4.7: Domain and Range of a Function
The domain of a function is all possible values of x that can be used as input to the function, which will result in a real number as the output. The range of a function is the set of all possible output values of a function.
Find the domain and range of the following function:
\(f(x) = 5x + 3 \)
Solution
Any real number, negative, positive or zero can be replaced with x in the given function. Therefore, the domain of the function \(f(x) = 5x + 3 \) is all real numbers, or as written in interval notation, is: \(D:(−\infty , \infty )\). Because the function \(f(x) = 5x + 3\) is a polynomial of degree 1, it is a straight line (without any breaks or holes).
The range of any polynomial of degree 1 is all real numbers or written in interval notation, is: \(R:(−\infty , \infty )\).
Find the domain and range of the following function:
\(g(x) = 2\sqrt{ x − 4}\)
Solution
Pay attention to the square root part of this function. The radicand (what’s inside the square root) must be nonnegative. Set the radicand greater than or equal to zero to find the domain:
\(\begin{aligned} x − 4 &\geq 0 && \text{Set the radicand greater than or equal to 0 }\\ x &\geq 4 &&\text{ Solve the inequality } \\ D&:[4, \infty ) &&\text{Write the solution in interval notation }\end{aligned}\)
Therefore, the domain of the function \(g(x) = 2\sqrt{ x − 4}\) is all real numbers in the interval from \([4, \infty )\), which is written \(D:[4, \infty )\).
To find the range of \(g(x) = 2\sqrt{ x − 4}\), let’s observe the behavior of the function for different values of x that are in the domain.
Let \(x = 4\), \(g(4) = 2\sqrt{ 4 − 4}\), so \(g(4) = 0\).
Let \(x = 5\), \(g(5) = 2\sqrt{ 5 − 4}\), so \(g(5) = 2\).
Let \(x = 8\), \(g(8) = 2\sqrt{ 8 − 4}\), so \(g(8) = 4\).
Any non-negative value chosen for x will result in a non-negative value for \(g(x)\). The function values for the range (the output from the function \(g(x)\)) are non-negative numbers, written as \(R:[0, \infty )\).
Find the domain and range of the following function:
\(h(x) = −2x^2 + 4x − 9\)
Solution
Any real number, negative, positive or zero can replace x in the given function.
Therefore, the domain of the function \(h(x) = 2x^2 + 4x − 9\) is all real numbers, or as written in interval notation, is: \(D:(−\infty , \infty )\).
Because the function \(h(x) = 2x^2 + 4x − 9\) is a quadratic of degree 2, when graphed, it is a parabola (without any breaks or holes). Identify two things about this parabola:
- Which way does it open, up or down? and
- Where is the vertex?
The sign of the coefficient of the leading term of the quadratic function (\(2x^2\) ) shows which way the parabola opens. The coefficient is 2, and since it’s positive, the quadratic function opens upward.
Now find the vertex. The y-value of the vertex ordered pair will show where the range begins.
The vertex is \(\left(−\dfrac{b}{2a} , f\left( −\dfrac{b}{2a} \right)\right)\), with \(a = 2\) and \(b = 4\).
The vertex is \(\left(− \dfrac{4 }{2∗2} , f \left(− \dfrac{4 }{2∗2}\right)\right)\)
The vertex is \((− 1, f(− 1))\), which is \((− 1, 2 ∗ (−1)^2 − 9))\) or \((− 1, −11)\)
The range will start at −11, and continue to increase, since the parabola opens upward. \(R:[-11, \infty)\)
Find the domain and range of the following function:
\(j(x) = \vert z − 6 \vert − 3\)
Solution
This function contains an absolute value. Any value can be chosen for \(z\), so the domain of the function is all real numbers, or as written in interval notation, is: \(D:(−\infty , \infty )\)
To find the range, examine inside the absolute value symbols. This quantity, \(\vert z−6 \vert\) will always be either 0 or a positive number, for any values of z. First, find what makes the expression z−6 equal to zero, which is the number 6.
\(\begin{aligned} j(x) &= \vert z − 6 \vert − 3 &&\text{ Original function } \\ j(x) &= \vert 6 − 6 \vert − 3 && \text{Replace z with 6 } \\j(x) &= \vert 0 \vert − 3 && \text{Simplify } \\ j(x) &= −3 && j(x) \text{ is } −3 \end{aligned}\)
Therefore, the range of the function \(j(x) = \vert z − 6 \vert − 3\) is −3 or higher, or as written in interval notation, is: \(R:[-3, \infty)\)
Some types of functions are more difficult to work with. Here are some examples of functions where the domain can be found but the range would be too difficult to find, and out of the scope of this course:
Find the domain of the following functions:
\(f(x) = \dfrac{x − 4 }{x^2 − 2x − 15 }\)
Solution
With any rational function (a quotient of polynomials), be aware of division by 0. Set the denominator polynomial equal to 0 and solve.
