Aimathic
Login | English | Deutsch

Free math worksheets

Build your own math worksheets from 30,000+ problems for grades 3 to 12, from fractions to AP Calculus. Every problem comes with step-by-step solutions.

Domain and range

Click problems to add them to your worksheet.

5549859
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: \(\{2,5,8\}\)
5549869
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) \(\{1,4,7\}\)
5129409
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) \(\{2,3,4,6,12\}\)
5549879
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: \((-2,3]\)
5549889
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 \(0\).
5549899
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.
5129289
A classroom temperature was recorded once each hour during the morning. <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 changes between consecutive temperature values. - The range contains only the output values that occur in the cleaned data. - The domain contains only the times when measurements were taken. - Extend the constant hourly change to estimate the missing trend value.

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 the unusual value. The data alone do not prove that 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. Continuing the pattern gives \(68.0+1.5=69.5\), so the estimated temperature at \(11{:}00\) a.m. 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) \(69.5\,^\circ\text{F}\)
5129419
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 \(0\).
5246589
Let \(f(x)=\sqrt[4]{x-3}\) and \(g(x)=\sqrt[3]{x-3}\). a) Find the maximal real domain of each function. b) Explain why the domains are different. c) Find all real values of \(x\) for which \(f(x)=g(x)\).

Hints

- Compare the domain restrictions for even-index and odd-index radicals. - In part c, substitute a new variable for \(x-3\). - Remember to consider zero separately before dividing by a power of the new variable.

Solution

a) For \(f\), the fourth root requires \(x-3\geq0\), so \(x\geq3\). Thus the domain of \(f\) is \([3,\infty)\). A cube root is defined for every real radicand, so the domain of \(g\) is \((-\infty, \infty)\). b) Even-index roots require nonnegative radicands because even powers cannot be negative over the real numbers. Odd-index roots accept every real radicand. c) First require \(x\geq3\). Let \(u=x-3\), so \(u\geq0\). Then \(u^{\frac{1}{4}}=u^{\frac{1}{3}}\). If \(u=0\), then \(x=3\). If \(u>0\), divide by \(u^{\frac{1}{4}}\) to get \(u^{\frac{1}{12}}=1\), so \(u=1\) and \(x=4\). Both values check in the original equation.

Answer

a) \(D_f=[3,\infty)\); \(D_g=(-\infty, \infty)\) b) An even-index root requires a nonnegative radicand, while an odd-index root is defined for every real radicand. c) \(x=3\) or \(x=4\)
5288089
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 \((-\infty,0)\cup(0,\infty)\)
5324649
The graph of \(f\) is defined on \([-3, 3]\) and consists of two line segments. A new function is defined by \(g(x)=1.5f(x-1)+1\). a) Describe the transformations that produce the graph of \(g\) from the graph of \(f\). b) Find the domain and range of \(g\).
Figure for problem 532464

Hints

- Changes inside the input affect horizontal position. - Factors and constants outside the function affect outputs. - Transform the original domain by solving an inequality for the new input. - Transform the original minimum and maximum output values.

Solution

1. The input \(x-1\) shifts the graph right \(1\) unit. 2. The factor \(1.5\) vertically stretches the graph by a factor of \(1.5\). 3. The outside \(+1\) shifts the graph up \(1\) unit. 4. For the domain, require \(-3\le x-1\le3\). Adding \(1\) gives \(-2\le x\le4\), so the domain is \([-2, 4]\). 5. From the graph, the range of \(f\) is \([-2, 2]\). Transforming the endpoint output values by \(y\mapsto1.5y+1\) gives \(-2\) and \(4\), so the range of \(g\) is \([-2, 4]\).

Answer

a) Shift right \(1\) unit, vertically stretch by a factor of \(1.5\), and shift up \(1\) unit. b) Domain: \([-2, 4]\); range: \([-2, 4]\)
5324669
The graph of \(f\) is shown with domain \([-3, 3]\). A new function is defined by \(g(x)=-f(x+1)+2\). a) Describe the transformations that produce the graph of \(g\) from the graph of \(f\). b) State the domain and range of \(g\).
Figure for problem 532466

Hints

- A change inside the input affects horizontal position. - A negative outside the function reflects output values. - An outside constant creates a vertical shift. - Apply the transformations to the endpoints of the domain and the extreme values of the range.

