5 Chapter 5
5.10 Combining and Transforming Functions
Learning Objectives: In this section, you will learn about transformations, combinations, and compositions of functions. Upon completion you will be able to:
• Identify vertical and horizontal shifts of parent functions.
• Identify reflections of parent functions about the x-axis.
• Identify vertical expansions and contractions of parent functions.
• Recognize the appropriate parent function and, then, state the series of transformations to be performed on the parent function which would result in the graph of a given function.
• Write the function that results from a given series of transformations to be performed on a stated parent function.
• Graph a function based on the transformations to the corresponding parent or given function.
• Compute the sum, difference, product, and quotient of functions.
• Compute the composition of two or more functions, using function notation, tables, and graphs.
• Recognize the difference in the mathematical notation for combining and composing functions.
Transforming Parent Functions
Often when given a problem, we try to model the scenario using mathematics in the form of words, tables, graphs, or equations. One method we can employ is to adapt the basic graphs of ‘parent’ functions to build new models for the given scenario. There are systematic ways to alter functions to construct appropriate models for the problems we are trying to solve.
In this section, we discuss how the graphs of parent functions change, or transform, when certain specialized modifications are made to their rules.
Suppose the graph in Figure 5.7.2, below, is a complete graph of f(x). For the sake of our discussions, we will consider f (x) to be a ‘parent’ function.
Figure 5.7.2: The function f(x) is drawn on a coordinate plane.
The graph includes the points (0, 1), (2, 3), (4, 3), and (5, 5), an increasing line between the
ordered pairs (0, 1) and (2, 3), a horizontal line connecting the ordered pairs (2, 3) and (4, 3),
and another increasing line between (4, 3) and (5, 5).
The transformations of our parent functions will fall into three broad categories: shifts, reflections, and scalings. We will present the transformations in this order.
Vertical and Horizontal ShiftsOur first kind of transformation involves shifting the entire graph of a function up, down, right, or left. The simplest shift is a vertical shift, moving the graph up or down, because this transformation only involves adding a positive or negative constant to the function. In other words, we add the same constant to the output value of the function, regardless of the input.
Suppose we wanted to graph the function g(x) = f(x)+2, where f(x) is the ‘parent’ function graphed in Figure 5.7.2. Using the points indicated on Figure 5.7.2, we can create Table
5.21 representing the ordered pairs on the graph of the function g(x).
| X | (x, f(x)) | f(x) | g(x) = f(x) + 2 | (x,g(x)) |
| 0 | (0, 1) | 1 | 3 | (0,3) |
| 2 | (2,3) | 3 | 5 | (2,5) |
| 4 | (4,3) | 3 | 5 | (4,5) |
| 5 | (5, 5) | 5 | 7 | (5, 7) |
Table 5.21: A chart of ordered pairs on the graphs of f(x) and g(x) = f(x) + 2.
In general, if (a, b) is on the graph of y = f(x), then f(a) = b, so g(a) = f(a) + 2 = b + 2. Hence, (a, b + 2) is on the graph of g(x). In other words, to obtain the graph of g(x), we add 2 to the y-coordinate of each point on the graph of f (x). Geometrically, adding 2 to the
y-coordinate of a point moves the point 2 units above its previous location, and is “shifting the graph up 2 units.” Notice, by comparing the graph of f(x) in Figure 5.7.3 and the graph
of g(x) in Figure 5.7.4, that the graph retains the same basic shape as before, it is just 2 units above its original location. In other words, we connect the four points we moved in the same manner in which they were connected before.
Figure 5.7.3: The graph of J(x).
shift up 2 units
add 2 to each y-coordinate
Figure 5.7.4: The graph of g(x).
You can easily imagine what would happen if we wanted to graph the function j (x) = f (x )-2. Instead of adding 2 to each of the y-coordinates on the graph of f (x), we would be subtract ing 2. Geometrically, we would be moving the graph down 2 units.
What we have just discussed is generalized in the following theorem.
Theorem 7 Vertical Shifts
In the language of inputs and outputs, Theorem 5.7 can be paraphrased as “adding to, or subtracting from, the output of a function causes the graph to shift up or down, respectively.”
So what happens if we add to or subtract from the input of the function? A change to the input results in a movement of the graph left or right in what is known as a horizontal shift.
Keeping with the graph of f(x) in Figure 5.7.2, suppose we wanted to graph h(x) = f(x+2). In words, we are looking to see what happens when we add 2 to the input of the function. Let’s try to generate a table of values of h(x), based on those we know for f(x). We quickly find that we run into some difficulties. (See Table 5.22.)
| X | (x, f(x)) | f(x) | h(x) = f(x + 2) | (x,h(x)) |
| 0 | (0, 1) | 1 | f(0 + 2) = f(2) = 3 | (0,3) |
| 2 | (2,3) | 3 | f(2 + 2) = f(4) = 3 | (2,3) |
| 4 | (4,3) | 3 | f(4 + 2) = f(6) =? | |
| 5 | (5,5) | 5 | f(5 + 2) = f(7) =? |
Table 5.22: A chart of ordered pairs of f(x) and the work to determine the ordered pars of h( x).
When we substitute x = 4 into the formula h(x) = f(x+2), we are asked to find f(4+2) = f(6), which does not exist because the domain of f(x) is only [0,5]. The same thing happens when we attempt to find h(5). What we need here is a new strategy.
