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## 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 $g(6.75$).

Solution

According 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.

1. $s=6$

2. $s=-1$

Solution
1. $f(6)=\dfrac{\sqrt{(6)+3}}{(6)}= \dfrac{\sqrt{9}}{6}=\dfrac{3}{6}=\dfrac{1}{2}\text{.}$ Thus, $f(6)=\dfrac{1}{2}\text{.}$

2. $f(-1)=\dfrac{\sqrt{(-1)+3}}{(-1)}= \dfrac{\sqrt{2}}{-1}=-\sqrt{2}\text{.}$ 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.