Aimathic – Domain and range – Free Math Worksheets

Domain and range

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Difficulty: 1 – 5

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) \(\{2,3,4,6,12\}\)

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 \(0\).

5288089

Decide whether each equation defines \(y\) as a function of \(x\). Justify your answer by checking whether each x-value in the relation has a unique y-value. When it is a function, write its equation and domain. a) \(y^3=x+8\) b) \(3y+0x=12\) c) \(x^2+y^2=16\) d) \(xy-y=x\) e) \(\frac{y}{x}=4\)

Hints

- A missing y-value at one input can restrict the domain without preventing the relation from being a function. - Look for an x-value that produces two different y-values. - When dividing by an expression, exclude values that make that expression zero. - Keep domain restrictions from the original equation.

Solution

1. For a), \(y=\sqrt[3]{x+8}\). Every real x-value has exactly one real cube root, so this is a function with domain \((-\infty, \infty)\). 2. For b), \(3y=12\), so \(y=4\). This constant relation is a function for all real x-values. 3. For c), choosing \(x=0\) gives \(y^2=16\), so \(y=4\) or \(y=-4\). Therefore, this is not a function of \(x\). 4. For d), factor to get \(y(x-1)=x\). When \(x\ne1\), \(y=\frac{x}{x-1}\), which is unique. At \(x=1\), the equation becomes \(0=1\), so that input is excluded. This is a function with domain \((-\infty, 1)\cup(1, \infty)\). 5. For e), the original quotient requires \(x\ne0\). Multiplying by \(x\) gives \(y=4x\). This is a function 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 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 \(g\).

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 \(g\).

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 \(f\) has domain \([-2, 4]\) and contains the marked points \(A(-2, -2)\), \(B(0, 2)\), \(C(2, 2)\), and \(D(4, -2)\). a) 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.

Hints

- Outside operations change y-coordinates. - Inside operations change x-coordinates and the domain. - A factor of \(0.5\) inside the input doubles horizontal coordinates. - Transform each point separately and check the new domain.

Solution

1. For \(g\), the input is unchanged, so the domain remains \([-2, 4]\). Transform each output by \(y\mapsto-0.5y+1\). 2. This gives \(A_g(-2, 2)\), \(B_g(0, 0)\), \(C_g(2, 0)\), and \(D_g(4, 2)\). 3. 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. 4. 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) 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: \(A_h(-4, -3)\), \(B_h(0, 1)\), \(C_h(4, 1)\), \(D_h(8, -3)\)

5340459

The graph of \(f\) is defined on \([-4, 4]\) and has range \([-1, 1]\). For each function, state the domain and range and describe the transformation from the graph of \(f\). a) \(g(x)=f(x+2)+2\) b) \(h(x)=-2f(x)\)

Hints

- Separate changes inside the input from changes to the output. - A horizontal shift changes the domain interval. - A vertical shift changes the range interval. - Track the minimum and maximum outputs when multiplying by a negative number.

Solution

1. For \(g\), replacing \(x\) with \(x+2\) shifts the graph left \(2\) units. Adding \(2\) shifts it up \(2\) units. 2. The original input interval \([-4, 4]\) shifts left to \([-6, 2]\). The original output interval \([-1, 1]\) shifts up to \([1, 3]\). 3. For \(h\), multiplying the output by \(-2\) reflects the graph across the x-axis and stretches it vertically by a factor of \(2\). 4. The domain stays \([-4, 4]\). Multiplying every value in \([-1, 1]\) by \(-2\) gives the range \([-2, 2]\).

Answer

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 \(p\).

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 \(g\).

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 \(h\) has domain \([-1, 4]\) and passes through \((-1, 1)\), \((1, -1)\), and \((4, 2)\). a) Find the transformed vertices of \(k(x)=-h(0.5x)+2\). b) State the domain of \(k\) and describe the transformations.

Hints

- Determine how \(0.5x\) changes x-coordinates. - Apply the negative sign and vertical shift to each y-coordinate. - Transform the endpoints and vertex one at a time. - Check that the transformed endpoints match the new domain.

Solution

1. 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. 2. Each point \((x, y)\) on \(h\) maps to \((2x, -y+2)\) on \(k\). 3. The transformed vertices are \((-2, 1)\), \((2, 3)\), and \((8, 0)\). 4. The original domain \([-1, 4]\) is stretched horizontally by a factor of \(2\), so the domain of \(k\) is \([-2, 8]\).

Answer

a) 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.

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