]> Functions and Their Graphs

Functions and Their Graphs

1.1 Functions

Functions

A relation is a set of coordinate pairs—some matching between two variables (say, x and y). Any equation in two variables is a relation.

Examples:

x	1	0	-1	-1	0	1
y	1	2	1	0	-1	0

Here's a set of points—they automatically make a relation, even if you can't find an equation that they satisfy.

x	0	1	1	4	4	9
y	0	1	-1	2	-2	3

Here's another set of points—another relation. This time, you can easily find an equation that matches the variables: $x = y^{2}$

Some relations are more important than others…in particular, if the relation assigns a single y-value (output) to every x-value (input), then the relation is a function. Every x is matched with a single y —that's not to say that every y is matched with a single x!

Examples:

The previous example is a relation, but not a function—there is a value of x (two, actually—1 and 4) that is assigned (matched, paired) to two different y-values.

x	0	1	-1	2	-2	3
y	0	1	1	4	4	9

In this relation, every x-value is paired with a single y-value. This is a function.

We call x the independent variable, and y the dependent variable.

To determine if an equation (which is automatically a relation) is a function, try to solve the equation for y. If you can do it without any absolute value symbols, or ±, then you've got a function.

Examples: Try $x^{2} + y^{2} = 4$ . Solving for y : $y^{2} = 4 - x^{2}$ , which means that $y = \pm \sqrt{4 - x^{2}}$ . Since there's a ± in that, this isn't a function.

Try $\frac{x + 1}{2 y} = 6$ . Multiply: $x + 1 = 6 (2 y)$ . Now divide: $\frac{x + 1}{12} = y$ . This has no absolute value, nor any ±, so we've got a function!

Function Notation

Once you've solved the equation for y, then you're ready to convert it to function notation. Replace y with $f (x)$ .

Example: Our last example could be written $f (x) = \frac{x + 1}{12}$ .

There's no reason to always use f—you could use g, or h, or any letter that suites your fancy. Thus, there is no reason to become confused by $t (x) = 5 x + 1$ , or $z (x) = x^{2}$ , or anything!

The text makes a brief mention of piecewise defined functions, but that comes later.

To evaluate a function, just plug in!

Example: If $f (x) = 3 x - 1$ , then f(4) = 3(4) - 1 = 11.

Note that f(5) = f(2 + 3) ≠ f(2) + f(3).

The Domain of a Function

The domain of a function is the set of x-values that can be used in the function. The largest possible domain for functions (for our functions) is the set of real numbers, $ℝ$ . We exclude any values that cause problems—there are many ways for an x-value to cause problems, but so far we've only encountered two: division by zero, and square roots (even indexed roots) of negative numbers.

Examples: Find the domain of $g (x) = \frac{x + 1}{2 x - 3}$ . To do this, we'll look for the two problems mentioned above. There is a denominator; any x-value that causes it to be zero must be excluded. What value of x causes the denominator to be zero? Solve: 2x - 3 = 0 (you get x = 1.5). Thus, 1.5 must be excluded from the domain. There isn't a radical, so the second problem isn't an issue. The domain is all real numbers except for 1.5.

Find the domain of $h (x) = \sqrt{x + 5}$ . First, there is no denominator, so that won't be an issue. There is an even indexed radical—the radicand (The stuff under the vinculum) can't be negative—what x values would cause that? Solve: x + 5 < 0 (you get x < -5). The domain is x ≥ -5 (the opposite of x < -5).

1.2 Analyzing Graphs of Functions

The Graph of a Function

The graph of a relation is the locus (collection) of points that satisfy the equation (make the equation true). If you're looking at the graph of a relation, and you want to know if it is the graph of a function, then use the vertical line test.

The Vertical Line Test : if any vertical line passes through the graph at no more than one point, then the graph passes the test. Any graph that passes the vertical line test is the graph of a function.

This makes sense, really—the definition of a function says that every x-value has only one y-value. Vertical lines are really just x-values—if a vertical line crossed a graph at more than one point, then you'd have two y-values for a single x-value!

Zeros of a Function

z is a zero of the function f if f(z) = 0.

