Subsection1.6Evaluating a Function
Finding the value of the output variable that corresponds to a particular value of the input variable is called evaluating the function.
Example1.12
Let \(g\) be the name of the postage function defined by Table1.3. Find \(g(1)\text{,}\) \(g(3)\text{,}\) and \(g(6.75\)).
SolutionAccording to the table,
when \(w=1\text{,}\) 

\(p=0.47\) 
so 
\(g(1)=0.47\) 
when \(w=3\text{,}\) 

\(p=0.89\) 
so 
\(g(3)=0.89\) 
when \(w=6.75\text{,}\) 

\(p=1.73\) 
so 
\(g(6.75)=1.73\) 
Thus, a letter weighing 1 ounce costs $0.47 to mail, a letter weighing 3 ounces costs $0.89, and a letter weighing 6.75 ounces costs $1.73.
If a function is described by an equation, we simply substitute the given input value into the equation to find the corresponding output, or function value.
Example1.13
The function \(H\) is defined by \(H=f(s) = \dfrac{\sqrt{s+3}}{s}\text{.}\) Evaluate the function at the following values.
\(s=6\)
\(s=1\)
Solution
\(f(6)=\dfrac{\sqrt{(6)+3}}{(6)}=
\dfrac{\sqrt{9}}{6}=\dfrac{3}{6}=\dfrac{1}{2}\text{.}\) Thus, \(f(6)=\dfrac{1}{2}\text{.}\)
\(f(1)=\dfrac{\sqrt{(1)+3}}{(1)}=
\dfrac{\sqrt{2}}{1}=\sqrt{2}\text{.}\) Thus, \(f(1)=\sqrt{2}\text{.}\)
To simplify the notation, we sometimes use the same letter for the output variable and for the name of the function.