We know, for instance, f(0) = 1. To determine the corresponding point on the graph of h(x), we need to figure out what value of x we must substitute into h(x) = f(x + 2) so that the quantity x + 2 works out to be 0. Solving x + 2 = 0 gives x = -2, and h(-2) = f(-2 + 2) = f(0) = 1, so (-2, 1) is on the graph of h(x). Similarly, we know f(2) = 3, and if we set x+2 = 2, we get x = 0. Substituting, gives h(0) = f(0+2) = f(2) = 3. Continuing in this fashion, we construct Table 5.23.
| X | x+2 | h(x) = f(x + 2) | (x, h(x)) |
| -2 | 0 | h(-2) = f(0) = 1 | (-2, 1) |
| 0 | 2 | h(0) = f(2) = 3 | (0,3) |
| 2 | 4 | h(2) = f(4) = 3 | (2,3) |
| 3 | 5 | h(3) = f(5) = 5 | (3, 5) |
Table 5.23: A chart of ordered pairs on the graph of h(x) = f(x + 2).
In summary, the points (0, 1), (2, 3), (4, 3), and (5, 5) on the graph of f(x) give rise to the
points (-2, 1),(0, 3), (2, 3), and (3, 5) on the graph of h(x), respectively. In general, if (a, b) is on the graph of f(x), then f(a) = b. Solving x + 2 = a gives x = a – 2 so that h(a – 2) = f((a – 2) + 2) = f(a) = b. As such, (a – 2, b) is on the graph of h(x). The point (a – 2, b) is exactly 2 units to the left of the point (a, b), so the graph of h(x) is obtained by shifting the graph f(x) to the left 2 units, as shown in Figure 5.7.6.
Figure 5.7.5: The graph of f(x).
Shift left two units Subtract 2 from each x-coordinate
Figure 5.7.6: The graph of h(x).
If we set out to graph p(x) = f(x-2), we would find ourselves adding 2 to all of the x-values of the points on the graph of f (x) to produce a shift to the right 2 units. Generalizing these notions produces the following theorem.
In other words, Theorem 5.8 says “adding to, or subtracting from, the input to a function amounts to shifting the graph left or right, respectively.”
Theorems 5.7 and 5.8 present a theme which will run common throughout this section: changes to the outputs from a function affect the y-coordinates of the graph, resulting in some kind of vertical change; changes to the inputs to a function affect the x-coordinates of the graph, resulting in some kind of horizontal change.
Example 1: (a) Graph f(x) = Jx. Plot at least three points.
(b) Use your graph of f(x) to graph g(x) = Jx – 1.
(c) Use your graph of f(x) to graph j(x) = Jx – 1.
Solution: (a) We know the domain of f(x) is [0, oo), as it is the parent even root function. We choose non-negative numbers which are perfect squares to build Table 5.24, in order to graph f(x) in Figure 5.7.7, below.
| X | f(x) = VX | (x, J(x)) |
| 0 | 0 | (0,0) |
| 1 | 1 | (1,1) |
| 4 | 2 | (4, 2) |
Table 5.24: A chart of ordered pairs on the graph of f(x) = Jx.
Figure 5.7.7: The graph of j(x) = .Ji.
(b) g(x) = Jx-1 = f(x) -1 indicates a vertical shift of -1, moving all points on the graph of f(x) down 1 unit. The point (0, 0) is transformed by subtracting 1 from the y-coordinate.
Similarly,
(0, 0)—+ (0, -1)
(1, 1) —+ (1, 0)
(4, 2)—+ (4, 1)
Plotting the new points and connecting them in the same manner as the graph of
f(x) results in Figure 5.7.8.
Figure 5.7.8: The graph of g(x) = – 1.
(c) j(x) = vx-=-I = f(x -1) indicates a horizontal shift of 1, such that all points on the graph of f(x) move right 1 unit. The point (0, 0) is transformed by adding 1 to the x-coordinate.
(0, 0) –+ (1, 0)
Similarly,
(1, 1) –+ (2, 1)
(4,2)–+ (5,2)
Plotting the new points and connecting them in the same manner as the graph of
f(x) results in Figure 5.7.9.
Figure 5.7.9: The graph of j(x) = vx-=-I.
For g(x) the domain remains [0, oo), but the range changes to [-1,oo), as changes were made only to the outputs. For j (x) the domain changes to [1, oo), but the range remains [0, oo), as changes were made only to the inputs.
Reflections
We now turn our attention to reflections. Imagine plotting the point (2, 3) on a coordinate plane and folding the paper along the axes.
If the fold is along the x-axis, then where would the point (2, 3) be transferred to below the x-axis?
If the fold is along the y-axis, then where would the point (2, 3) be transferred to on the left of the y-axis?
Figure 5.7.10: The coordinate plane with the points (2, 3) and (2, -3) labeled on the graph.
Figure 5.7.11: The coordinate plane with the points (2, 3) and (-2, 3) labeled on the graph.
Using Figure 5.7.10, we can see (2, 3) is transferred across the x-axis to the point (2, -3), a mirror image across the x-axis.
Using Figure 5.7.11, we can see (2, 3) is transferred across the y-axis to the point (-2, 3), a mirror image across the y-axis.
To reflect a point ( x, y) across the x-axis, we replace y with -y. If (x, y) is on the graph of f(x), then y = f(x), so replacing y with -y is the same as replacing f(x) with – f(x). Hence, the graph of y = – f (x) is the graph of f (x) reflected across the x-axis. Similarly, the graph of y = f(-x) is the graph of f(x) reflected across the y-axis.
Returning to the language of inputs and outputs, multiplying the output from a function by
-1 reflects its graph across the x-axis, while multiplying the input to a function by -1 reflects the graph across the y-axis.
