1.1 Functions and function notation  (Page 4/21)

[link] and [link] define functions. In both, each input value corresponds to exactly one output value. [link] does not define a function because the input value of 5 corresponds to two different output values.

When a table represents a function, corresponding input and output values can also be specified using function notation.

The function represented by [link] can be represented by writing

Similarly, the statements

represent the function in [link] .

[link] cannot be expressed in a similar way because it does not represent a function.

Replace the $x$ in the function with each specified value.

1. Because the input value is a number, 2, we can use simple algebra to simplify.
   $$\begin{array}{l}f\left(2\right)=2^2+3\left(2\right)-4\hfill \\ =4+6-4\hfill \\ =6\hfill \end{array}$$
2. In this case, the input value is a letter so we cannot simplify the answer any further.
   $$f\left(a\right)=a^2+3a-4$$
3. With an input value of $a+h,$ we must use the distributive property.
   $$\begin{array}{l}f(a+h)={(a+h)}^2+3(a+h)-4\hfill \\ =a^2+2ah+h^2+3a+3h-4\hfill \end{array}$$
4. In this case, we apply the input values to the function more than once, and then perform algebraic operations on the result. We already found that
   $$f\left(a+h\right)=a^2+2ah+h^2+3a+3h-4$$

   and we know that

   $$f\left(a\right)=a^2+3a-4$$

   Now we combine the results and simplify.

   $$\begin{array}{l}\begin{array}{l}\hfill \\ \frac{f(a+h)-f(a)}{h}=\frac{(a^2+2ah+h^2+3a+3h-4)-(a^2+3a-4)}{h}\hfill \end{array}\hfill \\ \begin{array}{ll}=\frac{2ah+h^2+3h}{h}\hfill & \hfill \\ =\frac{h(2a+h+3)}{h}\hfill & \begin{array}{cc}\begin{array}{cc}& \end{array}& \end{array}\text{Factor out }h.\hfill \\ =2a+h+3\hfill & \begin{array}{cc}\begin{array}{cc}& \end{array}& \end{array}\text{Simplify}.\hfill \end{array}\hfill \end{array}$$