Section 2.1 Functions
A function will often be given by an equation such as \(y=2x+3\text{.}\) Notice that by the way this equation is written, \(y\) depends on \(x\text{.}\) In this case \(y\) is called the dependent, or output variable, and \(x\) is called the independent, or input variable. We can express this using function notation where we replace the output variable with some other letter representing the function, like \(f\text{,}\) and express its dependency on the input variable using parentheses. In this case we replace \(y\) with \(f(x)\) (read \(f\) of \(x\)) and we have
\begin{equation*}
y = f(x) = 2x+3.
\end{equation*}
This makes evaluation easy to express. In this case, for domain (or input) values \(x=1\) and \(x=2\) we have the corresponding range (or output) values
- \(\displaystyle f(1)=2(1)+3=5\)
- \(\displaystyle f(2)=2(2)+3=7\)
So, for input value \(x=1\text{,}\) we have an output value \(y=f(1)=5\text{,}\) and for input value \(x=2\text{,}\) we have an output value \(y=f(2)=7\text{.}\)
Example 2.1.2.
We will use the following notation. For a funciton \(f(x)\text{,}\)
- dom(\(f\)) = the domain of the function \(f\)
- rng(\(f\)) = the range of the function \(f\)
Typcially, we will express these sets using interval notation. Note that the set of all real numbers can expressed using the notation \(\mathbb{R} = (-\infty, \infty)\text{.}\)
Example 2.1.3.
- \(f(x) = -5x + 2 \) Solution.Because all real numbers are permissible as input values,\begin{equation*} \textrm{dom}(f) = \mathbb{R} = (-\infty, \infty) \end{equation*}
- \(g(x) = 6x^2 - \dfrac{x}{2} + 71.4\) Solution.Because all real numbers are permissible as input values,\begin{equation*} \textrm{dom}(g) = \mathbb{R} = (-\infty, \infty) \end{equation*}
- \(p(t) = \dfrac{1}{t+3}\) Solution.Here, it is easier to think about what values of \(t\) are NOT permissible as input values. Since division by 0 is NOT allowed, \(t=-3\) is the only value NOT permitted as an input value and we have\begin{equation*} \textrm{dom}(p) =(-\infty,-3) \cup (-3,\infty) \end{equation*}where the symbol \(\cup\) is called union and means join the sets together.
- \(r(x) = \sqrt{x+1} + 2 \) Solution.Here, again, it easier to think about values of \(x\) that are NOT allowed. Since we cannot take the sqaure root of a negative number (and get another real number), we must solve\begin{align*} x+1 \amp \geq 0 \\ x \geq -1 \end{align*}so that\begin{equation*} \textrm{dom}(r) = [-1, \infty) \end{equation*}