Applying Theorem 5.9 to the graph of the ‘parent’ function f(x) given at the beginning of the section, we can graph j(x) = – f(x) by reflecting the graph of f(x) about the x-axis. (See Figure 5.7.13.)
Figure 5.7.12: The graph of J(x).
Multiply each y-coordinate by -1 reflect across x-axis
Figure 5.7.13: The graph of j(x).
By reflecting the graph of J(x) across the y-axis, we obtain the the graph of k(x) = J(-x),
shown in Figure 5.7.15.
Figure 5.7.14: The graph of J(x).
Multiply each y-coordinate by -1 reflect across y-axis
Figure 5.7.15: The graph of k(x).
Example 2: (a) Graph J(x) = 2x. Plot at least three points.
(b) Use your graph of J(x) to graph g(x) = -2x.
(c) Use your graph of J(x) to graph j(x) = 2-x.
Solution: (a) We know the domain of J(x) is (-oo, oo), as it is an exponential growth function. We choose integer values from -2 to 2 to build Table 5.25, in order to graph J(x) in Figure 5.7.16, below.
| X | J(x) = 2x | (x, J(x)) |
| -2 | 1/4 | (-2, 1/4) |
| -1 | 1/2 | (-1, 1/2) |
| 0 | 1 | (0, 1) |
| 1 | 2 | (1, 2) |
| 2 | 4 | (2, 4) |
Table 5.25: A chart of ordered pairs on the graph of f(x) = 2x.
Figure 5.7.16: The graph of J(x) = 2x.
|
(b) g (x) = -2x = -1 · 2x = -1 · f (x) indicates a reflection of f (x) across the x-axis. The point (-2, ¾) is transformed by multiplying the y-coordinate by -1.
Similarly,
Plotting the new points and connecting them in the same manner as the graph of
f(x) results in Figure 5.7.17.
Figure 5.7.17: The graph of g(x) = -2x.
(c) j (x) = 2-x = f (-x) indicates a reflection of f (x) across the y-axis. The point (-2, ¾) is transformed by multiplying the x-coordinate by -1.
Similarly,
| (-1,t) | (1,t) |
| (0, 1) | (0, 1) |
| (1,2) | (-1,2) |
| (2,4) | (-2,4) |
Plotting the new points and connecting them in the same manner as the graph of
J(x) results in Figure 5.7.18.
Figure 5.7.18: The graph of j(x) = 2-x_
Note: For g(x) the domain remains (-oo, oo), but the range changes to (-oo, 0), as changes were made to the outputs. For j(x) the domain remains (-oo, oo) and the range remains the same; although changes were made to the inputs, the set of real numbers (-oo, oo) does not change when multiplied by -1.
Scaling (Stretching and Compression)
Our last class of transformations are known as scalings. The transformations discussed thus far are known as rigid transformations. Simply put, they do not change the shape of the graph, only its position and orientation in the plane. If, however, we wanted to make a new graph twice as tall as a given graph, or one-third as wide, we would be changing the shape of the graph. This type of transformation is called non-rigid, where not only will it be important for us to differentiate between modifying inputs versus outputs, we must also pay attention to the magnitude of the changes we make. As you will see shortly, the mathematics turns out to be easier than the associated grammar.
Suppose we wish to graph the function m(x) = 2f(x), where f(x) is the ‘parent’ function
given at the beginning of the section. (See Figure 5.7.2.) From its graph, we can build a table of values for m(x), similarly as before.
|
X |
(x, f(x)) | f(x) | m(x) = 2f(x) | (x,m(x)) |
| 0 | (0, 1) | 1 | 2 | (0,2) |
| 2 | (2,3) | 3 | 6 | (2,6) |
| 4 | (4,3) | 3 | 6 | (4,6) |
| 5 | (5,5) | 5 | 10 | (5, 10) |
Table 5.26: A chart of ordered pairs on the graphs of f(x) and m(x) = 2f(x).
In general, if (a,b) is on the graph of f(x), then f(a) = b, so m(a) = 2f(a) = 2b puts (a,2b) on the graph of m(x). In other words, to obtain the graph of m(x), we multiply all of the y-coordinates of the points on the graph of f (x) by 2. Multiplying all of the y-coordinates
of all of the points on the graph off (x) by 2 causes what is known as a “vertical scaling by a factor of 2,” and the results are shown in Figure 5.7.20.
Figure 5.7.19: The graph of f(x).
Vertical scaling by a factor of 2 multiply each y-coordinate by 2
Figure 5.7.20: The graph of m(x).
If we wish to graph n(x) = ½f (x), we multiply all of the y-coorclinates of the points on the graph of f(x) by½- This creates a “vertical scaling by a factor of½,” as seen in Figure 5.7.22.
Figure 5.7.21: The graph of f(x).
Figure 5.7.22: The graph of n(x).
These results are generalized in the following theorem.
A few remarks about Theorem 5.10 are in order. First, a note about the verbiage. To the authors, the words ‘stretching,’ ‘expansion,’ and ‘dilation’ all indicate something getting bigger. Hence, “stretched by a factor of 2” makes sense if we are scaling something by multi plying it by 2. Similarly, we believe words like ‘shrinking,’ ‘compression,’ and ‘contraction’
all indicate something getting smaller, so if we scale something by a factor of ½, we would say
it “shrinks by a factor of 2,” not “shrinks by a factor of ½.” This is why we have written the descriptions “stretching by a factor of a” and “shrinking by a factor of ¼” in the statement
of the theorem. Second, in terms of inputs and outputs, Theorem 5.10 says multiplying the outputs from a function by positive number a causes the graph to be vertically scaled by a factor of a.
