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.

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]

[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].

To express the relationship in this form, we need to be able to write the relationship where [latex]p[/latex] is a function of [latex]n[/latex], which means writing it as p = expression involving n.

[latex]\begin{cases}2n+6p=12\hfill & \hfill \\ 6p=12 - 2n\hfill & \begin{cases}& & \end{cases}\text{Subtract }2n\text{ from both sides}.\hfill \\ p=\frac{12 - 2n}{6}\hfill & \begin{cases}& & \end{cases}\text{Divide both sides by 6 and simplify}.\hfill \\ p=\frac{12}{6}-\frac{2n}{6}\hfill & \hfill \\ p=2-\frac{1}{3}n\hfill & \hfill \end{cases}[/latex]

Therefore, [latex]p[/latex] as a function of [latex]n[/latex] is written as

[latex]p=f\left(n\right)=2-\frac{1}{3}n[/latex]

Analysis of the Solution

It is important to note that not every relationship expressed by an equation can also be expressed as a function with a formula.

First we subtract [latex]{x}^{2}[/latex] from both sides.

[latex]{y}^{2}=1-{x}^{2}[/latex]

We now try to solve for [latex]y[/latex] in this equation.

[latex]\begin{cases}y=\pm \sqrt{1-{x}^{2}}\hfill \\ \text{ }=+\sqrt{1-{x}^{2}}\text{ and }-\sqrt{1-{x}^{2}}\hfill \end{cases}[/latex]

We get two outputs corresponding to the same input, so this relationship cannot be represented as a single function [latex]y=f\left(x\right)[/latex].

• 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].

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.

1. To evaluate [latex]f\left(2\right)[/latex], locate the point on the curve where [latex]x=2[/latex], then read the y-coordinate of that point. The point has coordinates [latex]\left(2,1\right)[/latex], so [latex]f\left(2\right)=1[/latex].
2. To solve [latex]f\left(x\right)=4[/latex], we find the output value [latex]4[/latex] on the vertical axis. Moving horizontally along the line [latex]y=4[/latex], we locate two points of the curve with output value [latex]4:[/latex] [latex]\left(-1,4\right)[/latex] and [latex]\left(3,4\right)[/latex]. These points represent the two solutions to [latex]f\left(x\right)=4:[/latex] [latex]x=-1[/latex] or [latex]x=3[/latex]. This means [latex]f\left(-1\right)=4[/latex] and [latex]f\left(3\right)=4[/latex], or when the input is [latex]-1[/latex] or [latex]\text{3,}[/latex] the output is [latex]\text{4}\text{.}[/latex]See Figure 8.