\(\begin{aligned} x^2 − 2x − 15 &= 0 &&\text{Set the denominator function equal to } 0 \\ (x − 5)(x + 3) &= 0 &&\text{Factor the quadratic equation } \\ x − 5 &= 0 && \text{Set the first binomial factor equal to zero } \\ x &= 5 &&\text{Solve the first binomial factor } \\ x + 3 &= 0 &&\text{Set the second binomial factor equal to zero } \\ x &= −3 &&\text{Solve the second binomial factor} \end{aligned}\)
There are two solutions to the quadratic equation, 5 and −3.
These values must be excluded from the domain, because if \(x\) is either 5 or −3, the denominator will equal zero.
Division by zero is undefined. The domain of the function \(f(x) = x − 4 x^2 − 2x − 15\) is \((−\infty , −3) \cup (−3, −5) \cup (−5, \infty )\).
Find the domain of the following function:
\(g(x) = \dfrac{x }{x^2 − 9}\)
Solution
Once again this is a rational function, and the concern is to avoid division by 0. Set the denominator function equal to 0 and solve.
\(\begin{aligned} x^{2}-9&=0 && \text { Set the denominator function equal to } 0 \\ (x-3)(x+3)&=0 &&\text{Factor the quadratic equation} \\ x-3&=0 && \text{Set the first binomial factor equal to zero} \\ x&=3 &&\text{Solve the first binomial factor}\\ x+3&=0 && \text{Set the second binomial factor equal to zero} \\ x&=-3 && \text{Solve the second binomial factor} \end{aligned}\)
There are two solutions to the quadratic equation, 3 and −3. These values must be excluded from the domain, because if \(x\) is either 3 or −3, the denominator will equal zero. Division by zero is undefined. The domain of the function \(g(x) =\dfrac{ x}{ x^2 − 9 }\) is \((−\infty , −3) \cup (−3, 3) \cup (3, \infty )\).
Find the domain of the following function:
\(g(t) = \sqrt{6 + t − t^2}\)
Solution
The radicand of this square root function must be non-negative. Set the radicand greater than or equal to 0 and solve.
\(\begin{aligned} 6 + t − t^2 &\geq 0 &&\text{Set the radicand equal to }0 \\ −t^2 + t + 6 &\geq 0 &&\text{Rewrite the function with the leading term first } \\ (−t + 3)(t + 2) &= 0 && \text{Factor the quadratic equation } \\−t + 3 &= 0 && \text{Set the first binomial factor equal to zero } \\ t &= 3 && \text{Solve the first binomial factor } \\ t + 2 &= 0 &&\text{Set the second binomial factor equal to zero } \\ t &= −2 &&\text{Solve the second binomial factor} \end{aligned}\)
There are two values that will make the radicand of this square root function zero, 3 and −2.
Since the radicand must be non-negative, test the regions between the found solutions.
If \(x < −2\), for example, −4, \(g(−4) = \sqrt{6 + (−4) − (−4)^2}\) is negative, which is not allowed for the radicand.
If \(x\) is between −2 and 3, for example, 0, \(g(0) = \sqrt{6 + (0) − (0)^2}\) is positive. This region between −2 and 3 will be in the domain of the function.
There is one more region to check, where \(x > 3\). Let \(x = 4\). \(g(4) = \sqrt{ 6 + (4) − (4)^2}\) is negative, which is not allowed for the radicand. The domain of the function \(g(t) = \sqrt{6 + t − t^2}\) is \([−2, 3]\)
Find the domain and range of the following functions:
- \(f(x) = x ^2 − 8x + 12\)
- \(g(x) = \sqrt{x + 10}\)
- \(h(x) = \vert − 2x + 1\vert\)
Find the domain of the following functions:
- \(f(x) = \dfrac{6x + 7 }{5x + 2}\)
- \(f(x) = \dfrac{2x }{2x^2 + 3x − 20}\)
- \(f(x) =\dfrac{ 4x + 11 }{x^2 + 6x + 9}\)
- \(f(x) = \dfrac{3x }{x^2 − 5x − 14}\)
- \(f(x) = \dfrac{2x + 1}{ 6x^2 − x − 2}\)
- \(f(x) = \dfrac{−6 }{25x^2 − 4}\)