Example 3: (a) Graph f(x) = x3. Plot at least three points.
(b) Use the graph of f(x) to graph g(x) = 4×3.
(c) Use the graph of f(x) to graph j(x) = ¼x3.
Solution: (a) We know the domain of j(x) is (-oo, oo), as it is the parent cubic polynomial function. We choose real number values from -1 to 1 to build Table 5.27, in order to graph f(x) in Figure 5.7.23, below.
| X | f(x) | (x, J(x)) |
| -1 | -1 | (-1,-1) |
| -1/2 | -1/8 | (-1/2, -1/8) |
| 0 | 0 | (0,0) |
| 1/2 | 1/8 | (1/2, 1/8) |
| 1 | 1 | (1, 1) |
Table 5.27: A chart of ordered pairs on the graph of j(x) = x3.
Figure 5.7.23: The graph of f(x) = x3.
(b) g (x) = 4x3 = 4f ( x) indicates a vertical scaling by a factor of 4. Because a = 4 > 1, then the graph of f (x) undergoes a vertical stretch by a factor of 4. The point (-1, -1) is transformed by multiplying they-coordinate by 4.
(-1, -1)–+ (-1, -4)
Similarly,
(-1,-1)–+ (-1,-1)(0, 0) –+(0, 0)
(1,1)–(1+,1)(1, 1) –+ (1, 4)
Plotting the new points and connecting them in the same manner as the graph of
J(x) results in Figure 5.7.24.
Figure 5.7.24: The graph of g(x) = 4x3.
(c) j (x) = ¼x3 = ¼ f (x) indicates a vertical scaling by a factor of ¼. Because 0 < a =
¼ < 1, then the graph of f(x) undergoes a vertical compression by a factor of 4. The point (-1, -1) is transformed by multiplying they-coordinate by ¼-
(-1, -1)–+ (-1,-l)
Similarly,
(-1′-1) –+(-1,-31
|
(0, 0) –+(0, 0)
(11,)–+(1,31
2)
(1, 1)–+ ( 1, l)
Plotting the new points and connecting them in the same manner as the graph of
f(x) results in Figure 5.7.25.
Figure 5.7.25: The graph of j(x) = ¾x3.
Note: Although changes were made to the outputs in both parts b and c, the set of real numbers (-oo, oo) does not change when multiplying by a nonzero scalar. Thus, the range remains the same in the graphs of both transformations.
It is natural to ask what would happen if we multiply the inputs of a function by a positive number. As previously mentioned changes to inputs result in horizontal changes, and, in this instance, the result is a horizontal scaling. The authors leave the topic of horizontal scalings to the reader to explore outside of this text.
Combining Transformations
As with evaluating functions, where order of operations is important, so is the order of trans formations.
Order of Transformations
Due to the fact that the authors have chosen to exclude the discussion of horizontal scalings from this text, reflections across the y-axis are not considered in the above order of trans formations.
Note: The steps above are logical, as they follow the order in which we would use to evaluate a function at a given value of x.
When asked to graph g(x) where g(x) = Af(x + H) + K, the reader is strongly encouraged to graph the series of functions which shows the gradual transformation of the graph of f(x) into the graph of g( x), in the order outlined above.
If asked to graph the function g(x) = – Ix + 3 I – 8, how would you use the previously dis cussed procedure?
First, we identify the parent function, f(x), and graph it. The parent function for g(x) is
f(x) = lxl.
Second, we identify the horizontal shift, if one exists. There is a horizontal shift of f(x) left 3 units (H = 3). Thus, the next graph in the series would be fi(x) = f(x + 3) =Ix+ 31.
Third, we identify the scalar, if one exists. The scalar is A= -1. Given that IAI = 1, there is no vertical scaling, but because A < 0, Ji(x) is reflected across the x-axis. The subsequent graph in the series would be h(x) = – f1(x) =-Ix+ 31.
Last, we identify the vertical shift, if one exists. There is a vertical shift of h(x) down 8 units (K = -8). Thefinal graph in the series would be h(x) = h(x) – 8 =-Ix+ 31 – 8.
Example 4: Given g(x) = ¾(x – 1)2 + 5,
1. Identify the parent function, f (x).
2. Describe, step-by-step, how the graph of f(x) transforms into the graph of g(x).
Solution:
(a) f(x) = x2
(b) Notice from the rule of g(x), A= ¾, H = -1, and K = 5. Thus, in order, the list of transformations to the graph of f (x) are:
(a) Shift right 1 unit, as H = -1.
(b) As A = ¾, vertically compress by a factor of !.
(c) Shift up 5 units, as K = 5.
Example 5: Given g(x) = 2 – ex,
1. Identify the parent function, f (x).
2. Describe, step-by-step, how the graph of f(x) transforms into the graph of g(x).
Solution:
(a) f(x) = ex
(b) Before we determine the list of transformations, let’s rewrite g(x) in a more standard form: g(x) =-ex+ 2. Notice from the rule of g(x), A= -1, H = 0, and K = 2. Thus, in order, the list of transformations to the graph of f ( x) are:
(a) Reflection across the x-axis, as A= -1. (There is no vertical scaling, as IAI = 1.)
(b) Shift up 2 units, as K = 2.
Try It 4
Now let’s assume we are given a list of transformations to be performed on the graph of a specific ‘parent’ function. We can use the order of transformations to write the equation of the resulting function.
Example 6: Let f(x) = x3. Find the equation of the function 9(x) whose graph is the result of
f (x) undergoing the following transformations.
(a) Horizontal shift right 1 unit.
(b) Reflection across the x-axis.
