NOTES
77. On a standard piano, the A below middle C produces a sound wave with frequency 220 Hz (cycles per second). The
0.6 TRANSFORMATIONS OF FUNCTIONS
You are now familiar with a long list of functions including polynomials, rational functions, trigonometric functions, exponentials and logarithms. Many students at this stage expect that calculus and higher-level mathematics courses will consist of defining even more functions and solving more equations. Actually, one important goal for this course is to more fully understand the properties of the functions we have already defined. To a large extent, you
will build your understanding by examining a few key properties of functions. To briefly get an idea of where we are headed, recall what the function f (x)= x2 represents. To understand this function, you first had to understand how to multiply two numbers together.
From practice multiplying “3 times 3” and “6 times 6” and so on, the idea of multiplying any number by itself became natural. The function x2is, in this sense, simply an abstraction. We have put multiplication into the broader context of functions. Now that we have functions, we need to make the manipulations of functions as natural as possible.
The first few function manipulations we define are straightforward.
DEFINITION 6.1
Suppose that f (x) and g(x) are functions with domains D1and D2, respectively. The functions f + g, f − g and f · g are defined by
( f + g)(x) = f (x) + g(x), ( f − g)(x) = f (x) − g(x),
and
( f · g)(x) = f (x) · g(x), for all x in D1∩ D2(i.e., x ∈ D1, and x∈ D2). The function f
g is defined by
f g
(x)= f (x) g(x), for all x in D1∩ D2such that g(x)= 0.
In example 6.1, we examine various combinations of several simple functions.
EXAMPLE 6.1 Combinations of Functions If f (x)= x − 3 and g(x) =√
x− 1, determine the functions f + g, 3 f − g and f g, stating the domains of each.
Solution First, note that the domain of f is the entire real line and the domain of g is the set of all x≥ 1. Now,
( f + g)(x) = x − 3 +√ x− 1 and
(3 f − g)(x) = 3(x − 3) −√
x− 1 = 3x − 9 −√ x− 1.
Notice that the domain of both ( f + g) and (3 f − g) is {x|x ≥ 1}. For
f g
(x)= f (x)
g(x) =√x− 3 x− 1,
the domain is {x|x > 1}, where we have added the restriction x = 1 to avoid dividing by 0.
Definition 6.1 and example 6.1 show us how to do arithmetic with functions. An operation on functions that does not directly correspond to arithmetic is the composition of two functions.
g(x)
f (g(x)) x
g
f
( f◦g)(x) = f (g(x)).
DEFINITION 6.2
The composition of functions f (x) and g(x), written f◦g, is defined by ( f◦g)(x) = f (g(x)),
for all x such that x is in the domain of g and g(x) is in the domain of f .
The composition of two functions is a two-step process, as indicated in the margin schematic. Be careful to notice what this definition is saying. In particular, for f (g(x)) to be defined, you first need g(x) to be defined, so x must be in the domain of g. Next, f must be defined at the point g(x), so that the number g(x) will need to be in the domain of f .
EXAMPLE 6.2 Finding the Composition of Two Functions For f (x)= x2+ 1 and g(x) =√
x− 2, find the compositions f ◦g and g◦ f and identify the domain of each.
Solution First, we have
( f◦g)(x) = f (g(x)) = f (√ x− 2)
= (√
x− 2)2+ 1 = x − 2 + 1 = x − 1.
It’s tempting to write that the domain of f◦g is the entire real line, but look more carefully.
Note that for x to be in the domain of g, we must have x≥ 2. The domain of f is the whole real line, so this places no further restrictions on the domain of f◦g. Even though the final expression x− 1 is defined for all x, the domain of ( f ◦g) is {x|x ≥ 2}. For the second composition,
(g◦ f )(x) = g( f (x)) = g(x2+ 1)
=
(x2+ 1) − 2 = x2− 1.
The resulting square root requires x2− 1 ≥ 0 or |x| ≥ 1. Since the “inside” function f is defined for all x, the domain of g◦ f is {x ∈ R|x| ≥ 1}, which we write in interval notation as (−∞, −1] ∪ [1, ∞).
As you progress through the calculus, you will often find yourself needing to recognize that a given function is a composition of simpler functions. For now, it is an important skill to practice.
EXAMPLE 6.3 Identifying Compositions of Functions
Identify functions f and g such that the given function can be written as ( f◦g)(x) for each of (a)√
x2+ 1, (b) (√
x+ 1)2, (c) sin x2 and (d) cos2x. Note that more than one answer is possible for each function.
Solution (a) Notice that x2+ 1 is inside the square root. So, one choice is to have g(x)= x2+ 1 and f (x) =√
x.
(b) Here,√
x+ 1 is inside the square. So, one choice is g(x) =√
x+ 1 and f (x) = x2.
(c) The function can be rewritten as sin (x2), with x2clearly inside the sine function.
