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Domain and range

Learning goals

  • Determines domain and range from equations, graphs, tables, or contexts.
  • Expresses domain and range using appropriate notation.
  • Accounts for contextual or algebraic restrictions.
  • Distinguishes input restrictions from output restrictions.

Click problems to add them to your worksheet.

Domain and range 1
The table shows all input-output pairs of a function \(q\). <table><tr><td>\(x\)</td><td>\(-4\)</td><td>\(-1\)</td><td>\(0\)</td><td>\(3\)</td></tr><tr><td>\(q(x)\)</td><td>\(2\)</td><td>\(2\)</td><td>\(5\)</td><td>\(8\)</td></tr></table> State the domain and the range of \(q\).

Hints

- The domain comes from the input values. - The range comes from the output values. - A set lists each distinct value only once.

Solution

1. The domain is the set of all listed inputs: \(\{-4,-1,0,3\}\). 2. The range is the set of output values that occur: \(\{2,5,8\}\). 3. The output \(2\) is listed only once in the range even though two inputs produce it.

Answer

Domain: \(\{-4,-1,0,3\}\) Range: \(\{2,5,8\}\)
Domain and range 2
The function \(m\) is defined by these input-output mappings: \(-2\to4\), \(0\to1\), \(3\to4\), \(5\to7\). a) State the domain of \(m\). b) State the range of \(m\).

Hints

- Read each arrow from input to output. - Collect the values on the left for the domain and the values on the right for the range. - In a set, do not list a repeated output more than once.

Solution

1. The inputs are \(-2\), \(0\), \(3\), and \(5\), so the domain is \(\{-2,0,3,5\}\). 2. The outputs are \(4\), \(1\), \(4\), and \(7\). A set lists repeated values only once, so the range is \(\{1,4,7\}\).

Answer

a) \(\{-2,0,3,5\}\) b) \(\{1,4,7\}\)
Domain and range 3
A relation contains the points \((-2,5)\), \((0,1)\), and \((3,5)\). List its domain and its range. Which set contains the inputs?

Hints

- Which coordinate is the input in an ordered pair? - List the output values without repeating any number.

Solution

1. The inputs are the x-coordinates, so the domain is \(\{-2,0,3\}\). 2. The outputs are the y-coordinates. The value \(5\) is repeated, so the range is \(\{1,5\}\). 3. The domain is the set of inputs.

Answer

Domain: \(\{-2,0,3\}\). Range: \(\{1,5\}\). The domain contains the inputs.
Domain and range 4
This table lists a function's outputs. <table><tr><th>Input</th><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td></tr><tr><th>Output</th><td>\(6\)</td><td>\(9\)</td><td>\(6\)</td><td>\(12\)</td></tr></table> How many different values are in the range? Write the range as a set.

Hints

- The range contains output values, not input values. - Does a value appear twice in a set?

Solution

1. Read the output row: \(6,9,6,12\). 2. The repeated \(6\) belongs in the set only once. 3. The range is \(\{6,9,12\}\), containing three values.

Answer

Three different outputs; range \(\{6,9,12\}\).
Domain and range 5
A school club can buy up to four notebooks. Each notebook costs \(\$2\). Let \(C(n)=2n\) be the total cost in dollars. List the possible inputs \(n\) and the possible outputs \(C(n)\), assuming the club buys a whole number of notebooks.

Hints

- Can the club buy part of a notebook? - Include the case where no notebooks are purchased.

Solution

1. The number of notebooks can be any whole number from \(0\) through \(4\), so the domain is \(\{0,1,2,3,4\}\). 2. Multiply each input by \(2\). The range is \(\{0,2,4,6,8\}\) dollars.

Answer

Domain: \(\{0,1,2,3,4\}\). Range: \(\{0,2,4,6,8\}\) dollars.
Domain and range 6
Let \(f(x)=\frac{12}{x}\). a) Find the maximum domain \(D\) when the input values are real numbers. b) Use the equation to explain why \(0\) cannot be in the range. c) Restrict the domain to \(D_{\text{new}}=\{1,2,3,4,6\}\). State the corresponding range.

