Example: Evaluating Functions

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

Answer:

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

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

Example: Evaluating Functions at Specific Values

Evaluate $f\left(x\right)={x}^{2}+3x - 4$ at

1. $2$
2. $a$
3. $a+h$
4. $\frac{f\left(a+h\right)-f\left(a\right)}{h}$

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

1. Because the input value is a number, 2, we can use algebra to simplify.
   $\begin{cases}f\left(2\right)={2}^{2}+3\left(2\right)-4\hfill \\ =4+6 - 4\hfill \\ =6\hfill \end{cases}$
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.
   $\begin{cases}f\left(a+h\right)={\left(a+h\right)}^{2}+3\left(a+h\right)-4\hfill \\ ={a}^{2}+2ah+{h}^{2}+3a+3h - 4\hfill \end{cases}$
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.

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

Example: Solving Functions

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

Answer:

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

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

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

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

Example: Evaluating and Solving a Tabular Function

Using the table below,

1. Evaluate $g\left(3\right)$.
2. Solve $g\left(n\right)=6$.

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

Answer:

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

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

When we input 2 into the function $g$, our output is 6. When we input 4 into the function $g$, our output is also 6.