text{Mbps}\) from about noon to \(2{:}30\) p.m. and from about \(4{:}30\) p.m. to \(11{:}00\) p.m. 3. The claim is false. During the overnight and early-morning hours, the graph is below \(5\,\text{Mbps}\). 4. The reverse relation is not a function because the same data rate occurs at several different times during the day. One input data rate would be paired with more than one output time.

Answer

a) About \(8{:}00\) p.m.; about \(42\,\text{Mbps}\) b) About noon–\(2{:}30\) p.m. and \(4{:}30\) p.m.–\(11{:}00\) p.m. c) False; the rate falls below \(5\,\text{Mbps}\) overnight. d) No; one data rate can correspond to several times.
Function definition and input-output rules 60
A parking garage charges \(\$1.50\) for the first started hour and \(\$1.00\) for each additional started hour. a) Find the fees for parking \(30\), \(60\), \(61\), and \(120\) minutes. b) A driver claims, “If I know the fee, I can always tell exactly how many minutes the car was parked.” Give an example showing why this is false. c) Suggest a simple charge per minute, with no rounding, so that knowing the fee tells you exactly how many minutes the car was parked. Explain why your rule works.

Hints

- A started hour counts even when only part of that hour is used. - To check the driver's claim, look for two different parking times with the same fee. - Think of a rule whose fee changes every time the parking time changes.

Solution

1. A stay of \(30\) or \(60\) minutes costs \(\$1.50\), since either lasts no more than the first hour. 2. A stay of \(61\) or \(120\) minutes costs \(\$1.50+\$1.00=\$2.50\). 3. The claim is false: the same fee of \(\$1.50\) applies to both \(30\) and \(60\) minutes. The fee does not determine just one parking time. 4. For example, charge \(\$0.05\) per minute with no rounding. Then \(P(t)=0.05t\), and each fee determines exactly one time through \(t=P(t)\div0.05\).

Answer

a) \(30\) min: \(\$1.50\); \(60\) min: \(\$1.50\); \(61\) min: \(\$2.50\); \(120\) min: \(\$2.50\) b) Both \(30\) and \(60\) minutes cost \(\$1.50\), so a fee need not determine one parking time. c) For example, \(P(t)=0.05t\), charging \(\$0.05\) for every exact minute. Different times then have different fees, so the time can be found from the fee.

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