Solution

1. The input \(x+1\) shifts the graph left \(1\) unit. 2. The negative sign outside the function reflects the graph across the x-axis. 3. Adding \(2\) shifts the reflected graph up \(2\) units. 4. The original domain \([-3, 3]\) shifts left \(1\) unit, so the domain of \(g\) is \([-4, 2]\). 5. The graph shows that the range of \(f\) is \([-2, 2]\). Reflection keeps this interval unchanged, and shifting up \(2\) units gives the range \([0, 4]\).

Answer

a) Shift left \(1\) unit, reflect across the x-axis, and shift up \(2\) units. b) Domain: \([-4, 2]\); range: \([0, 4]\)
5324729
The graph of \(f\) is shown with marked points \(A\), \(B\), \(C\), and \(D\). a) Read the domain and the coordinates of the four marked points from the graph. Then, for \(g(x)=-0.5f(x)+1\), find the domain and the transformed coordinates of \(A\), \(B\), \(C\), and \(D\). b) For \(h(x)=f(0.5x)-1\), find the domain and the transformed coordinates of the four points.
Figure for problem 532472

Hints

- Read the endpoints and marked coordinates directly from the graph before transforming anything. - Outside operations change y-coordinates. - Inside operations change x-coordinates and the domain. - Transform each point separately and use the same horizontal rule on the domain endpoints.

Solution

1. From the graph, the domain of \(f\) is \([-2, 4]\), with \(A(-2, -2)\), \(B(0, 2)\), \(C(2, 2)\), and \(D(4, -2)\). 2. For \(g\), the input is unchanged, so the domain remains \([-2, 4]\). Transform each output by \(y\mapsto-0.5y+1\). 3. This gives \(A_g(-2, 2)\), \(B_g(0, 0)\), \(C_g(2, 0)\), and \(D_g(4, 2)\). 4. For \(h\), the input factor \(0.5\) creates a horizontal stretch by a factor of \(2\), so the domain becomes \([-4, 8]\). Then shift outputs down \(1\) unit. 5. Each point \((x, y)\) maps to \((2x, y-1)\), giving \(A_h(-4, -3)\), \(B_h(0, 1)\), \(C_h(4, 1)\), and \(D_h(8, -3)\).

Answer

a) Original domain: \([-2, 4]\); original points: \(A(-2, -2)\), \(B(0, 2)\), \(C(2, 2)\), \(D(4, -2)\). For \(g\), domain: \([-2, 4]\); points: \(A_g(-2, 2)\), \(B_g(0, 0)\), \(C_g(2, 0)\), \(D_g(4, 2)\). b) Domain: \([-4, 8]\); points: \(A_h(-4, -3)\), \(B_h(0, 1)\), \(C_h(4, 1)\), \(D_h(8, -3)\).
5334979
Match graphs \(f_1\), \(f_2\), and \(f_3\) to the correct function rules. - \(a(x)=\frac{1}{x^2}\) - \(b(x)=-\frac{1}{x}\) - \(c(x)=\sqrt{x}\) Justify each match using domain, sign, or symmetry.
Figure for problem 533497

Hints

- Compare the domains of the three functions. - Determine which graphs have only positive outputs. - Use y-axis or origin symmetry to distinguish the reciprocal functions.

Solution

1. Graph \(f_3\) is defined only for nonnegative inputs and begins at the origin, so it represents \(c(x)=\sqrt{x}\). 2. Graph \(f_1\) has only positive outputs and is symmetric about the y-axis, so it represents \(a(x)=\frac{1}{x^2}\). 3. Graph \(f_2\) lies in Quadrants II and IV. This matches \(b(x)=-\frac{1}{x}\), which is the graph of \(\frac{1}{x}\) reflected across the x-axis.

Answer

\(f_1: a(x)=\frac{1}{x^2}\); \(f_2: b(x)=-\frac{1}{x}\); \(f_3: c(x)=\sqrt{x}\)
5340459
The graph of \(f\) is shown. First read the domain and range of \(f\) from the graph. Then, for each function below, state the domain and range and describe the transformation from the graph of \(f\). a) \(g(x)=f(x+2)+2\) b) \(h(x)=-2f(x)\)
Figure for problem 534045

Hints

- Use the graph to identify the leftmost and rightmost inputs and the lowest and highest outputs. - Separate changes inside the input from changes to the output. - A horizontal shift changes the domain interval; a vertical change acts on the range. - When outputs are multiplied by a negative number, track both reflection and scale.