(c) Vertical shift up 8 units.
Solution: (a) A horizontal shift right 1 unit means H = -1 and
91(x) = J(x -1) = (x -1)3
(b) A reflection across the x-axis means A < 0. Seeing as no vertical scaling is described, IAI = 1 and thus, A= -1.
92(x) = -91(x) = -(x – 1)3
(c) A vertical shift up 8 units means K = 8, and so the equation of 9(x) is
9(x) = 92(x) + 8
9(x) = -(x – 1)3 + 8
Example 7: If the graph of f(x) = ¾ is shifted left 5 units, vertically stretched by a factor of 9, reflected across the x-axis, and then shifted down ½ a unit, what is the equation of the resulting graph?
Solution: We begin by interpreting the meaning of each transformation described. A shift left 5
units means H = 5 and
91 = f ( X + 5) = - 1
x + 5
A vertical stretch by a factor of 9, and then a reflection across the x-axis means A = -9, so
92 = -991(x) = -9 ( –1 ) = — 9
x + 5 x+5
Last, a shift down ½ a unit means K = -½ and so the equation of 9(x) is
1
9(x) = 92(x) – 2
9 1
9(x) = – X + 5 – 2
Try It 5
of real numbers. The following definitions allow us to add, subtract, multiply, and divide functions, using the arithmetic we already know for real numbers.
Function Arithmetic
In other words, to add two functions, we add their outputs; to subtract two functions, we subtract their outputs, and so on. Notice that while the formula (f +g)(x) = f(x)+g(x) looks suspiciously like some kind of distributive property, it is nothing of the sort; the addition on the left hand side of the equation is function addition, and we are using this equation to
define the output of the new function, f(x) + g(x), as the sum of the real number outputs of f(x) and g(x).
Example 8: Given f(x) = 6×2 – 2x and g(x) = 3 – ¾, compute the following.
(a) (f + g)(-1)
(b) (f g)(2)
(c) (g – J)(x)
(cl) (7)(x)
Solution: The domain of f(x) is (-oo, oo), because f(x) is a polynomial function. The domain of g(x) is (-oo,0) U (0,oo), as g(x) has a denominator which cannot be zero. Thus, the intersection of these domains is (-oo, 0) U (0, oo).
(a) x = -1 is in the intersection of the domains off (x) and g( x), therefore (f +g)(-1)
is defined.
(f + g)(-1) = f(-1) + g(-1)
= [6(-1)2 – 2(-1)] +
1
[3 – – -]
=8+4
=12
(-1)
(b) x = 2 is in the intersection of the domains of f (x) and g(x), therefore (f g)(2) is defined.
(f g)(2) = J(2) • g(2)
= [6(2)2 – 2(2)] • [3 – ]
(2)
= [20] • []
= 50
(c) In order for (g – f) (x) to be defined, x must be in the intersection of the domains of f(x) and g(x).
(g – J)(x) = g(x) – f(x)
|
= [3 – } ] – [6×2
= .3 – –1 – 6x2 + 2x
X
2x]
= (I) (
s)-
t-(6;2)(
s) + (
2:) ( s)
3x 1 6x3 2×2
= +
x X X X
3x 1 – 6x3 + 2×2
X
= -6×3 + 2×2 + 3x – 1 ,
X
where xis in
(-oo,0)
U (0,oo)
(cl) In order for ( 7) (x) to be defined, x must be in the intersection of the domains of f(x) and g(x) AND f(x) # 0. We know the intersection is (-oo,0) U (0,oo), but we need to find where f (x) =I=- 0. Thus,
6×2 – 2x =I=- 0
2x(3x – 1) =/ 0
2x =I=- 0
X =/=- 0
AND AND
3x -1 =/ 0
X =/=- ½
Therefore, for (7) (x) to be defined x must be in (-oo, 0) U (0, ½) U (½, oo).
|
g) g(x) 3 -½
f ( x)= f (x)= 6×2 – 2x
If simplified, we have
(y) (x)= 6:2 -=-tx
(;) (f) – (½) (½)
6×2 – 2x
3x-1
X
6×2 – 2x
3x- 1) ( 1 )
= ( x • 6×2 – 2x
(3x – 1)(1)
x (6×2 – 2x) 3x-1
2×2(3x – 1)
3x-1
2×2(3x – 1)
= 2 2, where xis in (-oo,0) U (o,}) U (},oo)
Please note the importance of identifying the domain of a function before simpli fying its expression. If we had waited to compute the domain of } : until after
simplifying, we would just have the formula 2;2 to go by, and we would (incor rectly!) state the domain as (-oo, 0) U (0, oo), because the other troublesome
number, x =½,was ‘divided out.’
Note: Because ( 7) (x) is a rational function, recall there will be a hole in the graph at
x = ½ and a vertical asymptote at x = 0.
Throughout this chapter we have seen function arithmetic in action. Thus, the authors leave it to the reader to practice function arithmetic with no further examples.
Finding the Composition of FunctionsSuppose we want to calculate how much it costs to heat a house on a particular day of the year. The cost to heat a house will depend on the average daily temperature, and in turn, the average daily temperature depends on the particular day of the year. Notice how we have just defined two relationships: the cost depends on the temperature, and the temperature depends on the day.
Using descriptive variables, we can notate these two relationships as functions. The function C(T) gives the cost, C, of heating a house for a given average daily temperature in T degrees Fahrenheit. The function T(d) gives the average daily temperature on day d of the year. Then, for any given day, Cost= C(T(d)) means that the cost depends on the temperature, which in turn depends on the day of the year. Thus, we can evaluate the cost function at the temperature T(d). For example, we could evaluate T(5) to determine the average daily temperature on the 5th day of the year. Then, we could evaluate the cost function at that temperature. We would write C(T(5)).