To find the zeros of a function, set the function equal to zero (this is the same as setting y equal to zero…)

Increasing and Decreasing Functions

A function is increasing on an interval if, for any x₁ and x₂ in the interval, x₁ < x₂ means that $f (x_{1}) < f (x_{2})$ .

Graphically, the function must be going up, as you move from left to right.

A function is decreasing on an interval if, for any x₁ and x₂ in the interval, x₁ < x₂ means that $f (x_{1}) > f (x_{2})$ .

Graphically, the function must be going down, as you move from left to right.

If a function is neither increasing nor decreasing (on an interval), then it must be constant (OK, technically, this isn't true—but it is highly unlikely that you would ever see one of these really weird functions).

When we report intervals of increasing and decreasing, we use interval notation, and we refer to the x-values where the function is increasing and decreasing.

Example: Determine the intervals where $h (x) = x^{3} - 3 x^{2} - 12 x$ is increasing and decreasing. This is definitely going to take a calculator. Here's the graph:

An Image

This function rises to a height of 7, but we say that it is increasing on the interval (-∞, -1), since those are the x-values where the graph is rising.

This function decreases to a value of -20, but we say that it is decreasing on the interval (-1, 2).

The function is increasing (again) on the interval (2, ∞).

Note that where the function changes from rising to falling, we get a bump—a crest. When the function switches from falling to rising, we get a valley—a trough. These are important features of function graphs.

$f (a)$ is a local maximum of f if there is some interval (probably small) where any x in that interval gives an $f (x)$ smaller than $f (a)$ .

$f (b)$ is a local maximum of f if there is some interval where any x in that interval gives an $f (x)$ bigger than $f (b)$ .

For now, you must use a calculator to find these points.

Linear Functions

Er…haven't we already talked about this?

Step Functions and Piecewise-Defined Functions

Step functions have breaks in them that look like stair steps. For example, the greatest integer function. The book denotes it like this: An Image . I would write it like this: ${}_{└}x_{┘}$ . In either case, the greatest integer function gives back the largest integer that is NOT greater than x.

Example: ${}_{└}{2.5}_{┘} = 2$ . ${}_{└}{1.999}_{┘} = 1$ . ${}_{└}{-3.01}_{┘} = -4$ . And the graph:

An Image

Piecewise defined functions are defined in pieces—the Frankensteins of Function World.

Example: Here's a lovely piecewise function: $f (x) = {\begin{array}{r} - x & x < - 2 \\ x^{3} - 3 x & - 2 \leq x < 2 \\ 3 & x \geq 2 \end{array}}$ . And here is its graph:

An Image

To graph a piecewise function, just draw each bit, but only for the x-values that are given!

Even and Odd Functions

A function is even if $f (- x) = f (x)$ . Graphically, this means that it has y-axis symmetry.

A function is odd if $f (- x) = - f (x)$ . Graphically, this means that it has origin symmetry.

1.3 Shifting, Reflecting, and Stretching Graphs

Summary of Graphs of Common Functions

Know them!

Shifting Graphs

Graphs can be shifted two ways—vertically, or horizontally.

If the function $g (x) = f (x) + d$ , then the graph of $g (x)$ is the same as $f (x)$ , except it is shifted d units vertically. If d > 0, then the graph is shifted up; if d < 0, then the graph is shifted down. This is typically very easy to spot.

Example: the graph of $m (x) = \sqrt{x} + 5$ is the same as the graph of $b (x) = \sqrt{x}$ , except that $m (x)$ is shifted up five units—take a look.

An Image

If the function $h (x) = f (x - c)$ , then the graph of $h (x)$ is the same as the graph of $f (x)$ , except that it is shifted horizontally c units. If c > 0, then the graph is shifted right; if c < 0, then the graph is shifted left. This is typically harder to see.

Example: the graph of $h (x) = {(x - 1)}^{3}$ is the same as the graph of $h (x) = x^{3}$ , except that $h (x)$ is shifted right one unit. Take a look:

An Image

It is possible for a function to have both horizontal and vertical shifts!

Reflecting Graphs

The graph of $- f (x)$ is the same as the graph of $f (x)$ , except that it is flipped vertically. This is typically easy to see.

