1.2 Domain and range  (Page 2/11)

First identify the input values. The input value is the first coordinate in an ordered pair . There are no restrictions, as the ordered pairs are simply listed. The domain is the set of the first coordinates of the ordered pairs.

The input value, shown by the variable $x$ in the equation, is squared and then the result is lowered by one. Any real number may be squared and then be lowered by one, so there are no restrictions on the domain of this function. The domain is the set of real numbers.

In interval form, the domain of $f$ is $\left(-\infty ,\infty \right).$

When there is a denominator, we want to include only values of the input that do not force the denominator to be zero. So, we will set the denominator equal to 0 and solve for $x.$

Now, we will exclude 2 from the domain. The answers are all real numbers where $x<2$ or $x>2.$ We can use a symbol known as the union, $\cup ,$ to combine the two sets. In interval notation, we write the solution: $\left(\mathrm{-\infty },2\right)\cup \left(2,\infty \right).$

In interval form, the domain of $f$ is $\left(-\infty ,2\right)\cup \left(2,\infty \right).$