Temperature on day 5
Performing algebraic operations on functions combines them into a new function, but we can also create functions by composing functions. The process of combining functions so that the output of one function becomes the input of another is known as a composition of functions. The resulting function is known as a composite function.
In this definition the left-hand side is read as “f composed with g at x,” and the right-hand side as “f of g of x.” The two sides of the equation have the same mathematical meaning and are equal. The open circle symbol o is called the composition operator. Composition is a binary operation that takes two functions and forms a new function, much as addition or multiplication takes two numbers and gives a new number.
It is also important to understand order of operations in evaluating a composite function. We follow the usual convention with parentheses by starting with the innermost parentheses first, and then working to the outside. In the equation above, the function g takes the input x first and yields an output g( x). Then, the function f takes g( x) as its input and yields an output f(g(x)).
g(x), the output of g
is the input of f
(f o g) (x) = f (g Cr) )
x is the input of g
In the expression f (g( x)), the function g is often called the ‘inside’ function, while the func tion f is often called the ‘outside’ function.
Abstractly, we have
Figure 5.7.26: A visual representation of the composition of two functions, (Jo g)(x).
Multiplication of functions is not the same as composition of functions. Pay close attention to the operator given.
(f • g)(x) =/= (f o g)(x)
For the purposes of this text, when composing functions x is assumed to be in the domain of the composition. The authors will not compute the domain of a composition at this time, as similar domains have been discussed in previous sections.
Example 9: Using the functions provided, compute f(g(x)) and g(f(x)).
f (X) = 2x + 1 g(X) = -X + 3
Solution: Let’s begin with J(g(x)), by substituting g(x) into f(x) for x.
g(x) = -x + 3
J(g(x)) = 2(-x + 3) + 1
= -2x + 6 + 1
= -2x + 7
Now, we move to g(f(x)), by substituting f(x) into g(x) for x.
f(x) = 2x + 1
g(f(x)) = -(2x + 1) + 3
= -2x -1 + 3
= -2x + 2
Note: In general, (f o g)(x) and (g o f)(x) are different. In other words, in many cases
f(g(x)) =/= g(f(x)) for all x.
Note: Because g(f(x)) =/= f(g(x)) for all x, the operation of function composition is not commutative.
Example 10: Given f(x) = x2 – 4x, g(x) = 2 – x + 3, and h(x) = x!i, compute the following.
(a) (go J)(l)
(b) (go g)(6)
(c) (ho g)(x)
(cl) (f o .f)(x)
Solution: (a) Using the definition, (go J)(l) = g(f(l)). We first calculate J(l) :
J(l) = (1)2 – 4(1)
=1-4
= -3
So,
(g O f) (1) = g(J (1))
= g(-3)
= 2- -3 + 3
= 2- 5
(gof)(l) = 2
(b) Again, the definition tells us (go g)(6) = g(g(6)). That is, we evaluate g(x) at
x = 6, and then plug the result back into g(x). Given
g(6) = 2 – 6 +3
=2-v’9
= -1,
then
(g O g)(6) = g(g(6))
= g(-1)
=2- -1+3
(gog)(6) = 2-vi2
(c) We know (ho g)(x) = h(g(x)). Thus, we insert the expression g(x) into h for x
to get
(ho g)(x) = h(g(x))
= h(2 – Jx + 3)
2(2 – Jx + 3)
(2 – Jx + 3) + 1
|
ho )(x)=4-2Jx+3
g 3- Jx +3
(cl) We know (J o f) (x) = f (J (x)). Thus, we insert the expression f (x) into f for x
to get
(Jo f)(x) = J(J(x))
= J (x2 – 4x)
= (x2 – 4x) 2 – 4 (x2 – 4x)
= (x2 – 4x) (x2 – 4x) – 4 (x2 – 4x)
= x4 – 4x3 – 4x3 + 16×2 – 4x2 + 16x
(Jo J)(x) = x4 – 8×3 + 12×2 + 16x
When working tables of function values, we read input and output values of the composition from the table entries. We evaluate the inside function first, and then use the output of the inside function as the input of the outside function.
Example 11: Using Table 5.28, evaluate f(g(3)) and g(f(3)).
| X | f(x) | g(x) |
| 0 | 2 | 11 |
| 1 | 6 | 3 |
| 2 | 8 | 5 |
| 3 | 4 | 2 |
| 4 | 1 | 7 |
| 5 | 9 | 0 |
Table 5.28: A chart of x-values, with corresponding output values from f(x) and g(x).
Solution: To evaluate f(g(3)), we start from the inside with the input value, x = 3. We then evaluate the inside expression g(3) using the table that defines the function g : g(3) = 2. We then use this result as the input to the function f, so g(3) is replaced by 2 and we get f (2). Last, using the table that defines the function f, we determine that f(2) = 8.
g(3) = 2
and
f(g(3)) = f(2) = 8
To evaluate g(f (3) ), we first evaluate the inside expression, f(3), using the table:
f(3) = 4. Then, using the table for function g, we can evaluate g(f(3)).
g(f(3)) = g(4) = 7
When we are given individual functions as graphs, the procedure for evaluating composite functions is similar to the process we used when given functions as tables. We read the input and output values, but this time, from the points on the coordinate plane.
Example 12: Using Figure 5.7.27, evaluate J(g(l)).
Figure 5.7.27: The coordinate plane with two curves J(x) and g(x).
