Example: Evaluating Functions

Answer:

To evaluate $h\left(4\right)$, we substitute the value 4 for the input variable $p$ in the given function.

\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}

Example: Evaluating Functions at Specific Values

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

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{align}f\left(2\right)&={2}^{2}+3\left(2\right)-4 \\ &=4+6 - 4 \\ &=6\hfill \end{align}
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{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}
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{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}

Example: Solving Functions

Answer:

\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}

\begin{align}&p+3=0, &&p=-3 \\ &p - 1=0, &&p=1\hfill \end{align}

Example: Finding an Equation of a Function

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

\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}

$p=f\left(n\right)=2-\frac{1}{3}n$

Analysis of the Solution

Example: Expressing the Equation of a Circle as a Function

Answer: First we subtract ${x}^{2}$ from both sides.

${y}^{2}=1-{x}^{2}$

\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}

Example: Evaluating and Solving a Tabular Function

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

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 $n=4$.

Example: Reading Function Values from a Graph

1. Evaluate $f\left(2\right)$.
2. Solve $f\left(x\right)=4$.

Answer:

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