1. 1 Copyright © Cengage Learning. All rights reserved. 3 Functions and Graphs 3.7 Operations on Functions

2. 2 Operations on Functions Functions are often defined using sums, differences, products, and quotients of various expressions. For example, if h(x) = x 2 + we may regard h(x) as a sum of values of the functions f and g given by f (x) = x 2 and g(x) = We call h the sum of f and g and denote it by f + g. Thus, h(x) = (f + g) (x) = x 2 +

3. 3 Operations on Functions In general, if f and g are any functions, we use the terminology and notation given in the following chart. Sum, Difference, Product, and Quotient of Functions

4. 4 Operations on Functions The domains of f + g, f – g, and fg are the intersection I of the domains of f and g—that is, the numbers that are common to both domains. The domain of f/g is the subset of I consisting of all x in I such that g (x) ≠ 0.

5. 5 Example 1 – Finding function values of f + g, f – g, fg, and f/g If f (x) = 3x – 2 and g (x) = x 3, find (f + g)(2), (f – g)(2), (fg)(2), and (f/g)(2). Solution: Since f(2) = 3(2) – 2 = 4 and g(2) = 2 3 = 8, we have (f + g)(2) = f(2) + g(2) = 4 + 8 = 12 (f – g)(2) = f(2) – g(2) = 4 – 8 = –4 (fg)(2) = f(2)g(2) = (4)(8) = 32

6. 6 Operations on Functions A function f is a polynomial function if f(x) is a polynomial—that is, if f(x) = a n x n + a n –1 x n –1 + …. + a 1 x + a 0, where the coefficients a 0, a 1,..., a n are real numbers and the exponents are nonnegative integers. A polynomial function may be regarded as a sum of functions whose values are of the form cx k, where c is a real number and k is a nonnegative integer. Note that the quadratic functions considered earlier are polynomial functions.

7. 7 Operations on Functions Illustration: Algebraic Function Functions that are not algebraic are transcendental. The exponential and logarithmic functions are examples of transcendental functions.

8. 8 Operations on Functions In the remainder of this section we shall discuss how two functions f and g may be used to obtain the composite functions f  g and g  f (read “f circle g” and “g circle f,” respectively). Functions of this type are very important in calculus. The function f  g is defined as follows.

9. 9 Operations on Functions Figure 1 is a schematic diagram that illustrates relationships among f, g, and f  g. Note that for x in the domain of g, first we find g (x) (which must be in the domain of f ) and then, second, we find f (g (x)). Figure 1

10. 10 Operations on Functions For the composite function g  f, we reverse this order, first finding f (x) and second finding g (f (x)). The domain of g  f is the set of all x in the domain of f such that f (x) is in the domain of g. Since the notation g(x) is read “g of x,” we sometimes say that g is a function of x. For the composite function f  g, the notation f(g(x)) is read “f of g of x,” and we could regard f as a function of g (x). In this sense, a composite function is a function of a function or, more precisely, a function of another function’s values.

11. 11 Example 3 Let f (x) = x 2 – 1 and g (x) = 3x + 5. (a) Find (f  g) (x) and the domain of f  g. (b) Find (g  f ) (x) and the domain of g  f. (c) Find f(g(2)) in two different ways: first using the functions f and g separately and second using the composite function f  g.

12. 12 Example 3(a) – Solution (f  g) (x) = f (g (x)) = f (3x + 5) = (3x + 5) 2 – 1 = 9x 2 + 30x + 24 The domain of both f and g is. Since for each x in (the domain of g), the function value g (x) is in (the domain of f ), the domain of (f  g) is also. Note that both g(x) and f (g(x)) are defined for all real numbers. definition of (f  g) definition of g definition of f Simplify

13. 13 Example 3(b) – Solution (g  f ) (x) = g (f (x)) = g(x 2 – 1 ) = 3(x 2 – 1) + 5 = 3x 2 + 2 Since for each x in (the domain of f), the function value f (x) is in (the domain of g), the domain of g  f is. Note that both f (x) and g (f (x)) are defined for all real numbers. definition of (g  f ) definition of f definition of g Simplify cont’d

14. 14 Example 3(c) – Solution To find f(g(2)) using f (x) = x 2 – 1 and g(x) = 3x + 5 separately, we may proceed as follows: g(2) = 3(2) + 5 = 11 f(g(2)) = f(11) = 11 2 – 1 = 120 To find f(g(2)) using (f  g), we refer to part (a), where we found (f  g) (x) = f (g(x)) = 9x 2 + 30x + 24. Hence, f(g(2)) = 9(2) 2 + 30(2) + 24 = 36 + 60 + 24 = 120. cont’d

15. 15 Operations on Functions Note that in Example 3, f (g(x)) and g (f (x)) are not always the same; that is, f  g ≠ g  f. In some applied problems it is necessary to express a quantity y as a function of time t. The next example illustrates that it is often easier to introduce a third variable x, express x as a function of t (that is, x = g(t)), express y as a function of x (that is, y = f (x)), and finally form the composite function given by y = f (x) = f (g(t)).