Solution: To evaluate f(g(l)), we start with the inside evaluation. We evaluate g(l) using the graph of g(x), finding the point where x = 1, and noting the output value of the graph at that input value. Here, g(l) = 3. (See Figure 5.7.28.)
Figure 5.7.28: The coordinate plane with the curve
g(x). The point (1, 3) is indicated.
We use this function value as the input for function f : J(g(l)) = f(3). We evaluate f(3) using the graph of f(x), finding the point where x = 3, and noting the output value of the graph at that input value. Here, f(3) = 6. (See Figure 5.7.29.)
Figure 5.7.29: The coordinate plane with the curve
f(x). The point (3, 6) is indicated.
Thus, f(g(l)) = 6.
Example 13: Suppose f(x) gives miles that can be driven in x hours and g(y) gives the gallons of gas used in driving y miles. Which of these expressions is meaningful: f (g(y)) or g(J (x))?
Solution: The function y = f(x) is a function whose output is the number of miles driven corresponding to the number of hours driven. This means:
number of miles = f (number of hours)
The function g(y) is a function whose output is the number of gallons used correspond ing to the number of miles driven. This means:
number of gallons = g( number of miles )
f (g(y)) : The expression g(y) takes miles as the input and gives a number of gallons as the output. The function f(x) requires a number of hours as the input, so trying to input a number of gallons does not make sense. Therefore, the expression f(g(y)) is meaningless.
g(J(x)) : The expression f(x) takes hours as the input and gives a number of miles driven as the output. The function g(y) requires a number of miles as the input. Using f(x) (miles driven) as an input value for g(y), where gallons of gas depends on miles driven, does make sense. Therefore, the expression g(J(x)) makes sense, and will yield the number of gallons of gas used, g, driving a certain number of miles, f (x), in x hours.
Try It 10
In calculus, it is sometimes necessary to decompose a complicated function. In other words, we need to write the function as a composition of two simpler functions. There may be more than one way to decompose a composite function, so we may choose the decomposition that appears to be most expedient.
Example 14: Write f(x) = J5 – x2 as the composition of two functions.
Solution: We are looking for two functions, g(x) and h(x), such that f(x) = g(h(x)). To do this, we look for a function inside of another function in the formula for f (x). As one possibility, we might notice that the expression 5 – x2 is inside of the square root. We could then decompose the function as
h(x) = 5 – x2 and g(x) = Jx
We can check our answer by recomposing the functions.
Try It Answers
1. (a)
(b)
(c)
2. (a)
(b)
(c)
3. (a)
(b)
(c)
4. (a) J(x) = VX
(b) • Horizontal shift right 2 units.
• Vertically stretch by a factor of 3, and then reflect across the x-axis.
• Vertical shift down 1 unit.
5. g(x) = R –
6. (a) -8
(b) -v13
(c) 2x – 1 + x +3, where xis in [-3, oo)
(cl) (2x- l)( x+3), where xis in [-3,oo)
7. (a) 113
(b) 4×2 + 12x
8. (a) 2
(b) 3
(c) -31
(cl) x4 – 18×2 + 72
(c) 8
(cl) 0
9. 3
10. The number of calories burned by doing sit-ups for 3 minutes.
11. One decomposition is f(x) = g(h(x)) where h(x) = 3 – 4 +x2 and g(x) = –
Exercises
Basic Skills Practice
For Exercises 1 – 4, write the equation for the parent function described.
1. Linear Function
2. Absolute Value Function
3. Square Root Function
4. Exponential Decay Function Vertical and Horizontal Shifts
For Exercises 5 – 8, given the graph of f(x) below, sketch a graph of each of the following.
5. f(x) + 3
6. f(x) – 2
7. J(x+3)
8. f(x – 2)
For Exercises 9 – 12, state the appropriate parent function, f (x), and the list of transforma tions (in the correct order) needed to graph the function.
9. g(x)=ijx-6
10. j(x) = (x – 7)3
11. h( X) = ex + 1
12. k(x) = Ix+ 41
For Exercises 13 – 16, write the equation of the function, g(x), whose graph is the result of
f(x) undergoing the given transformation.
13. f(x) = x2 shifted down 8 units.
14. f(x) = 3x shifted right 5 units.
15. J(x) = lxl shifted up 20 units.
16. f(x) = x3 shifted left 8 units. Reflections
For Exercises 17 – 18, given the graph of f (x) below, sketch a graph of each of the following.
17. f(-x) 18. -f(x)
For Exercises 19 – 20, state the appropriate parent function, f (x), and the list of transfor mations (in the correct order) needed to graph the function.
19. g(x) = -x2 20. h(x) =
For Exercises 21 – 22, write the equation of the function, g(x), whose graph is the result of
f (x) undergoing the given transformation.
21. J(x) = ;2 reflected across the x-axis
22. f(x) = 10x reflected across the y-axis Vertical Scalings
For Exercises 23 – 26, given the graph of f(x) below, sketch a graph of each of the following.
23. 2f (x)
24. 0.25j(x)
25. ½J(x)
26. f(x)
For Exercises 27 – 30, state the appropriate parent function, f (x), and the list of transfor mations (in the correct order) needed to graph the function.
27. g(x) = 6lxl
28. j(x) = 3.8×3
29. h(x) = i· 10x
30. k(x) = tx
For Exercises 31 – 32, write the equation of the function, g(x), whose graph is the result of
f(x) undergoing the given transformation.
31. f (x) = ¾ vertically compressed by a factor of 6.