Hints

- Identify the input that makes the denominator zero. - Recall when a fraction can equal zero. - Evaluate the function at each value in the restricted domain.

Solution

1. Division by zero is undefined, so \(x\neq0\). Therefore, \(D=\mathbb{R}\setminus\{0\}\). 2. A fraction equals zero only when its numerator is zero. Because the numerator is always \(12\), \(f(x)\) can never equal \(0\). 3. The function values are \(f(1)=12\), \(f(2)=6\), \(f(3)=4\), \(f(4)=3\), and \(f(6)=2\). 4. Therefore, the restricted range is \(\{2,3,4,6,12\}\).

Answer

a) \(D=\mathbb{R}\setminus\{0\}\) b) The numerator is always \(12\), so \(\frac{12}{x}\) cannot equal \(0\). c) \(\{2,3,4,6,12\}\)
Domain and range 7
Let \(f(x)=\sqrt[4]{x-3}\) and \(g(x)=\sqrt[3]{x-3}\). a) Find the largest real domain of each function. b) Explain why the domains are different. c) Is \(x=1\) allowed as an input for either function? Explain.

Hints

- Think about which real numbers can be raised to an even power. - Compare that with raising a negative number to an odd power. - Check whether each radicand is allowed at the requested input.

Solution

1. The fourth root in \(f\) requires \(x-3\ge0\), giving domain \([3,\infty)\). A cube root accepts any real radicand, so the domain of \(g\) is \((-\infty,\infty)\). 2. Even powers of real numbers cannot be negative, but odd powers can. Therefore, even-index and odd-index roots have different input restrictions. 3. At \(x=1\), the radicand is \(-2\). A real fourth root is not defined, but a real cube root is. Thus only \(g(1)\) is defined.

Answer

a) \(D_f=[3,\infty)\); \(D_g=(-\infty,\infty)\). b) An even-index root needs a nonnegative radicand, but an odd-index root does not. c) Only \(g\) accepts \(x=1\); \(f(1)\) is not defined as a real number.
Domain and range 8
Casey says the range of \(f\) is \([-2,0]\) because those are the y-values at the endpoints. Find the domain and range of \(f\). Explain what Casey missed.
Figure for problem 532464

Hints

- Could an output between the endpoints be higher than both endpoint outputs? - Trace the whole graph, not only its ends, when looking for the highest value.

Solution

1. The graph begins at \(x=-3\) and ends at \(x=3\), so the domain is \([-3,3]\). 2. Its lowest output is \(-2\) at the right endpoint, and its highest output is \(2\) at \(x=-1\), an interior point. The range is \([-2,2]\). 3. Casey missed the interior peak. Reading only the endpoints does not give the full range.

Answer

Domain: \([-3,3]\). Range: \([-2,2]\). Casey missed the peak at \((-1,2)\), so the range reaches above both endpoint outputs.
Domain and range 9
The graph shows a function \(g\). State the domain and range of \(g\) using interval notation.
Figure for problem 554987

Hints

- For the domain, scan the graph from left to right and pay attention to the endpoint symbols. - For the range, scan from the lowest y-value to the highest y-value. - An open endpoint is excluded; a filled endpoint is included.

Solution

1. The graph begins at \((-4,-2)\) with an open endpoint, so \(x=-4\) is excluded. It ends at \((5,1)\) with a closed endpoint, so \(x=5\) is included. The domain is \((-4,5]\). 2. The lowest y-value is \(-2\), but it occurs only at the open endpoint, so it is excluded. The highest y-value is \(3\) at an included point. The range is \((-2,3]\).

Answer

Domain: \((-4,5]\) Range: \((-2,3]\)
Domain and range 10
Let \(r(x)=\frac{x+2}{x-5}\). State the domain of \(r\) in interval notation and explain why one input must be excluded.

Hints

- Ask whether every real input can be substituted into the formula. - Focus on the denominator and identify when the expression would be undefined. - Write all remaining real inputs as two intervals.

