Review of Functions: Fresh Take

Lumen Learning

Review of Functions: Fresh Take

Identify zeros of functions using equations, graphs, and tables
Interpret graphs and tables to describe their behavior including properties of symmetry
Make new functions from two or more given functions

Functions

The Main Idea

A function is a special type of relation where each input is associated with exactly one output.
Functions are used to describe relationships between two sets.
The input of a function is called the independent variable, often denoted as $x$ .
The output of a function is called the dependent variable, often denoted as $y$ .
Function notation: $y = f(x)$ is read as “ $y$ equals $f$ of $x$ .”
Functions can be represented using tables, graphs, or formulas (algebraic expressions).

Consider the function $g(x) = x² - 4x + 3$

Evaluate $g(3)$
Create a small table of values for $x = 0, 1, 2, 3, 4$

$g(3) = 3² - 4(3) + 3 = 9 - 12 + 3 = 0$
x g(x)

$0$ $3$

$1$ $0$

$2$ $-1$

$3$ $0$

$4$ $3$

The Domain and Range of a Function

The Main Idea

The domain of a function is the complete set of possible input values ( $x$ -values).
The range of a function is the complete set of possible output values ( $y$ -values).
Domain restrictions can occur due to mathematical limitations (e.g., division by zero, square roots of negative numbers).
Set notation and interval notation are used to describe domains and ranges.
The union symbol ( $\cup$ ) is used to describe domains and ranges with gaps or breaks.

Determine the domain and range of the function $f(x) = |x + 1| - 3$ .

Determine the domain:
The absolute value function is defined for all real numbers meaning there are no restrictions on the input. So the domain is:

$(-\infty, \infty)$ or ${x | x \text{ is any real number}}$
Determine the range:
The absolute value $|x + 1|$ is always non-negative. The minimum value of $|x + 1|$ is $0$ , which occurs when $x = -1$ .

Subtracting $3$ shifts the entire function down by $3$ units. The minimum value of $f(x)$ is therefore $-3$ .

There is no upper limit to the function’s value. Making the range:

$[-3, ∞)$ or ${y | y ≥ -3}$

Intercepts of a Function

The Main Idea

$x$ -intercepts are points where a function’s graph crosses the $x$ -axis ( $f(x) = 0$ ).
$x$ -intercepts are also called zeros or roots of the function.
The $y$ -intercept is the point where a function’s graph crosses the $y$ -axis ( $x = 0$ ).
A function can have multiple $x$ -intercepts but at most one y-intercept.
$x$ -intercepts provide information about the shape of the function’s graph.
The $y$ -intercept represents the function’s output when the input is zero.

Consider the function $f(x)=\sqrt{x+3}+1$ .

Find all zeros of $f$ .
Find the $y$ -intercept (if any).
Sketch a graph of $f$ .

To find the zeros, solve $\sqrt{x+3}+1=0$ . This equation implies $\sqrt{x+3}=-1$ . Since $\sqrt{x+3}\ge 0$ for all $x$ , this equation has no solutions, and therefore $f$ has no zeros.
The $y$ -intercept is given by $(0,f(0))=(0,\sqrt{3}+1)$ .
To graph this function, we make a table of values. Since we need $x+3\ge 0$ , we need to choose values of $x\ge -3$ . We choose values that make the square-root function easy to evaluate.

$x$ $-3$ $-2$ $1$

$f(x)$ $1$ $2$ $3$

Making use of the table and knowing that, since the function is a square root, the graph of $f$ should be similar to the graph of $y=\sqrt{x}$ , we sketch the graph (Figure 9).

An image of a graph. The y axis runs from -2 to 4 and the x axis runs from -3 to 2. The graph is of the function “f(x) = (square root of x + 3) + 1”, which is an increasing curved function that starts at the point (-3, 1). There are 3 points plotted on the function at (-3, 1), (-2, 2), and (1, 3). The function has a y intercept at (0, 1 + square root of 3). — Figure 9. The graph of $f(x)=\sqrt{x+3}+1$ has a $y$ -intercept but no $x$ -intercepts.

Find the zeros of $f(x)=x^3-5x^2+6x$ .

$x=0,2,3$

Symmetry of Functions

The Main Idea

Function Symmetry:
- Symmetry about the $y$ -axis: $f(x) = f(-x)$ (even functions)
- Symmetry about the origin: $f(-x) = -f(x)$ (odd functions)
- Some functions are neither even nor odd
Absolute Value Function:
- Defined as $f(x) = |x| = x$ if $x ≥ 0$ , and $-x$ if $x < 0$
- Always outputs non-negative values
- Symmetric about the $y$ -axis (even function)

Figure 13. (a) A graph that is symmetric about the $y$ -axis. (b) A graph that is symmetric about the origin.

If we are given the graph of a function, it is easy to see whether the graph has one of these symmetry properties. But without a graph, how can we determine algebraically whether a function $f$ has symmetry? Looking at Figure 13(a) again, we see that since $f$ is symmetric about the $y$ -axis, if the point $(x,y)$ is on the graph, the point $(−x,y)$ is on the graph. In other words, $f(−x)=f(x)$ . If a function $f$ has this property, we say $f$ is an even function, which has symmetry about the $y$ -axis. For example, $f(x)=x^2$ is even because

$f(−x)=(−x)^2=x^2=f(x)$ .