32. J(x) = ex vertically stretched by a factor of 8.
Function Arithmetic
For Exercises 33 – 40, use the given functions to calculate each operation and simplify, if possible.
f(x) = 3x + 9 g(x) = -x2 + 4 h(x) = x3 + x – l j(x) = lxl
33. (J+g)(3) 37. (gh )(0)
34. (g + h)(x) 38. (Jg)(x)
35. (h – j)(-2) 39. (tf)(-1)
36. (j – f)(x) 40. G) (x)
Compositions of Functions
For Exercises 41 – 44, given J(x) = 2x and g(x) = x – 8, compute each of the following.
41. (Jog)(7)
42. (JO 1)(2)
43. (g O f) (3)
44. (g O g)(-5)
For Exercises 45 – 50, given the values in the table, compute each of the following.
| X | f(x) | g(x) |
| -2 | 2 | -1 |
| -1 | 0 | 1 |
| 0 | 1 | -2 |
| 1 | -2 | 2 |
| 2 | -1 | 0 |
45. u O g)(-2)
46. (go f)(l)
47. (f O f)(-1)
48. (fog)(O)
49. (g O !)(2)
50. (gog)(l)
For Exercises 51 – 54, write f(x) as the composition of two functions.
51. f(x) = vx – 9
52. f(x) = e7x+l0
53. f(x) = x!B
54. J(x) = (2x l)2
Intermediate Skills Practice
For Exercises 55 – 58, write the equation and name for the graphed parent function.
55.
56.
57.
58.
For Exercises 59 – 62, given the graph of f(x) below, sketch a graph of each of the following.
59. f(x + 1) – 2
60. 2f(x) – 1
61. f(x – 3) + 1
62. -0.5f(x)
For Exercises 63 – 66, state the appropriate parent function, f (x), and the list of transfor mations (in the correct order) needed to graph the function.
63. g(x) = -2x + 3
64. j(x) = 3ex-5
65. h(x) = x + 4 – 7
66. k(x) = 0.25(x – 1)3
For Exercises 67 – 70, write the equation of the function, g(x), whose graph is the result of
f(x) undergoing the given transformation.
67. f(x) = lxl shifted left 3 units and then shifted clown 8 units
68. f(x) = 5x shifted right 7 units and then reflected over the x-axis
69. f(x) = vertically stretched by a factor of 9 and then shifted up 1 unit
70. f(x) = x3 reflected over the x-axis and then vertically compressed by a factor of 2
For Exercises 71 – 78, use the given functions to calculate each operation and simplify, if possible.
f(x) = e-x g(x) = 3ex h(x) = ij6 – x j(x) = vx+T
71. (J+g)(0) 75. (gh)(0)
72. (g + h)(x) 76. (f g)(x)
73. (h – j)(3) 77. (1) (-2)
74. (j – J)(x) 78. G) (x)
For Exercises 79 – 82, given f (x) = 5×2 + x and g(x) = x + 1, compute each of the following.
79. (Jo g)(x)
80. (Jo J)(x)
81. (go J)(x)
82. (gog)(x)
For Exercises 83 – 88, given the graphs of f(x) and g(x), compute each of the following.
83. u O g)(3)
84. (g O !)(4)
85. (fog)(O)
86. (g O f) (0)
87. (fof)(-2)
88. (gog)(-3)
For Exercises 89 – 92, write f (x) as the composition of two functions.
89. f(x) = 3Js=”x
90. f(x) = ex2 + 100
91. f(x) = 7×2:11
92. f(x) = 1} (x – 2)3 – 11
Mastery Practice
93. Given the graph of f(x) shown below, sketch the graph of – f(x – 1) + 2.
94. State the appropriate parent function, f(x), and the list of transformations (in the correct order) needed to graph the function g(x) = -½ C 5) + 6.
95. State the appropriate parent function, f(x), and the list of transformations (in the correct order) needed to graph the function g(x) = 2 • 10x+4 – 1. 96. Write the equation of the function, g(x), whose graph is the result of f(x) = }2 shifted left 3 units, vertically stretched by a factor of 6, reflected over the x-axis, and then shifted down 2 units.
96. Write the equation of the function, g(x), whose graph is the result of f(x) = lxl, shifted right 2 units, vertically compressed by a factor of 12, and then shifted up 1 unit.
For Exercises 98 – 109, use the given functions to calculate each operation and simplify, if
possible.
f(x) = 7 – 4x g(x) = -3×2 + 2x h(x) = 5ex-2 j(x) = - 1
x + 6
| 98. (j + f)(-1) | 104. | (f +g)(x) | |
| 99. (h – g)(2) | 105. | (j – h)(x) | |
| 100. | (f j)(19) | 106. | (f h)(x) |
| 101. | (*) (0) | 107. | G) (x) |
| 102. | f(h(3)) | 108. | (go f)(x) |
| 103. | (j o j)(0) | 109. | g(g(x)) |
- Given J(x) = x2 + 3 and g(x) is given by the graph below, compute g(f(0)) and
(f o g)(0).
- Write J(x) = 1×7:t! as the composition of two functions.
|
- The function A(d) gives the pain level on a scale of 0 to 10 experienced by a patient with d milligrams of a pain-reducing drug in their The milligrams of the drug in the patient’s system after t minutes is modeled by m(t). Which of the following would determine when the patient will be at a pain level of 4?
- Evaluate A(m(4)).
- Evaluate m(A(4)).
- Evaluate A(rn(t)) = 4.
- Evaluate m(A(d)) = 4.
Communication Practice
- Explain when the order matters in graphed transformations.
- Explain the difference between (f · g)(x) and (Jo g)(x), mathematically.