Then, g(x)= x2and f (x)= sin x is one choice.
(d) The function as written is shorthand for (cos x)2. So, one choice is g(x)= cos x and f (x)= x2.
y
x 2
4 6 8 10
2 4
2
4
FIGURE 0.70a y= x2.
y
x 2
4 6 8 10
2 4
2
4
FIGURE 0.70b y= x2+ 3.
In general, it is quite difficult to take the graphs of f (x) and g(x) and produce the graph of f (g(x)). If one of the functions f and g is linear, however, there is a simple graphical procedure for graphing the composition. Such linear transformations are explored in the remainder of this section.
The first case is to take the graph of f (x) and produce the graph of f (x)+ c for some constant c. You should be able to deduce the general result from the following example.
EXAMPLE 6.4 Vertical Translation of a Graph Graph y= x2and y= x2+ 3; compare and contrast the graphs.
Solution You can probably sketch these by hand. You should get something like those in Figures 0.70a and 0.70b. Both figures show parabolas opening up. The main obvious difference is that x2has a y-intercept of 0 and x2+ 3 has a y-intercept of 3. Some thought should convince you that this is an important clue. For any given value of x, the point on the graph of y = x2+ 3 will be plotted exactly 3 units higher than the corresponding point on the graph of y= x2. This is shown in Figure 0.71a.
In Figure 0.71b, the two graphs are shown on the same set of axes. To many people, it does not look like the top graph is the same as the bottom graph moved up 3 units. This is an unfortunate optical illusion. Humans usually mentally judge distance between curves as the shortest distance between the curves. For these parabolas, the shortest distance is vertical at x = 0 but becomes increasingly horizontal as you move away from the y-axis.
The distance of 3 between the parabolas is measured vertically.
x 4
2 2
4
5 10 15 20
25 Move graph up 3 units y
x 4
2 2
4
5 10 15 20 25 y
FIGURE 0.71a Translate graph up.
FIGURE 0.71b
y= x2and y= x2+ 3.
In general, the graph of y = f (x) + c is exactly the same as the graph of y = f (x) shifted up (if c> 0) or down (if c < 0) by |c| units. We usually refer to f (x) + c as a vertical translation (up or down, by|c| units).
In example 6.5, we explore what happens if a constant is added to x.
EXAMPLE 6.5 A Horizontal Translation
Compare and contrast the graphs of y= x2and y= (x − 1)2.
Solution The graphs are shown in Figures 0.72a and 0.72b, respectively.
y
x 2
4 6 8 10
2 4
2
4
y
x 4
6 8 10
2 4
2
4 FIGURE 0.72a
y= x2.
FIGURE 0.72b y= (x − 1)2.
x 4
4 2 2 4 6 8 10
y Move graph to the right one unit
FIGURE 0.73
Translation to the right.
Notice that the graph of y= (x − 1)2appears to be the same as the graph of y = x2, except that it is shifted 1 unit to the right. This should make sense for the following reason.
Pick a value of x, say x = 13. The value of (x − 1)2at x = 13 is 122, the same as the value of x2at x = 12, 1 unit to the left. Observe that this same pattern would hold for any x you choose. A simultaneous plot of the two functions (see Figure 0.73) shows this.
In general, for c> 0, the graph of y = f (x − c) is the same as the graph of y = f (x) shifted c units to the right. Likewise (again, for c> 0), you get the graph of f (x + c) by moving the graph of y= f (x) to the left c units. We usually refer to f (x − c) and
f (x+ c) as horizontal translations (to the right and left, respectively, by c units).
y
x
2 3
3 1
15
10
5 5 10 15
FIGURE 0.74a y= f (x).
To avoid confusion on which way to translate the graph of y= f (x), focus on what makes the argument (the quantity inside the parentheses) zero. For f (x), this is x= 0, but for f (x− c) you must have x = c to get f (0) [i.e., the same y-value as f (x) when x = 0].
This says that the point on the graph of y= f (x) at x = 0 corresponds to the point on the graph of y= f (x − c) at x = c.
EXAMPLE 6.6 Comparing Vertical and Horizontal Translations Given the graph of y= f (x) shown in Figure 0.74a, sketch the graphs of y = f (x) − 2 and y= f (x − 2).
Solution To graph y= f (x) − 2, simply translate the original graph down 2 units, as shown in Figure 0.74b. To graph y= f (x − 2), simply translate the original graph to the right 2 units (so that the x-intercept at x = 0 in the original graph corresponds to an x-intercept at x= 2 in the translated graph), as seen in Figure 0.74c.
y
x
2 3
2 1
3 1
15
10
5 5 10 15
y
x 5 4
2 3
1 1
15 5 10 15
FIGURE 0.74b y= f (x) − 2.
FIGURE 0.74c y= f (x − 2).
Example 6.7 explores the effect of multiplying or dividing x or y by a constant.