Solution

1. A rational expression is undefined when its denominator equals \(0\). 2. Solve \(x-5=0\), which gives \(x=5\). 3. Therefore, the domain is all real numbers except \(5\): \((-\infty,5)\cup(5,\infty)\).

Answer

Domain: \((-\infty,5)\cup(5,\infty)\). The input \(5\) is excluded because it makes the denominator \(0\).
Domain and range 11
A school play has \(12\) unsold seats in one row. On the ticketing screen, a user may select any whole number of those seats, including selecting none before checkout. Let \(t\) be the number of seats selected, and let \(C(t)\) be the displayed order cost. What is the domain of \(C\) in this situation? Explain why an interval such as \([0,12]\) is not the most precise description.

Hints

- Use the meaning of the input variable, not just its minimum and maximum. - Decide whether values between consecutive whole numbers make sense for this quantity. - Include every allowable selection count from the smallest possible value to the largest.

Solution

1. The number of selected seats cannot be negative and cannot exceed the \(12\) seats available. 2. Seats are counted in whole numbers, so fractional values are not possible. The screen also allows \(0\) selected seats before checkout. 3. The contextual domain is \(\{0,1,2,\ldots,12\}\). The interval \([0,12]\) incorrectly includes noninteger values.

Answer

Domain: \(\{0,1,2,\ldots,12\}\). The domain is discrete because the number of selected seats must be a whole number.
Domain and range 12
Find the domain and range of the graphed segment. An open circle means that endpoint is not included.
Figure for problem 560221

Hints

- Look for the smallest and largest x-coordinates shown. - Then look for the lowest and highest y-values. - Which endpoints are open, and which are filled?

Solution

1. The segment starts at x-coordinate \(-2\) with an open endpoint and ends at x-coordinate \(3\) with a filled endpoint. The domain is \((-2,3]\). 2. Its y-values go from \(1\) (not included) to \(4\) (included). The range is \((1,4]\).

Answer

Domain: \((-2,3]\). Range: \((1,4]\).
Domain and range 13
Two functions have the same allowed inputs: \(x\in\{-1,0,1\}\). \(f(x)=x+1\) and \(g(x)=2x+1\). Do they also have the same range? Show the outputs for both functions.

Hints

- Substitute all three allowed inputs into each rule. - Compare the sets of outputs, not the formulas alone.

Solution

1. For \(x=-1,0,1\), the outputs of \(f\) are \(0,1,2\). Its range is \(\{0,1,2\}\). 2. For the same inputs, the outputs of \(g\) are \(-1,1,3\). Its range is \(\{-1,1,3\}\). 3. The two ranges differ even though the domains are equal.

Answer

For inputs \(-1,0,1\), the outputs of \(f\) are \(0,1,2\), and the outputs of \(g\) are \(-1,1,3\), in that order. No: their ranges are \(\{0,1,2\}\) and \(\{-1,1,3\}\).
Domain and range 14
A classroom temperature was recorded once each hour during the school day. <table><thead><tr><th>Time \(t\)</th><th>\(8{:}00\) a.m.</th><th>\(9{:}00\) a.m.</th><th>\(10{:}00\) a.m.</th><th>\(11{:}00\) a.m.</th><th>\(12{:}00\) p.m.</th><th>\(1{:}00\) p.m.</th></tr></thead><tbody><tr><td>Temperature \(T(t)\) in \(^\circ\text{F}\)</td><td>\(65.0\)</td><td>\(66.5\)</td><td>\(68.0\)</td><td>\(92.0\)</td><td>\(71.0\)</td><td>\(72.5\)</td></tr></tbody></table> a) One value does not fit the linear pattern in the other data. Identify the value and justify your choice mathematically. b) After excluding the unusual value, state the range of the remaining time-temperature relation. c) State the domain of the original relation, using hour numbers. d) Estimate the temperature at \(11{:}00\) a.m. if the pattern in the other measurements had continued.

