Example: Evaluating Functions

Given the function [latex]h\left(p\right)={p}^{2}+2p[/latex], evaluate [latex]h\left(4\right)[/latex].

Answer:

To evaluate [latex]h\left(4\right)[/latex], we substitute the value 4 for the input variable [latex]p[/latex] in the given function. [latex]\begin{cases}\text{ }h\left(p\right)={p}^{2}+2p\hfill \\ \text{ }h\left(4\right)={\left(4\right)}^{2}+2\left(4\right)\hfill \\ \text{ }=16+8\hfill \\ \text{ }=24\hfill \end{cases}[/latex]

Therefore, for an input of 4, we have an output of 24.

Example: Evaluating Functions at Specific Values

Evaluate [latex]f\left(x\right)={x}^{2}+3x - 4[/latex] at

1. [latex]2[/latex]
2. [latex]a[/latex]
3. [latex]a+h[/latex]
4. [latex]\frac{f\left(a+h\right)-f\left(a\right)}{h}[/latex]

Answer: Replace the [latex]x[/latex] in the function with each specified value.

1. Because the input value is a number, 2, we can use algebra to simplify.
   [latex]\begin{cases}f\left(2\right)={2}^{2}+3\left(2\right)-4\hfill \\ =4+6 - 4\hfill \\ =6\hfill \end{cases}[/latex]
2. In this case, the input value is a letter so we cannot simplify the answer any further.
   [latex]f\left(a\right)={a}^{2}+3a - 4[/latex]
3. With an input value of [latex]a+h[/latex], we must use the distributive property.
   [latex]\begin{cases}f\left(a+h\right)={\left(a+h\right)}^{2}+3\left(a+h\right)-4\hfill \\ ={a}^{2}+2ah+{h}^{2}+3a+3h - 4\hfill \end{cases}[/latex]
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
   [latex]f\left(a+h\right)={a}^{2}+2ah+{h}^{2}+3a+3h - 4[/latex]
   and we know that
   [latex]f\left(a\right)={a}^{2}+3a - 4[/latex]
   Now we combine the results and simplify.

   [latex]\begin{cases}\begin{cases}\hfill \\ \frac{f\left(a+h\right)-f\left(a\right)}{h}=\frac{\left({a}^{2}+2ah+{h}^{2}+3a+3h - 4\right)-\left({a}^{2}+3a - 4\right)}{h}\hfill \end{cases}\hfill \\ \begin{cases}\text{ }=\frac{2ah+{h}^{2}+3h}{h}\hfill & \hfill \\ \text{ }=\frac{h\left(2a+h+3\right)}{h}\hfill & \begin{cases}{cc}\begin{cases}{cc}& \end{cases}& \end{cases}\text{Factor out }h.\hfill \\ \text{ }=2a+h+3\hfill & \begin{cases}{cc}\begin{cases}{cc}& \end{cases}& \end{cases}\text{Simplify}.\hfill \end{cases}\hfill \end{cases}[/latex]

Example: Solving Functions

Given the function [latex]h\left(p\right)={p}^{2}+2p[/latex], solve for [latex]h\left(p\right)=3[/latex].

Answer:

[latex]\begin{cases}\text{ }h\left(p\right)=3\hfill & \hfill & \hfill & \hfill \\ \text{ }{p}^{2}+2p=3\hfill & \hfill & \hfill & \text{Substitute the original function }h\left(p\right)={p}^{2}+2p.\hfill \\ \text{ }{p}^{2}+2p - 3=0\hfill & \hfill & \hfill & \text{Subtract 3 from each side}.\hfill \\ \text{ }\left(p+3\text{)(}p - 1\right)=0\hfill & \hfill & \hfill & \text{Factor}.\hfill \end{cases}[/latex]

If [latex]\left(p+3\right)\left(p - 1\right)=0[/latex], either [latex]\left(p+3\right)=0[/latex] or [latex]\left(p - 1\right)=0[/latex] (or both of them equal 0). We will set each factor equal to 0 and solve for [latex]p[/latex] in each case.

[latex]\begin{cases}\left(p+3\right)=0,\hfill & p=-3\hfill \\ \left(p - 1\right)=0,\hfill & p=1\hfill \end{cases}[/latex]

This gives us two solutions. The output [latex]h\left(p\right)=3[/latex] when the input is either [latex]p=1[/latex] or [latex]p=-3[/latex]. We can also verify by graphing as in Figure 5. The graph verifies that [latex]h\left(1\right)=h\left(-3\right)=3[/latex] and [latex]h\left(4\right)=24[/latex].

Example: Evaluating and Solving a Tabular Function

Using the table below,

1. Evaluate [latex]g\left(3\right)[/latex].
2. Solve [latex]g\left(n\right)=6[/latex].

n	1	2	3	4	5
g(n)	8	6	7	6	8

Answer:

• Evaluating [latex]g\left(3\right)[/latex] means determining the output value of the function [latex]g[/latex] for the input value of [latex]n=3[/latex]. The table output value corresponding to [latex]n=3[/latex] is 7, so [latex]g\left(3\right)=7[/latex].
• Solving [latex]g\left(n\right)=6[/latex] means identifying the input values, [latex]n[/latex], that produce an output value of 6. The table below shows two solutions: [latex]n=2[/latex] and [latex]n=4[/latex].

n	1	2	3	4	5
g(n)	8	6	7	6	8

When we input 2 into the function [latex]g[/latex], our output is 6. When we input 4 into the function [latex]g[/latex], our output is also 6.