In contrast, looking at Figure 13(b) again, if a function $f$ is symmetric about the origin, then whenever the point $(x,y)$ is on the graph, the point $(−x,−y)$ is also on the graph. In other words, $f(−x)=−f(x)$ . If $f$ has this property, we say $f$ is an odd function, which has symmetry about the origin. For example, $f(x)=x^3$ is odd because

$f(−x)=(−x)^3=−x^3=−f(x)$ .

Determine whether $f(x)=4x^3-5x$ is even, odd, or neither.

Compare $f(−x)$ with $f(x)$ and $−f(x)$ .

$f(x)$ is odd.

Absolute Value Functions

For the function $f(x)=|x+2|-4$ , find the domain and range.

$|x+2|\ge 0$ for all real numbers $x$ .

Domain = $(−\infty ,\infty )$ , range = $\{y|y\ge -4\}$ .

Composing Functions

The Main Idea

Combining Functions with Mathematical Operators:
- Sum: $(f + g)(x) = f(x) + g(x)$
- Difference : $(f - g)(x) = f(x) - g(x)$
- Product: $(f · g)(x) = f(x) · g(x)$
- Quotient: $(\frac{f}{g})(x) = \frac{f(x)}{g(x)}\text{, where }g(x) ≠ 0$
Composite Functions:
- $(f ∘ g)(x) = f(g(x))$
- The output of g becomes the input of f
- Order matters: $(f ∘ g)(x) ≠ (g ∘ f)(x)$ in general

In general $f\circ g$ and $g\circ f$ are different functions. In other words in many cases $f\left(g\left(x\right)\right)\ne g\left(f\left(x\right)\right)$ for all $x$ .

For example if $f\left(x\right)={x}^{2}$ and $g\left(x\right)=x+2$ , then

$\begin{align}f\left(g\left(x\right)\right)&=f\left(x+2\right) \\[2mm] &={\left(x+2\right)}^{2} \\[2mm] &={x}^{2}+4x+4\hfill \end{align}$

but

$\begin{align}g\left(f\left(x\right)\right)&=g\left({x}^{2}\right) \\[2mm] \text{ }&={x}^{2}+2\hfill \end{align}$

Consider the functions $f(x)=x^2+1$ and $g(x)=\frac{1}{x}$ .

Find $(g\circ f)(x)$ and state its domain and range.
Evaluate $(g\circ f)(4)$ and $(g\circ f)\left(-\frac{1}{2}\right)$ .

We can find the formula for $(g\circ f)(x)$ in two different ways. We could write
$(g\circ f)(x)=g(f(x))=g(x^2+1)=\dfrac{1}{x^2+1}$ .

Alternatively, we could write

$(g\circ f)(x)=g(f(x))=\dfrac{1}{f(x)}=\dfrac{1}{x^2+1}$ .

Since $x^2+1\ne 0$ for all real numbers $x$ , the domain of $(g\circ f)(x)$ is the set of all real numbers.

Since $0<\frac{1}{(x^2+1)}\le 1$ , the range is, at most, the interval $(0,1]$ . To show that the range is this entire interval, we let $y=\frac{1}{(x^2+1)}$ and solve this equation for $x$ to show that for all $y$ in the interval $(0,1]$ , there exists a real number $x$ such that $y=\frac{1}{(x^2+1)}$ .

Solving this equation for $x$ , we see that $x^2+1=\frac{1}{y}$ , which implies that

$x=±\sqrt{\frac{1}{y}-1}$ .

If $y$ is in the interval $(0,1]$ , the expression under the radical is nonnegative, and therefore there exists a real number $x$ such that $\frac{1}{(x^2+1)}=y$ . We conclude that the range of $g\circ f$ is the interval $(0,1]$ .
$(g\circ f)(4)=g(f(4))=g(4^2+1)=g(17)=\frac{1}{17}$
$(g\circ f)(-\frac{1}{2})=g(f(-\frac{1}{2}))=g((-\frac{1}{2})^2+1)=g(\frac{5}{4})=\frac{4}{5}$

Let $f(x)=2-5x$

Let $g(x)=\sqrt{x}$

Find $(f\circ g)(x)$

$(f\circ g)(x)=2-5\sqrt{x}$ .

Consider the functions $f$ and $g$ described below.

$x$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$	$4$
$f(x)$	$0$	$4$	$2$	$4$	$-2$	$0$	$-2$	$4$

$x$	$-4$	$-2$	$0$	$2$	$4$
$g(x)$	$1$	$0$	$3$	$0$	$5$

Evaluate $(f\circ f)(3)$ and $(f\circ f)(1)$ .
State the domain and range of $(f\circ f)(x)$ .

$(f\circ f)(3)=f(f(3))=f(-2)=4$
$(f\circ f)(1)=f(f(1))=f(-2)=4$
The domain of $f\circ f$ is the set $\{-3,-2,-1,0,1,2,3,4\}$ . Since the range of $f$ is the set $\{-2,0,2,4\}$ , the range of $f\circ f$ is the set $\{0,4\}$ .

If items are on sale for $10\%$ off their original price, and a customer has a coupon for an additional $30\%$ off, what will be the final price for an item that is originally $x$ dollars, after applying the coupon to the sale price?

The sale price of an item with an original price of $x$ dollars is $f(x)=0.90x$ . The coupon price for an item that is $y$ dollars is $g(y)=0.70y$ .

$(g\circ f)(x)=0.63x$