Finding the Domain of Functions

Pre-Calculus Tutorial

Finding the Domain of Functions

Intro

In this tutorial I will demonstrate the process of finding the domain of a function. The domain of a function is all the allowable values of real numbers that the independent variable (usually x) can be. The restrictions (what we cannot do with real numbers) are

1) Can’t divide by zero.
2) Can’have a negative under an even root radical.
3) The argument of a logarithmic function must be greater than zero.

A free online Precalculus book can be obtained by going to www.stitz-zeager.com. Note that the Precalculus book is parted into two books, College Algebra and College trigonometry. The College Algebra which I used as a reference contains the material for this topic.

Sample Problem

Analytically find the domain of the following functions. Express the answer in interval notation.

1) \quad f(x) = x^2 + 4
2) \quad k(w) = \frac{3w}{w^2 + w - 2}
3) \quad h(x) = \sqrt{2x + 5}
4) \quad s(t) = \frac{6}{4 - \sqrt{6t - 2}}
5) \quad g(v) = \frac{1}{4 - \frac{1}{v^2}}
6) \quad f(x) = \ln(4x - 20) + \ln(x^2 + 9x + 18)
7) \quad f(x) = \sqrt[4]{\log_{4}(x)}
8) \quad j(z) = \frac{\sqrt{-1 - z}}{\log_{\frac{1}{2}}(z)}

Solution

1) \quad f(x) = x^2 + 4

To find the domain of this function we look for restrictions. We see that there are no even root radicals, no logarithms and no possibility of division by zero because the x is not the denominator of a fraction. Therefore there are no restrictions and the domain is all real numbers.

D = (-\infty, \infty)
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2) \quad k(w) = \frac{3w}{w^2 + w - 2}

To find the domain of a rational function we take the intersection of the domain of the numerator function and the denominator function if necessary. We see that the numerator function of 3w has the domain of all reals. To find the domain of the denominator function we factor the bottom and set it not equal to zero and solve for w. These values will be what w cannot be.

    \begin{equation*}      \begin{align}           & \textit{The domain for 3w is} \\           & D_1 = (-\infty, \infty) \\           &{} \\           & \textit{Factoring $w^2 + w - 2$ we have} \\           & (w + 2)(w - 1) \ne 0 \\           & \textit{Set each individual factor not equal to zero and solve for w} \\           & w + 2 \ne 0 \to w \ne -2 \quad or \quad w - 1 \ne 0 \to w \ne 1 \\           & D_2 = (-\infty, -2) \cup (-2, 1) \cup (1, \infty) \\           &{} \\           & \textit{The domain of the function is the intersection of $D_1 \thickspace and \thickspace D_2$} \\           & D = (-\infty, -2) \cup (-2, 1) \cup (1, \infty) \\      \end{align} \end{equation*}

A visual way of determining and writing the domain in interval notation is to graph each domain on a number line using the notation:

1) Place the pair )( at each \ne value.
2) Place a ( for >.
3) Place ) for <.
4) Place a [ for \ge.
5) Place a ] for \le.
6) The ends of the number line are always \pm \infty so place a ( or ).
7) Union the intersection of the intervals as needed. The intersection is the overlap of the intervals or whats common between the them.

The parenthesis means the endpoint is not included, equivalent to an open dot, and the bracket means the endpoint is included, equivalent to a solid dot.

Graphing we have

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From the graph the solution in interval notation should be obviuos.
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3) \quad h(x) = \sqrt{2x + 5}

To find the domain of this function we set the radicand (the expression under the radical sign) greatrer than or equal to zero then solve for x which will be the domain.