Hints

- Compare the hourly changes in temperature. - The range lists output values from the retained data. - The domain lists the hours when measurements were taken. - Extend the consistent hourly pattern to estimate the unusual reading.

Solution

1. Except for the \(11{:}00\) a.m. value, the temperature increases by \(1.5\,^\circ\text{F}\) each hour. Therefore, \(92.0\,^\circ\text{F}\) is unusual. The data alone do not prove it was a measurement error. 2. Excluding that value, the range is \(\{65.0,66.5,68.0,71.0,72.5\}\). 3. The original domain is \(D=\{8,9,10,11,12,13\}\). 4. Extending the regular change gives \(68.0+1.5=69.5\), so the estimated temperature is \(69.5\,^\circ\text{F}\).

Answer

a) \(92.0\,^\circ\text{F}\) at \(11{:}00\) a.m.; the other values increase by \(1.5\,^\circ\text{F}\) per hour. b) \(\{65.0,66.5,68.0,71.0,72.5\}\). c) \(D=\{8,9,10,11,12,13\}\). d) \(69.5\,^\circ\text{F}\).
Domain and range 15
Consider \(f(x)=3x\) and \(g(x)=\frac{3}{x}\), with real-number inputs. a) Explain the difference between the domains of \(f\) and \(g\). b) For each function, find the input \(x\) that produces an output of \(12\). c) Explain why the graph of \(g\) never touches or crosses the x-axis, while the graph of \(f\) has exactly one x-intercept.

Hints

- Determine whether either formula is undefined for any real input. - Set each function equal to \(12\) and solve. - An x-intercept occurs where the output equals \(0\).

Solution

1. Every real number can be used in \(f\), so \(D_f=\mathbb{R}\). For \(g\), \(x=0\) would cause division by zero, so \(D_g=\mathbb{R}\setminus\{0\}\). 2. For \(f\), solve \(3x=12\) to get \(x=4\). 3. For \(g\), solve \(\frac{3}{x}=12\). This gives \(3=12x\), so \(x=\frac{1}{4}\). 4. An x-intercept requires an output of \(0\). The equation \(3x=0\) has the single solution \(x=0\). The equation \(\frac{3}{x}=0\) has no solution because its numerator is nonzero.

Answer

a) \(D_f=\mathbb{R}\), while \(D_g=\mathbb{R}\setminus\{0\}\). b) For \(f\), \(x=4\). For \(g\), \(x=\frac{1}{4}\). c) \(f\) has the x-intercept \((0, 0)\). The function \(g\) has no x-intercept because \(\frac{3}{x}\) can never equal \(0\).
Domain and range 16
For each relation below, decide whether it defines \(y\) as a function of \(x\). For every relation that is a function, solve for \(y\) and state its domain in interval notation. For any relation that is not a function, give one specific \(x\)-value that corresponds to two different \(y\)-values. a) \(y^3=x+8\) b) \(3y+0x=12\) c) \(x^2+y^2=16\) d) \(xy-y=x\) e) \(\frac{y}{x}=4\) Keep any domain restrictions from the original equation.

Hints

- For each relation, ask whether one allowed input can produce two different outputs. - After solving for \(y\), inspect any denominator or other operation that restricts the original input. - A relation can be a function even when one real input must be excluded from its domain. - When you rewrite an equation, keep restrictions that came from the original form.

Solution

1. a) Solving gives \(y=\sqrt[3]{x+8}\). Every real \(x\) produces exactly one real cube root, so the domain is \((-\infty,\infty)\). 2. b) The equation simplifies to \(y=4\), so every real \(x\) is allowed. The domain is \((-\infty,\infty)\). 3. c) The relation does not define \(y\) as a function of \(x\). For example, when \(x=0\), both \(y=4\) and \(y=-4\) satisfy the equation. 4. d) Factor \(y\): \(y(x-1)=x\), so \(y=\frac{x}{x-1}\). The input \(x=1\) is excluded, so the domain is \((-\infty,1)\cup(1,\infty)\). 5. e) The original quotient requires \(x\ne0\). Multiplying by \(x\) gives \(y=4x\), with domain \((-\infty,0)\cup(0,\infty)\).