16. 16 Example 6 – Using a composite function to find the volume of a balloon A meteorologist is inflating a spherical balloon with helium gas. If the radius of the balloon is changing at a rate of 1.5 cm/sec, express the volume V of the balloon as a function of time t (in seconds). Solution: Let x denote the radius of the balloon. If we assume that the radius is 0 initially, then after t seconds x = 1.5t To illustrate, after 1 second, the radius is 1.5 centimeters; after 2 seconds, it is 3.0 centimeters; after 3 seconds, it is 4.5 centimeters; and so on. radius of balloon after t seconds

17. 17 Example 6 – Solution Next we write This gives us a composite function relationship in which V is a function of x, and x is a function of t. By substitution, we obtain volume of a sphere of radius x cont’d

18. 18 Example 6 – Solution Simplifying, we obtain the following formula for V as a function of t: cont’d

19. 19 Operations on Functions If f and g are functions such that y = f (u)and u = g (x), then substituting for u in y = f(u) yields y = f (g (x)). For certain problems in calculus we reverse this procedure; that is, given y = h (x) for some function h, we find a composite function form y = f (u) and u = g(x) such that h(x) = f(g (x)).

20. 20 Example 7 – Finding a composite function form Express y = (2x + 5) 8 as a composite function form. Solution : Suppose, for a real number x, we wanted to evaluate the expression (2x + 5) 8 by using a calculator. We would first calculate the value of 2x + 5 and then raise the result to the eighth power. This suggests that we let u = 2x + 5 and y = u 8, which is a composite function form for y = (2x + 5) 8.

21. 21 Operations on Functions In general, suppose we are given y = h (x). To choose the inside expression u = g (x) in a composite function form, ask the following question: If a calculator were being used, which part of the expression h (x) would be evaluated first? This often leads to a suitable choice for u = g (x). After choosing u, refer to h (x) to determine y = f (u).

22. 22 Operations on Functions The following illustration contains typical problems. Illustration: Composite Function Forms Function value Choice for u = g(x) Choice for y = f (u)

23. 23 Operations on Functions The composite function form is never unique. For example, consider the first expression in the preceding illustration: y = (x 3 – 5x + 1) 4 If n is any nonzero integer, we could choose u = (x 3 – 5x + 1) n and y = u 4/n. Thus, there are an unlimited number of composite function forms.

24. 24 Example 8 – Graphically analyzing a composite function Let f (x) = x 2 – 16 and g(x) = (a) Find f(g(3)). (b) Sketch y = (f  g) (x), and use the graph to find the domain of f  g.

25. 25 Example 8(a) – Solution We begin by making the assignments Y 1 = and Y 2 = (Y 1 ) 2 – 16 Note that we have substituted Y 1 for x in f (x) and assigned this expression to Y 2. Next we store the value 3 in the memory location for x and then query the value of Y 2. We see that the value of Y 2 at 3 is –13; that is, f(g(3)) = –13. cont’d

26. 26 Example 8(b) – Solution To determine a viewing rectangle for the graph of f  g, we first note that f (x)  –16 for all x and therefore choose Ymin less than –16; say, Ymin = –20. If we want the rectangle to have a vertical dimension of 40, we must choose Ymax = 20. If your screen is in 1 : 1 proportion (horizontal : vertical), then a reasonable choice for [Xmin, Xmax] would be [–10, 30], a horizontal dimension of 40. If your screen is in 3 : 2 proportion, choose [Xmin, Xmax] to be [–10, 50], a horizontal dimension of 60. cont’d

27. 27 Example 8(b) – Solution Selecting Y 2 and then displaying the graph of Y 2 using the viewing rectangle [–10, 50, 5] by [–20, 20, 5] gives us a graph similar to Figure 2. We see that the graph is a half-line with endpoint (0, –16). Thus, the domain of Y 2 is all x  0. Figure 2 [–10, 50, 5] by [–20, 20, 5] cont’d