The graph of $f (- x)$ is the same as the graph of $f (x)$ , except that it is flipped horizontally. This is not as easy to see.

Examples:: Let $f (x) = x^{3} - 5 x + 1$ . $- f (x) = - x^{3} + 5 x - 1$ , and $f (- x) = {(- x)}^{3} - 5 (- x) + 1$ .

Here are the graphs of $f (x)$ and $- f (x)$ :

An Image

Here are the graphs of $f (x)$ and $f (- x)$ :

An Image

Again, note that you could have both transformations in a single graph!

Nonrigid Transformations

A rigid transformation doesn't change the shape of a graph—it just moves or flips it. If you imagine that the shape is made out of hard plastic, or thick metal wire, then lifting that and moving it around doesn't change the shape—because the shape is rigid.

A nonrigid transformation does change the shape—it distorts the image, either vertically or horizontally. For now, the only nonrigid transformation that we'll look at is the vertical.

The graph of $j (x) = a \cdot f (x)$ is stretched vertically from the graph of $f (x)$ . If a > 1, then the graph is stretched upwards; if a < 1, then the graph is squashed down.

Examples: Let $f (x) = \sqrt{x}$ . Here are the graphs of $f (x)$ and $3 \cdot f (x)$ :

An Image

Here are the graphs of $f (x)$ and $\frac{1}{2} f (x)$ :

An Image

You cannot have both of these types in a single graph.

1.4 Combinations of Functions

Arithmetic Combinations of Functions

Sum of Functions: $(f + g) (x) = f (x) + g (x)$ .

Difference of Functions: $(f - g) (x) = f (x) - g (x)$ .

Product of Functions: $(f g) (x) = f (x) g (x)$ .

Quotient of Functions: $(\frac{f}{g}) (x) = \frac{f (x)}{g (x)}$ .

Of particular interest is the domain of this combination—for sum, difference and product, it's easy: the overlap (intersection) of the domains of f and g. For quotients, take that same overlap, but then remove any values of x that cause the denominator (g) to be zero.

Example: Let $f (x) = \sqrt{x}$ and $g (x) = x^{3}$ .

$(f + g) (x) = \sqrt{x} + x^{3}$ .

$(f - g) (x) = \sqrt{x} - x^{3}$ .

$(f g) (x) = (\sqrt{x}) (x^{3}) = x^{\frac{7}{2}}$ .

The domain for these three is x ≥ 0.

$(\frac{f}{g}) (x) = \frac{\sqrt{x}}{x^{3}} = x^{\frac{-5}{2}} = \frac{1}{\sqrt{x^{5}}}$ .

The domain for the quotient is x > 0 (we can't let x³ = 0, which only happens when x = 0).

Composition of Functions

This is terribly important!

A composition of functions is where the result of one function becomes the input (x) for another function.

The symbol is (f ◦ g)(x), which means $f (g (x))$ — $g (x)$ is put into $f (x)$ .

Example: Let $f (x) = \frac{1}{x}$ and $g (x) = \sqrt{x}$ . Then $(f ◦ g) (x) = \frac{1}{g (x)} = \frac{1}{\sqrt{x}}$ .

The domain of a composition is important. To find the domain of (f ◦ g)(x), first find the domain of $g (x)$ —note that your final answer cannot be larger than this! Next, work out (f ◦ g)(x) and see what its domain looks like it ought to be. Finally, take the intersection (overlap) between the two domains you worked out—that's your answer.

Examples: Let's continue our previous example. The domain of g (x) is x ≥ 0. The composition itself appears to have two problems—an even indexed radical, and a denominator. Because of the radical, the domain can't be bigger than x ≥ 0. Because of the denominator, √x ≠ 0, which means that x ≠ 0. So, the domain of the composition seems to be x > 0. Now, intersect this against the domain of $g (x)$ ! x ≥ 0 intersected with x > 0 is just x > 0.

This happens quite often—the domain that we get from the composition is almost always correct. The intersection with $g (x)$ 's domain is just a check that rarely does anything.