Answer

a) Function: \(y=\sqrt[3]{x+8}\); domain \((-\infty,\infty)\) b) Function: \(y=4\); domain \((-\infty,\infty)\) c) Not a function; for example, \(x=0\) gives \(y=4\) and \(y=-4\) d) Function: \(y=\frac{x}{x-1}\); domain \((-\infty,1)\cup(1,\infty)\) e) Function: \(y=4x\); domain \((-\infty,0)\cup(0,\infty)\)
Domain and range 17
Read the domain and range of \(f\) from its graph. Let \(g(x)=-f(x+1)+2\). Find the domain and range of \(g\). Why does the negative sign by itself not change the range interval in this example?
Figure for problem 532466

Hints

- Find the original domain and range before considering the new function. - Which values of \(x\) make \(x+1\) an allowed input for \(f\)? - What happens to the original lowest and highest outputs when their signs change?

Solution

1. The original graph has domain \([-3,3]\) and range \([-2,2]\). 2. For \(g\), require \(-3\le x+1\le3\), giving domain \([-4,2]\). 3. Reflecting the outputs changes \(y\) to \(-y\). The interval \([-2,2]\) stays \([-2,2]\) because it is symmetric about zero. Adding \(2\) then gives range \([0,4]\).

Answer

\(f\): domain \([-3,3]\), range \([-2,2]\). \(g\): domain \([-4,2]\), range \([0,4]\). Reflection alone leaves this range interval unchanged because \([-2,2]\) is symmetric about zero.
Domain and range 18
Read the domain and range of \(f\) from the graph. A student says, “If we use only inputs from \(1\) to \(3\), the range will stay the same.” Is that correct? State the new domain and range and explain using the graph.
Figure for problem 532472

Hints

- First check the leftmost, rightmost, highest, and lowest points of the full graph. - Look only at the part whose x-values lie between the two new limits. - Which output values does that shorter piece still reach?

Solution

1. The graph of \(f\) begins at \(x=-2\), ends at \(x=4\), and its y-values run from \(-2\) to \(2\). Thus its domain is \([-2,4]\) and range is \([-2,2]\). 2. With inputs restricted to \([1,3]\), the graph stays at \(y=2\) from \(x=1\) to \(x=2\), then falls to \(y=0\) at \(x=3\). 3. The new domain is \([1,3]\) and range is \([0,2]\). The student is incorrect because negative outputs are no longer reached.

Answer

\(f\): domain \([-2,4]\), range \([-2,2]\). Restricted graph: domain \([1,3]\), range \([0,2]\). No, the range becomes smaller.
Domain and range 19
Match graphs 1, 2, and 3 to the correct function rules. - \(a(x)=\frac{1}{x^2}\) - \(b(x)=-\frac{1}{x}\) - \(c(x)=\sqrt{x}\) For each rule, give its domain and range for all real inputs where it is defined. Explain how the graph helps you match the rule. The picture shows only part of each graph.
Figure for problem 533497

Hints

- Which input would cause division by zero, or make a square root of a negative number? - Ask which signs each output can have, including whether it can equal zero. - Parts of a curve may be outside the picture. Use the rule to decide its full domain and range.

Solution

1. The rule \(a(x)=\frac{1}{x^2}\) excludes \(x=0\) because of division by zero. Its output is always positive, but never zero. Its domain is \((-\infty,0)\cup(0,\infty)\) and its range is \((0,\infty)\). This matches graph 1, with both branches above the x-axis. 2. The rule \(b(x)=-\frac{1}{x}\) also excludes \(x=0\), but its output can be positive or negative, never zero. Its domain and range are both \((-\infty,0)\cup(0,\infty)\). This matches graph 2, with branches in Quadrants II and IV. 3. The rule \(c(x)=\sqrt{x}\) needs \(x\ge0\), and its outputs are nonnegative. Its domain and range are both \([0,\infty)\). This matches graph 3, which starts at the origin.