Solution

1. From the graph, the domain of \(f\) is \([-4, 4]\) and its range is \([-1, 1]\). 2. For \(g\), replacing \(x\) with \(x+2\) shifts the graph left \(2\) units, and adding \(2\) shifts it up \(2\) units. 3. The domain becomes \([-6, 2]\), and the range becomes \([1, 3]\). 4. For \(h\), multiplying the output by \(-2\) reflects the graph across the x-axis and stretches it vertically by a factor of \(2\). 5. The domain stays \([-4, 4]\), and the range becomes \([-2, 2]\).

Answer

Original \(f\): domain \([-4, 4]\), range \([-1, 1]\). a) Domain: \([-6, 2]\); range: \([1, 3]\); shift left \(2\) units and up \(2\) units. b) Domain: \([-4, 4]\); range: \([-2, 2]\); reflect across the x-axis and stretch vertically by a factor of \(2\).
5340609
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. \(x=-4, 0, 3\)
5340839
The graph of \(f\) is defined on \([-3, 3]\) and has range \([-2, 2]\). a) Find the transformed vertices of \(g(x)=0.5f(x+2)-1\). b) Find the domain and range of \(g\).
Figure for problem 534083

Hints

- Transform each vertex one coordinate at a time. - The change inside the input affects x-coordinates and the domain. - The multiplier and constant outside the function affect y-coordinates and the range. - Keep the transformed vertices in the original order.

Solution

1. The vertices of \(f\) are \((-3, -2)\), \((-1, 2)\), \((1, 2)\), and \((3, 0)\). 2. The input \(x+2\) shifts the graph left \(2\) units. Multiplying outputs by \(0.5\) compresses vertically by a factor of \(0.5\), and subtracting \(1\) shifts the graph down \(1\) unit. 3. Each point \((x, y)\) maps to \((x-2, 0.5y-1)\). The transformed vertices are \((-5, -2)\), \((-3, 0)\), \((-1, 0)\), and \((1, -1)\). 4. The domain shifts from \([-3, 3]\) to \([-5, 1]\). 5. The range \([-2, 2]\) is multiplied by \(0.5\) and shifted down \(1\), giving \([-2, 0]\).

Answer

a) Transformed vertices: \((-5, -2)\), \((-3, 0)\), \((-1, 0)\), and \((1, -1)\) b) Domain: \([-5, 1]\); range: \([-2, 0]\)
5340849
The graph of \(h\) is shown with three vertices. a) Read the domain and the three vertex coordinates from the graph. Then find the transformed vertices of \(k(x)=-h(0.5x)+2\). b) State the domain of \(k\) and describe the transformations.
Figure for problem 534084

Hints

- Read the endpoint x-values and vertex coordinates from the graph before applying the transformation. - Determine how \(0.5x\) changes x-coordinates. - Apply the reflection and vertical shift to each y-coordinate. - Use the transformed endpoint x-values to check the new domain.

Solution

1. From the graph, the domain of \(h\) is \([-1, 4]\), and the vertices are \((-1, 1)\), \((1, -1)\), and \((4, 2)\). 2. Replacing \(x\) with \(0.5x\) stretches the graph horizontally by a factor of \(2\). Multiplying the output by \(-1\) reflects the graph across the x-axis, and adding \(2\) shifts it up \(2\) units. 3. Each point \((x, y)\) on \(h\) maps to \((2x, -y+2)\) on \(k\). 4. The transformed vertices are \((-2, 1)\), \((2, 3)\), and \((8, 0)\). 5. The domain of \(k\) is \([-2, 8]\).

Answer

a) Original domain: \([-1, 4]\); original vertices: \((-1, 1)\), \((1, -1)\), \((4, 2)\). Transformed vertices: \((-2, 1)\), \((2, 3)\), and \((8, 0)\). b) Domain: \([-2, 8]\); stretch horizontally by a factor of \(2\), reflect across the x-axis, and shift up \(2\) units.
5549909
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
5549919
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.

All problems may be used, copied and printed free of charge for school and tutoring, including paid tutoring. Commercial adaptations as well as publication or redistribution on the internet are not permitted.

Building free math worksheets