An *interval* is a connected subset of the real number line. In interval notation, an interval is specified by an ordered pair of numbers consisting of the left and the right endpoints of the interval. Since an interval might or might not contain its endpoints we use square brackets \([\cdot]\) when endpoints are included in the interval and parentheses \((\cdot)\) when endpoints are not included. For example, \((0,1]\) represents the interval of all positive numbers less than or equal to the number 1. Since \(0\) is not a positive number, it is not included in the interval whereas 1 is included since 1 is less than or equal to 1. The entire number line, written in interval notation is \((-\infty,\infty)\). Explain why we do not put square brackets by the infinity symbol \(\infty\). The set of non-negative numbers is \([0,\infty)\). A subset of the number line consisting or two or more intervals is written using the *set union* symbol \(\cup\). For example, the set of all real numbers lying at least one unit from \(0\) but less than two units from zero written in interval notation is \((-2,-1]\cup[1,2)\).

Consider the function \(f(x)=\dfrac{1}{x-2}\).

Notice that there is no output number when the input number equals \(2\), since division by zero is meaningless. Thus \(x\) can have any value except \(2\). We say that \(2\) is not in the *domain* of the function \(f\).

Here is another example. Let \(g(x)=\sqrt{x-4}\). In this course, we will only consider real-valued functions. This means that no output numbers may contain a multiple of the *imaginary* number \(i=\sqrt{-1}\). Thus we cannot allow an input number such as 0 for the function \(g\), since \(\sqrt{0-4}=2\sqrt{-1}=2i\). So zero is *not* in the domain of \(g\). Furthermore, no value of \(x\) for which \(x < 4\) is in the domain of \(g\). [Why is this true?]

**The implicit domain principle:** It will be assumed that the domain of a function contains every possible input number **unless otherwise specified**. Inputs which imply division by zero or which would produce an output containing the square root of −1 are excluded. Additional restrictions also apply to certain other functions we will encounter later such as logarithmic functions and certain of the trigonometric functions.

The implicit domain of g is the interval \([4,\infty)\). [Why is this true?]

When there is reason to do so, the domain of a function may be restricted to be something less than the implicit domain.

For example, one could define \(p(x)=2x+5\) for \(x > 3\). Then the **explicit** domain is the interval \((3,\infty)\). If it were not for the explicit restriction on the domain of \(p\) it would have an implicit domain of \((-\infty,\infty)\)).

For each of the following exercises find the implicit domain of the function. Write the domains in interval notation.

We call the set of all inputs of a function its domain, and we call the set of all outputs of a function its *range*.

It is much easier, in general, to look at the equation of a function and figure out its domain than it is to figure out its range.

For example, take \(f(x)=\dfrac{x+2}{x-3}\). We can see that its domain is all real numbers except \(3\). In interval notation that is written \((-\infty,3)\cup(3,\infty)\). It is not as easy to see what the the range must be. One technique which sometimes works is to replace the \(f(x)\)in the equation with \(y\) and solve the equation for \(x\). When we do so with this example, we find that \(x=\dfrac{3y+2}{y-1}\). Thus we see that the output number \(y\) can be anything except \(1\). Thus, the range of the function is \((-\infty,1)\cup(1,\infty)\).

Find the domain and the range of \( f(x)=\dfrac{2x-1}{x-1}\). Express the answers in interval notation.

See SolutionAnother technique for finding the range is to sketch the graph and see what kind of \(y\) coordinates points on the graph may have. For example, if we graph \(y=x^2\), we see that all the \(y\) coordinates of points on the graph are greater than or equal to zero since no part of the graph extends below the \(x\)-axis. So the range of \(f(x)=x^2\) is \([0,\infty)\).

Sketch the graph of \(f(x)=x^2-1\). Find its domain and its range. Graph paper

See Solution