Answer

Graph 1: \(a(x)=\frac{1}{x^2}\). Domain: \((-\infty,0)\cup(0,\infty)\). Range: \((0,\infty)\). Graph 2: \(b(x)=-\frac{1}{x}\). Domain: \((-\infty,0)\cup(0,\infty)\). Range: \((-\infty,0)\cup(0,\infty)\). Graph 3: \(c(x)=\sqrt{x}\). Domain: \([0,\infty)\). Range: \([0,\infty)\).
Domain and range 20
Read the domain and range of \(f\) from the graph. a) For which inputs is \(f(x)\ge0\)? What range do those inputs give? b) For which inputs is \(f(x)<0\)? What range do those inputs give?
Figure for problem 534045

Hints

- Where does the graph cross the x-axis? - Compare the parts on or above the axis with the parts strictly below it. - Pay attention to whether the points at height zero should be included.

Solution

1. The graph of \(f\) covers \(-4\le x\le4\) and \(-1\le y\le1\), so \(D_f=[-4,4]\) and \(R_f=[-1,1]\). 2. The graph crosses the x-axis at \(x=-3\) and \(x=3\). It is on or above the x-axis from \(-3\) to \(3\), so for part a) the input set is \([-3,3]\) and the range is \([0,1]\). 3. The graph is below the x-axis from \(-4\) up to but not including \(-3\), and from just after \(3\) to \(4\). For part b) the input set is \([-4,-3)\cup(3,4]\) and the range is \([-1,0)\).

Answer

\(f\): domain \([-4,4]\), range \([-1,1]\). a) Inputs \([-3,3]\); range \([0,1]\). b) Inputs \([-4,-3)\cup(3,4]\); range \([-1,0)\).
Domain and range 21
The graph of \(p\) is shown. Define \(q(x)=\frac{1}{2}p(x)-2\). 1. State the domain and range of \(p\). 2. Use the transformation to find the domain and range of \(q\). 3. For which values of \(x\) does \(q(x)=-1\)? Use the graph of \(p\).
Figure for problem 534060

Hints

- Read the smallest and largest input and output values from the graph. - An output transformation does not change the domain. - Apply the transformation to the endpoints of the original range. - Rewrite \(q(x)=-1\) as an equation involving \(p(x)\).

Solution

1. The graph extends from \(x=-5\) through \(x=3\), including both endpoints, so the domain of \(p\) is \([-5, 3]\). Its y-values range from \(1\) to \(4\), so the range is \([1, 4]\). 2. The transformation does not change the inputs, so the domain of \(q\) is also \([-5, 3]\). 3. The output transformation sends the range endpoints to \(\frac{1}{2}\cdot 1-2=-1.5\) and \(\frac{1}{2}\cdot 4-2=0\). Since the transformation is increasing, the range of \(q\) is \([-1.5, 0]\). 4. Solve \(\frac{1}{2}p(x)-2=-1\). This is equivalent to \(p(x)=2\). From the graph, \(p(x)=2\) at \(x=-4\), \(x=0\), and \(x=3\).

Answer

1. Domain: \([-5, 3]\); range: \([1, 4]\) 2. Domain: \([-5, 3]\); range: \([-1.5, 0]\) 3. \(x=-4, 0, 3\)
Domain and range 22
Read the domain and range of \(f\) from the graph. The function \(g\) uses the same rule as \(f\), but allows only inputs from \(-3\) to \(-2\) or from \(1\) to \(3\), including the endpoints. Find the domain and range of \(g\). Does its smaller domain give it a smaller range? Explain using the graph.
Figure for problem 534083

Hints

- Look only at the two allowed parts of the graph, not the gap between them. - What y-values occur on each allowed part? - Combine the two sets of outputs. Is there a missing output between the lowest and highest?

