1.3: Function Notation and Simplify Expressions

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- 1.2: FOIL Method and Special Products
- 2: Polynomials

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The goal of this section is to practice function notation. Functions have many interconnected mathematical concepts. Therefore, more detail about functions will emerge in your Precalculus course. This section will practice the correspondence of inputs to outputs using function notation, giving further practice with the properties introduced in previous sections.

The following three equations express three different mathematical rules. The rule creates a correspondence between $x$ -values and $y$ -values.

Equation	The Rule In Words
$y = 2x + 6$	Double the $x$ -value then add $6$ .
$y = \dfrac{x}{2} − 4$	Take half of the $x$ -value then subtract $4$ .
$y = x^2 -10$	Square the $x$ -value then subtract $10$ .

Definition: Function

A function is a rule that assigns each input to exactly one output

A table displays the correspondence between inputs ( $x$ -values) and outputs ( $y$ -values). The three functions above are shown below using $x$ - $y$ tables. An $x$ -value is chosen, the rule is followed, and a $y$ -value results. Each $y$ -value is computed according to the function’s rule.

$x$	$y = 2x + 6$
$-10$	$2(-10)+6 = -14$
$-3$	$2(-3)+6 = 0$
$0$	$2(0) + 6 = 6$
$2$	$2(2) + 6 =10$

$x$	$y = \dfrac{x}{2} − 4$
$-6$	$−6 ÷ 2 − 4 = −7$
$0$	$0 ÷ 2 − 4 = −4$
$6$	$6 ÷ 2 − 4 = −1$
$18$	$18 ÷ 2 − 4 = 5$

$x$	$y = x^2 -10$
$-1$	$(−1)^2 − 10 = −9$
$0$	$0^2 − 10 = −10$
$1$	$1^2 − 10 = −9$
$4$	$4^2 − 10 = 6$

Since the output of the function depends on the input, the $y$ -value is called the dependent variable. Function notation emphasizes this dependence by replacing the variable $y$ with $f(x)$ .

$x$ - $y$ notation	Function notation	Read out loud
$y = 2x + 6$	$f(x) = 2x+6$	“ $f$ of $x$ equals two $x$ plus six”

Example 1.3.1

Let $f(x) = 2x + 6$ . Evaluate of each of the following.

$f(−1)$
$f \left( \dfrac{1}{2} \right)$
$f(3a)$

Solution

In each problem, the input, or $x$ -value, is given inside the parentheses. We substitute the given $x$ -value into the function $f(x) = 2x + 6$ .

$f(−1) = 2(−1) + 6 = −2 + 6 = 4$ . Therefore, $f(−1) = 4$ .
$f \left( \dfrac{1}{2} \right) = 2 \left( \dfrac{1}{2} \right) + 6 = 1 + 6 = 7$ . Therefore, $f \left( \dfrac{1}{2} \right) = 7$ .
The $x$ -value is replaced by $3a$ . This will yield an expression rather than a numerical value: $f(3a) = 2(3a) + 6 = 6a + 6$ . Therefore, $f(3a) = 6a + 6$ .

Definition: Naming Functions f, g, and h

Functions are typically named with the letters $f$ , $g$ , and $h$ , but functions can take on other letter-names as well. Using different letter-names for functions can help us refer to several different functions in one problem. Example $1.3.2$ makes use of this concept.

Example 1.3.2

Let $h(x) = f(x) + g(x)$ where $f(x) = 3x^2$ and $g(x) = 5x − 9$ . Evaluate each of the following:

$h(−4)$
$h \left( \dfrac{1}{2} \right)$
$h(a + 1)$

Solution

The function $h$ is defined as the sum of $f$ and $g$ . Plug the given input-value into each function, evaluate to find each output, and take the sum.

$h(−4) = f(−4) + g(−4) = \underbrace{3(−4)^2}_{f(−4)} + \underbrace{5( − 4 ) − 9}_{g(−4)} = 3 \cdot 16 + (−20 − 9) = 48 − 29 = 19$

Therefore, $h(−4) = 19$ .

$h \left( \dfrac{1}{2} \right) = f \left( \dfrac{1}{2} \right) + g \left( \dfrac{1}{2} \right) = \underbrace{3 \left( \dfrac{1}{2} \right)^2}_{f(\frac{1}{2})} + \underbrace{5 \left( \dfrac{1}{2} \right) -9}_{g(\frac{1}{2})} = 3\left( \dfrac{1}{4} \right) + \dfrac{5}{2} - 9 = \dfrac{3}{4} + \dfrac{10}{4} - \dfrac{36}{4} = -\dfrac{23}{4}$

