Defining and Writing Functions

HOW TO

Given the formula for a function, evaluate.

1. Substitute the input variable in the formula with the value provided.
2. Calculate the result.

Example 6

Evaluating Functions at Specific Values

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

(a) 2

(b) $a$

(c) $a+h$

(d) Now evaluate $\dfrac{f(a+h)-f(a)}{h}$

Solution

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

(a) Because the input value is a number, 2, we can use simple algebra to simplify.

$\begin{aligned}f(2)=2^{2}+3(2)-& 4 \\&=4+6-4 \\&=6\end{aligned}$

(b) In this case, the input value is a letter so we cannot simplify the answer any further.

$f(a)=a^{2}+3 a-4$

With an input value of $a+h$, we must use the distributive property.

$\begin{aligned}f(a+h)=(a+h)^{2}+3(a+h)-4 & \\&=a^{2}+2 a h+h^{2}+3 a+3 h-4\end{aligned}$

(c) 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

$f(a+h)=a^{2}+2 a h+h^{2}+3 a+3 h-4$

and we know that

$f(a)=a^{2}+3 a-4$

Now we combine the results and simplify.

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

Example 7

Evaluating Functions

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

Solution

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

$\begin{aligned}h(p) &=p^{2}+2 p \\h(4) &=(4)^{2}+2(4) \\&=16+8 \\&=24\end{aligned}$

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

Try It #4

Given the function $g(m)=\sqrt{m-4}$, evaluate $g(5)$.

Example 8

Solving Functions

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

Solution

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

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

$\begin{aligned}&(p+3)=0, \quad p=-3 \\&(p-1)=0, \quad p=1\end{aligned}$

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

Figure 3

Try It #5

Given the function $g(m)=\sqrt{m-4}$, solve $g(m)=2$.