Try these: $f (x) = x^{2}$ and $g (x) = \sqrt{x}$ . The domain of $g (x)$ is x ≥ 0. The composition is $(f ◦ g) (x) = {(g (x))}^{2} = {\sqrt{x}}^{2} = x$ . It looks like the domain of the composition has no problems—all real numbers. Don't forget: check against the domain of g ! All real numbers intersected with x ≥ 0 is x ≥ 0.

Finally, it is important to be able to see when a function can be expressed as the composition of two others—I call this decomposing a function. There aren't any steps to this—you just have to see it. To make things more fun, there are always an infinite number of correct answers! Two of them are trivial, though—if you let either function be just a plain x, that's trivial.

Examples:: Find $f (x)$ and $g (x)$ so that $(f ◦ g) (x) = {(x + 1)}^{2}$ . How about $f (x) = x^{2}$ and $g (x) = x + 1$ ?

Try it again: $\frac{1}{x + 4}$ . Easy, right? $f (x) = \frac{1}{x}$ and $g (x) = x + 4$ .

Try this: $\frac{2 x}{2 x + 3}$ . One answer is $f (x) = \frac{x}{x + 3}$ and $f (x) = 2 x$ .

1.5 Inverse Functions

The Inverse of a Function

A function takes an input, and produces a single output. Sometimes, we'd like to take an output and figure out what the input was. Sometimes, this is even possible!

The inverse of a function "undoes" what the function did. Suppose $f (x) = x^{2}$ —this means that f takes an input and squares it. A function that undoes this must "unsquare" (square root).

If $f (x)$ and $g (x)$ are functions such that $f (g (x)) = x$ and $g (f (x)) = x$ , then f and g are inverse functions.

The inverse function of $f (x)$ is denoted $f^{- 1} (x)$ . It is important to realize that this does not mean $\frac{1}{f (x)}$ .

To find the inverse function, swap x and y, then solve for y (remember that $f (x) = y$ ).

Example: Find the inverse function for $b (x) = 2 x + 5$ . First, swap: x = 2 y + 5. Now let's solve for y : x - 5 = 2y; $y = \frac{x - 5}{2}$ . Ta da!

The Graph of the Inverse of a Function

Since we find the inverse function by swapping x and y, it is logical that their graphs are connected in the same way. Swapping x for y on a graph reflects (flips) the graph across the line y = x.

Now is the time to talk about when a function actually has an inverse function. Not every function does. Consider the simple squaring function: $f (x) = x^{2}$ . Here's its graph:

An Image

Now, when you flip that across the line y = x, you get this:

An Image

Wait! That new thing isn't a function! It doesn't pass the Vertical Line Test!

And there you have it—almost. If the new function passes the vertical line test, then the original had better pass the Horizontal Line Test (since we swapped x and y!).

The Horizontal Line Test : If any horizontal line crosses the graph in more than one spot, then the function fails the horizontal line test.

A function has an inverse function only if it passes the horizontal line test.

Example: Does the function $g (x) = x^{3} - 5 x - 2$ have an inverse function?

To answer this, check the graph:

An Image

I've thrown in a horizontal line to show that there is one that crosses more than once. $g (x)$ does not have an inverse function.

Finding the Inverse of a Functions Analytically

There is one more detail in finding inverse function algebraically—since we swapped x and y, we must also swap domains and ranges. The domain of the function must be the range of the inverse; the domain of the inverse must be the range of the function.

Occasionally, we must restrict the domain of one of the function to make this work.

Example: $h (x) = \sqrt{x - 1}$ . Does this even have an inverse?

An Image

Yes! It passes the HLT. Take note: the domain is x ≥ 1, and the range is y ≥ 0.

To find the inverse, swap x and y : $x = \sqrt{y - 1}$ . Now solve for y : x² = y - 1; x² + 1 = y. The domain of this inverse function is all reals, and its range is y ≥ 1.

Uh-oh. The domain of the original function is the range of the inverse; but the domain of the inverse isn't the domain of the function!

To fix this, restrict the bigger domain: the inverse has the bigger domain (all real numbers), so we'll restrict it to x ≥ 0.

So, out results are $h (x) = \sqrt{x - 1}$ , and $h^{- 1} (x) = x^{2} + 1$ (where x ≥ 0).

Page last validated 2010-08-15