2x + 5 \ge 0 \to 2x \ge -5 \to x \ge -\frac{5}{2}

Graphing we have

Rendered by QuickLaTeX.com

The domain of the function is [ -\frac{5}{2}, \infty )
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4) \quad s(t) = \frac{6}{4 - \sqrt{6t - 2}}

To find the domain of this function note that there are two restrictions and the domain must satisfy both of them at the same time which means we must find the intersection of both restrictions. We can’t have a negative under a radical and we can’t divide by zero.

    \begin{equation*}      \begin{align}           & \textit{For the radical we have} \\           & 6t - 2 \ge 0 \to 6t \ge 2 \to t \ge \frac{2}{6} \to t \ge \frac{1}{3} \\           & \textit{The domain for the radical is} \\           & D_1 = \left[ \frac{1}{3}, \infty \right) \\           & {} \\           & \textit{For the denominator we have} \\           & 4 - \sqrt{6t - 2} \ne 0 \\           & -\sqrt{6t -2} \ne -4 \\           & (-\sqrt{6t -2})^2 \ne (-4)^2 \\           & 6t - 2 \ne 16 \\           & 6t \ne 18 \\           & t \ne 3 \\           & \textit{The domain for the denominator is} \\           & D_2 = (-\infty, 3) \cup (3, \infty) \\           & {} \\           & \textit{The domain of the function is the intersection of $D_1 \thickspace and \thickspace D_2$} \\           & D = D_1 \cap D_2 = \left[ \frac{1}{3}, 3 \right) \cup \left( 3, \infty \right) \\      \end{align} \end{equation*}

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5) \quad g(v) = \frac{1}{4 - \frac{1}{v^2}}

To find the domain of this function note that we have two denominators.

    \begin{equation*}      \begin{align}           & \textit{For the fraction of $\frac{1}{v^2}$ we have} \\           & v^2 \ne 0 \to v \ne 0 \to D_1 = (-\infty, 0) \cup (0, \infty) \\           & \textit{For the denominator of $4 - \frac{1}{v^2}$ we have} \\           & 4 - \frac{1}{v^2} \ne 0 \\           & 4 \ne \frac{1}{v^2} \\           & \textit{Take reciprocal of each side} \\           & \frac{1}{4} \ne v^2 \\           & \textit{Square root each side} \\           & \pm \sqrt{\frac{1}{4}} \ne \sqrt{v^2} \\           & \pm \frac{1}{2} \ne v \to D_2 = (-\infty, \frac{-1}{2}) \cup (\frac{-1}{2}, \frac{1}{2}) \cup (\frac{1}{2}, \infty) \\           & \textit{The domain of the function is the intersection of $D_1 \thickspace and \thickspace D_2$} \\           & D = \left(-\infty, \frac{-1}{2} \right) \cup \left(\frac{-1}{2}, 0 \right) \cup \left(0, \frac{1}{2} \right) \cup \left(\frac{1}{2}, \infty \right)      \end{align} \end{equation*}

If you cannot see how I arrived at this domain let us graph it.

Graphing we have

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From the graph we can now easily determine the domain.
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6) \quad f(x) = \ln(4x - 20) + \ln(x^2 + 9x + 18)

To find the domain of this function we set each argument of the logarithm greater than zero and solve for x. Then we take in intersection of the two domains.

For 4x - 20 we have
4x - 20 > 0 \to 4x > 20 \to x > 5 \to D_1 = (5, \infty)

For x^2 + 9x + 18 we have
x^2 + 9x + 18 > 0 \to (x + 6)(x + 3) > 0
Do not solve the inequality as though it was an equal sign. We cannot do the following!
x + 6 > 0 \thickspace or \thickspace x + 3 > 0

To solve a nonlinear inequality use a sign diagram.

How to Create a Sign Diagram for a Polynomial Function
1) Move all terms to one side of the inequality.
2) Factor the polynomial.
3) Replace the inequality with an equal sign.
4) Find the zeros and plot them on a number line placing a )( if > or < or a ][ if \ge or \le above them.
5) Place a ( or ) at the ends of the number line.
6) Pick a test point in each interval and check them in the factored or original inequality. If the test point makes the inequality true then write a yes or true in that interval otherwise write no or false in that interval. We can also put plus or minus signs of the polynomial in the interval.
7) Union all the yes intervals. If you use plus or minus then union all the plus intervals if > or \ge; or union all the minus intervals if < or \le.