EXAMPLE 6.7 Comparing Some Related Graphs
Compare and contrast the graphs of y = x2− 1, y = 4(x2− 1) and y = (4x)2− 1.
Solution The first two graphs are shown in Figures 0.75a and 0.75b, respectively.
y
x
2 3
2 1
3 1
2 2 4 6 8 10
FIGURE 0.75a y= x2− 1.
y
x
2 3
2 1
3 1
8 8 16 24 32 40
FIGURE 0.75b y= 4(x2− 1).
y
x
2 3
2 1
3 1
4 2 4 6 8 10
y x2 1 y 4(x2 1)
y
0.75 0.75 x
2 2 4 6 8 10
0.25 0.25
FIGURE 0.75c
y= x2− 1 and y = 4(x2− 1).
FIGURE 0.75d y= (4x)2− 1.
These graphs look identical until you compare the scales on the y-axes. The values in Figure 0.75b are four times as large, resulting from the multiplication of the original function by 4. The effect looks different when the functions are plotted on the same scale, as in Figure 0.75c. Here, the parabola y= 4(x2− 1) looks thinner. Note that the x-intercepts remain the same but the y-intercepts are different. (Why would that be?) The graph of y= (4x)2− 1 is shown in Figure 0.75d.
y
x
2 3
2 1
3 1
2 4 6 8 10
y x2 1 y (4x)2 1
FIGURE 0.75e
y= x2− 1 and y = (4x)2− 1.
Can you spot the difference here? In this case, the x-scale has now changed, by the same factor of 4 as in the function. To see this, note that substituting x = 1/4 into (4x)2− 1 produces (1)2− 1, exactly the same as substituting x = 1 into the original function. When plotted on the same set of axes (as in Figure 0.75e), the parabola y = (4x)2− 1 looks thinner. Note that here, the x-intercepts are different, but the y-intercepts are the same. (Why would that be?)
We can generalize the observations made in example 6.7. Before reading our explana-tion, try to state a general rule for yourself. How are the graphs of the functions c f (x) and
f (cx) related to the graph of y= f (x)?
Based on example 6.7, notice that to obtain a graph of y= cf (x) for some constant c> 0, you can take the graph of y = f (x) and multiply the scale on the y-axis by c. To obtain a graph of y= f (cx) for some constant c > 0, you can take the graph of y = f (x) and multiply the scale on the x-axis by 1/c.
These basic rules can be combined to understand more complicated graphs.
y
x
2 4
2
4
10 20
FIGURE 0.76a y= x2.
y
x
2 4
2
4
20 40
FIGURE 0.76b y= 2x2− 3.
EXAMPLE 6.8 A Translation and a Stretching
Describe how to get the graph of y= 2x2− 3 from the graph of y = x2.
Solution You can get from x2to 2x2− 3 by multiplying by 2 and then subtracting 3.
In terms of the graph, this has the effect of multiplying the y-scale by 2 and then shifting the graph down by 3 units (see the graphs in Figures 0.76a and 0.76b).
EXAMPLE 6.9 A Translation Involving Both xand yScales Describe how to get the graph of y= x2+ 4x + 3 from the graph of y = x2.
Solution We can again relate this (and the graph of every quadratic) to the graph of y= x2. We must first complete the square. Recall that in this process, you take the coefficient of x(4), divide by 2(4/2 = 2) and square the result (22= 4). Add and subtract this number and then, rewrite the x terms as a perfect square. We have
y= x2+ 4x + 3 = (x2+ 4x + 4) − 4 + 3 = (x + 2)2− 1.
To graph this function, take the parabola y= x2 (see Figure 0.77a) and translate the graph 2 units to the left and 1 unit down (see Figure 0.77b).
y
x
2 4
2
4
10 20
y
x
4 2 2
6
10 20
FIGURE 0.77a y= x2.
FIGURE 0.77b y= (x + 2)2− 1.
The following table summarizes our discoveries in this section.
Transformations of f (x)
Transformation Form Effect on Graph
Vertical translation f (x)+ c |c| units up (c > 0) or down (c < 0) Horizontal translation f (x+ c) |c| units left (c > 0) or right (c < 0) Vertical scale c f (x)(c> 0) multiply vertical scale by c Horizontal scale f (cx)(c> 0) divide horizontal scale by c You will explore additional transformations in the exercises.
EXERCISES 0.6
WRITING EXERCISES
1. The restricted domain of example 6.2 may be puzzling. Con-sider the following analogy. Suppose you have an airplane flight from New York to Los Angeles with a stop for refueling in Minneapolis. If bad weather has closed the airport in Min-neapolis, explain why your flight will be canceled (or at least rerouted) even if the weather is great in New York and Los Angeles.
2. Explain why the graphs of y= 4(x2− 1) and y = (4x)2− 1 in Figures 0.75c and 0.75e appear “thinner” than the graph of y= x2− 1.
3. As illustrated in example 6.9, completing the square can be