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{align}h\left(p\right)&={p}^{2}+2p \\ h\left(4\right)&={\left(4\right)}^{2}+2\left(4\right) \\ &=16+8 \\ &=24 \end{align}[/latex]

Therefore, for an input of 4, we have an output of 24 or [latex]h(4)=24[/latex].

Example: Evaluating Functions at Specific Values

For the function, [latex]f\left(x\right)={x}^{2}+3x - 4[/latex], evaluate each of the following.

1. [latex]f\left(2\right)[/latex]
2. [latex]f(a)[/latex]
3. [latex]f(a+h)[/latex]
4. [latex]\dfrac{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{align}f\left(2\right)&={2}^{2}+3\left(2\right)-4 \\ &=4+6 - 4 \\ &=6\hfill \end{align}[/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{align}f\left(a+h\right)&={\left(a+h\right)}^{2}+3\left(a+h\right)-4 \\[2mm] &={a}^{2}+2ah+{h}^{2}+3a+3h - 4 \end{align}[/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{align}\dfrac{f\left(a+h\right)-f\left(a\right)}{h}&=\dfrac{\left({a}^{2}+2ah+{h}^{2}+3a+3h - 4\right)-\left({a}^{2}+3a - 4\right)}{h} \\[2mm] &=\dfrac{2ah+{h}^{2}+3h}{h}\\[2mm] &=\frac{h\left(2a+h+3\right)}{h}&&\text{Factor out }h. \\[2mm] &=2a+h+3&&\text{Simplify}.\end{align}[/latex]

[ohm_question]1647[/ohm_question]

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{align}&h\left(p\right)=3\\ &{p}^{2}+2p=3 &&\text{Substitute the original function }h\left(p\right)={p}^{2}+2p. \\ &{p}^{2}+2p - 3=0 &&\text{Subtract 3 from each side}. \\ &\left(p+3\text{)(}p - 1\right)=0 &&\text{Factor}. \end{align}[/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{align}&p+3=0, &&p=-3 \\ &p - 1=0, &&p=1\hfill \end{align}[/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: Finding an Equation of a Function

Express the relationship [latex]2n+6p=12[/latex] as a function [latex]p=f\left(n\right)[/latex], if possible.

Answer: 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 [latex]p=[/latex] expression involving [latex]n[/latex].

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

Example: Expressing the Equation of a Circle as a Function

Does the equation [latex]{x}^{2}+{y}^{2}=1[/latex] represent a function with [latex]x[/latex] as input and [latex]y[/latex] as output? If so, express the relationship as a function [latex]y=f\left(x\right)[/latex].

Answer: 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{align}y&=\pm \sqrt{1-{x}^{2}} \\[1mm] &=\sqrt{1-{x}^{2}}\hspace{3mm}\text{and}\hspace{3mm}-\sqrt{1-{x}^{2}} \end{align}[/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].

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].

[latex]n[/latex]	1	2	3	4	5
[latex]g(n)[/latex]	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].

[latex]n[/latex]	1	2	3	4	5
[latex]g(n)[/latex]	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.

Example: Reading Function Values from a Graph

Given the graph below,

1. Evaluate [latex]f\left(2\right)[/latex].
2. Solve [latex]f\left(x\right)=4[/latex].

Answer:

1. To evaluate [latex]f\left(2\right)[/latex], locate the point on the curve where [latex]x=2[/latex], then read the [latex]y[/latex]-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 the graph below.