Solution

1. The graph of \(f\) has domain \([-3,3]\) and range \([-2,2]\). 2. The allowed inputs for \(g\) give domain \([-3,-2]\cup[1,3]\). 3. On \([-3,-2]\), the graph's outputs run from \(-2\) to \(0\). On \([1,3]\), they run from \(2\) down to \(0\). These intervals together cover \([-2,2]\). 4. The domain of \(g\) is smaller, but its range is still \([-2,2]\) because the remaining graph pieces reach every original output.

Answer

\(f\): domain \([-3,3]\), range \([-2,2]\). \(g\): domain \([-3,-2]\cup[1,3]\), range \([-2,2]\). The domain shrinks, but the range does not.
Domain and range 23
Read the domain and range of \(h\) from the graph. A new function is defined by \(k(x)=-h(0.5x)+2\). Find the domain and range of \(k\). Explain why the order of the largest and smallest outputs reverses before adding \(2\).
Figure for problem 534084

Hints

- Find the original graph's extreme inputs and outputs. - What x-values make the inside input \(0.5x\) fall in the original domain? - Compare the negatives of two different original outputs before adding \(2\).

Solution

1. The graph of \(h\) runs from \(x=-1\) to \(x=4\), with y-values from \(-1\) to \(2\), so \(D_h=[-1,4]\) and \(R_h=[-1,2]\). 2. For \(k\), require \(-1\le0.5x\le4\). Multiplying by \(2\) gives domain \([-2,8]\). 3. Multiplication by \(-1\) changes the output interval \([-1,2]\) to \([-2,1]\): the old maximum becomes the new minimum. 4. Adding \(2\) gives range \([0,3]\).

Answer

\(h\): domain \([-1,4]\), range \([-1,2]\). \(k\): domain \([-2,8]\), range \([0,3]\). Negating an output changes its sign: the old maximum \(2\) becomes the new minimum \(-2\), and the old minimum \(-1\) becomes the new maximum \(1\). Adding \(2\) then shifts both up.
Domain and range 24
A tank contains \(120\,\text{gal}\) of water and drains at a constant rate of \(15\,\text{gal}\) per minute until it is empty. Let \(V(t)=120-15t\) be the volume of water, in gallons, after \(t\) minutes. State the contextual domain and range of \(V\), including units.

Hints

- The formula is used only while the physical draining process is happening. - Find the time interval from the start until the tank reaches zero volume. - Then identify the smallest and largest volumes attained during that interval.

Solution

1. Time starts at \(t=0\). The tank is empty when \(120-15t=0\), so \(t=8\) minutes. 2. Therefore, the contextual domain is \([0,8]\) minutes. 3. The volume starts at \(120\,\text{gal}\) and decreases continuously to \(0\,\text{gal}\), so the range is \([0,120]\) gallons.

Answer

Domain: \([0,8]\) minutes Range: \([0,120]\) gallons
Domain and range 25
Two representations use the same rule \(y=2x-1\). Representation A is the graph shown. Representation B is the table. <table><tr><td>\(x\)</td><td>\(-2\)</td><td>\(0\)</td><td>\(2\)</td></tr><tr><td>\(y\)</td><td>\(-5\)</td><td>\(-1\)</td><td>\(3\)</td></tr></table> a) State the domain of Representation A. b) State the domain of Representation B. c) Explain why the domains are different even though both representations follow the same rule.
Figure for problem 554991

Hints

- For the graph, look at every x-value covered from the left endpoint to the right endpoint. - For the table, use only the inputs that are actually listed. - The same algebraic rule does not force two representations to have the same allowed input set.

Solution

1. The graph includes every x-value from \(-2\) through \(2\), including both endpoints, so Representation A has domain \([-2,2]\). 2. The table defines only the three displayed input values, so Representation B has domain \(\{-2,0,2\}\). 3. A formula describes how outputs are related to allowed inputs, but the representation or context can restrict which inputs are actually included. The graph is continuous on an interval, while the table lists only three discrete inputs.

Answer

a) \([-2,2]\) b) \(\{-2,0,2\}\) c) The graph includes every real input from \(-2\) to \(2\), while the table includes only the three listed inputs.

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