(x + 6)(x + 3) > 0
(x + 6)(x + 3) = 0 \to x + 6 = 0 \to x = -6 \thickspace or \thickspace x + 3 = 0 \to x = -3

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From the graph we see that the intervals are (-\infty, -6), \thinspace (-6, -3), \thinspace (-3, \infty).

The test points I will use are x = {-7, -5, 0}.

    \begin{equation*}      \begin{align}           & For \thinspace x = -7 \to (-7 + 6)(-7 + 3) > 0 \to yes, \thinspace true, \thinspace + \\           & For \thinspace x = -5 \to (-5 + 6)(-5 + 3) > 0 \to no, \thinspace false, \thinspace - \\           & For \thinspace x = 0 \to (0 + 6)(0 + 3) > 0 \to yes, \thinspace true, \thinspace + \\           & \text{So the domain for the second logarithm expression is} \\           & D_2 = (-\infty, -6) \cup (-3, \infty) \\           & \textit{The domain of the function is the intersection of $D_1 \thickspace and \thickspace D_2$} \\ & D = D_1 \cap D_2 = (5, \infty)      \end{align} \end{equation*}

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7) \quad f(x) = \sqrt[4]{\log_{4}(x)}

To find the domain of this function note that we have an even root radical and a logarithm.

Set the radicand greater than or equal to zero, \log_4(x) \ge 0. From the properties of logs we know that \log_{b}(x) is negative when 0<x<1 and nonnegative when x \ge 1 (nonnegative means 0 or greater than 0). So the domain of the function is D = [1, \infty).
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8) \quad j(z) = \frac{\sqrt{-1 - z}}{\log_{\frac{1}{2}}(z)}

To find the domain of this funcdtion note that we have three restrictions, an even root radical in the numerator, a logarithm in the denominator and division by zero.

For the numerator
-1 - z \ge 0 \to -z \ge 1 \to z \le -1 \to D_1 = (-\infty, -1]

For the denominator
\log_{\frac{1}{2}}(z) \ne 0 \to z > 0 \thickspace and \thickspace z \ne 1 \to D_2 = (0, 1) \cup (1, \infty)

The domain of the function is the intersection of D_1 \thickspace and \thickspace D_2.
D = D_1 \cap D_2 \to No \thinspace Solution

If the domain is not obvious just graph it.

Graphing we have

Rendered by QuickLaTeX.com

The domain should now be obvious.
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9) \quad y(x) = \frac{1}{\sqrt{x - 2}}

Don’t let this simple looking problem fool you. Note that there are two restrictions; we cannot have a negative under an even root radical and we cannot divide by zero. When we have an even root radical as a denominator we set the radicand to greater than zero and solve for x which will be our domain.

x - 2 > 0 \to x > 2 \to D = (2, \infty)
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10) \quad y(x) = \frac{\sqrt{x}}{x^2 -3x + 2}

To find the domain of this function note that there are two restriction; cannot have a negative under an even root radical and cannot divide by zero.

For the numerator we have
x \ge 0 \to D_1 = [0, \infty)

For the denominator we have
x^2 - 3x + 2 \ne 0 \to (x - 2)(x - 1) \ne 0
x - 2 \ne 0 \to x \ne 2 \quad or \quad x - 1 \ne 0 \to x \ne 1
D_2 = ( -\infty, 1) \cup (1, 2) \cup (2, \infty)

The domain of the function is
D = D_1 \cap D_2 = [0, 1) \cup (1, 2) \cup (2, \infty)



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I am a graduate of Cleveland State University. After earning my Bachelors of Electrical Engineering I went on to earn a Master’s of Science in Electrical Engineering. Also I am a graduate of Lorain County Community College (LCCC) earning an Associates of Science Pre-Professional Engineering and ...
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