Therefore, $h \left( \dfrac{1}{2} \right) = -\dfrac{23}{4}$

Replace $x$ with $a + 1$ , then simplify. $h(a + 1) = f(a + 1) + g(a + 1)$

$\begin{array} &&=\underbrace{3(a+1)^2}_{f(a+1)} + \underbrace{5(a+1) -9}_{g(a+1)} \\ &= 3(a + 1)(a + 1) + 5a + 5 − 9 \\ &= 3(a^2 + 2a + 1) + 5a − 4 \\ &= 3a^2 + 6a + 3 + 5a − 4 \\ &= 3a^2 + 6a + 5a + 3 − 4 \\ h(a + 1) &= 3a^2 + 11a − 1 \end{array}$

Example 1.3.3

Let $f(x) = 2x^2 − 4x + 1$ . Evaluate each of the following:

$f(−3)$
$f(a + 2)$
$f(a + 2) − f(a)$

Solution

Replace each $x$ with $x = −3$ .

$f(−3) = 2(−3) 2 − 4(−3) + 1 = 2 \cdot 9 + 12 + 1 = 18 + 12 + 1 = 31$

Therefore, $f(−3) = 31$ .

Replace each $x$ with $a + 2$ . Then simplify.

$\begin{array} &f(a + 2) &= 2(a + 2) 2 − 4(a + 2) + 1 &\\ &= 2(a^2 + 4a + 4) − 4a − 8 + 1 &\text{Use Special Products and the Distributive Property} \\ & = 2a^2 + 8a + 8 − 4a − 8 + 1& \\ &= 2a^2 + 8a − 4a + 8 − 8 + 1 &\text{Rearrange terms to combine like terms} \\ f(a + 2) &= 2a^2 + 4a + 1 & \end{array}$

We will subtract $f(a)$ from the answer to part b.

$\begin{array} &&\underbrace{2a^2 + 4a + 1}_{f(a+2)} − \underbrace{(2a^2 − 4a + 1 )}_{f(a)} &\text{Use answer from part b: $f(a + 2) = 2a^2 + 4a + 1$} \\ &= 2a^2 + 4a + 1 − 2a^2 + 4a − 1 &\text{Subtract a quantity, so subtract each term.} \\ &= 2a^2 − 2a^2 + 4a + 4a + 1 − 1 &\text{Rearrange terms to combine like terms.} \end{array}$

Answer $f(a + 2) − f(a) = 8a$

Example 1.3.4

Let $f(x) = 3x − 4$ . Evaluate each of the following.

$f(a^2)$
$[f(a)]^2$

Solution

The order of operations will yield different results.

$f(a^2)$ indicates $x = a^2$ . Therefore, $f(a^2) = 3a^2 − 4$ .
$[f(a)]^2$ indicates $x = a$ . After substitution, square the result.

$\begin{array} &&[f(a)]^2 = (3a − 4)^2 &\text{Substitute $x = a$ into the function $f$.} \\ &[f(a)]^2 = 9a^2 − 24a + 16 &\text{Use the Special Products Formula.} \end{array}$

Try It! (Exercises)

For #1-13, evaluate the functions using $f$ and $g$ as defined below. Do not use a calculator.

$f(x) = – 2x + 1$

$g(x) = x^2 + 3x − 2$

$f(3)$
$g(1)$
$f \left( \dfrac{1}{2} \right)$
$g(−2)$
$f(−1) − g(3)$
$g(2a)$
$f(3a)$
$f(a − 4)$
$g(a − 1)$
$g(a^2)$
$[f(a)]^2$
$g(a) − f(a + 1)$
$g(a) − f(a^2)$

14. Let $h(x) = f(x) + g(x)$ . Evaluate each of the following:

$h(−5)$
$h(4a)$
$h(a − 3)$

For #15-34, evaluate the functions using $f$ and $g$ as defined below. Do not use a calculator.

$f(x) = \dfrac{x}{x+1}$

$g(x) = \dfrac{4x}{x+2}$

$f(3)$
$f(−3)$
$g(2)$
$g(−2)$
$f(2) + g(4)$
$f(0) − g(6)$
$[f(1)]^2$
$[g(3)]^2$
$f \left( \dfrac{1}{2} \right)$
$g \left( \dfrac{1}{2} \right)$
$f(a − 1)$
$g(a − 1)$
$[g(1)]^{−2}$
$[f(4)]^{−2}$
$[f(−2)]^{−1}$
$[g(−1)]^{−1}$
$\left[g \left( \dfrac{1}{4} \right) \right]^{−1}$
$[f(5)]^{−1} + [g(5)]^{−1}$
$[f(5) + g(5)]^{−1}$
$\dfrac{f(a)}{g(a)}$

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Definition: Function

Example 1.3.1

Definition: Naming Functions f, g, and h

Example 1.3.2

Example 1.3.3

Example 1.3.4

Try It! (Exercises)

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