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College Algebra Sixth Edition James Stewart  Lothar Redlin  Saleem Watson

College Algebra Sixth Edition James Stewart  Lothar Redlin  Saleem Watson. Functions. 2. Combining Functions. 2.6. Combining Functions. In this section, we study: Different ways to combine functions to make new functions. Sums, Differences, Products, and Quotients.

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College Algebra Sixth Edition James Stewart  Lothar Redlin  Saleem Watson

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  1. College Algebra Sixth Edition James StewartLothar RedlinSaleem Watson

  2. Functions 2

  3. Combining Functions 2.6

  4. Combining Functions • In this section, we study: • Different ways to combine functions to make new functions.

  5. Sums, Differences, Products, and Quotients

  6. Sums, Differences, Products, and Quotients • Two functions f and g can be combined to form new functions f + g, f – g, fg, f/gin a manner similar to the way we add, subtract, multiply, and divide real numbers.

  7. Sum of Functions • For example, we define the function f + gby: (f + g)(x) = f(x) + g(x) • The new function f + g is called the sum of the functions f and g. • Itsvalue at x is f(x) + g(x).

  8. Sum of Functions • Of course, the sum on the right-hand side makes sense only if both f(x) and g(x) are defined—that is, if x belongs to the domain of f and also to the domain of g. • So, if the domain of f is A and that of g is B, then the domain of f + g is the intersection of these domains. • That is, AB.

  9. Differences, Products, and Quotients • Similarly, we can define the difference f – g, the product fg, and the quotient f/gof the functions f and g. • Their domains are AB. • However, in the case of the quotient, we must remember not to divide by 0.

  10. Algebra of Functions • Let f and g be functions with domains A and B. • Then, the functions f + g, f – g, fg, and f/g are defined as follows.

  11. E.g. 1—Combinations of Functions & Their Domains • Let and . • Find the functions f + g, f – g, fg, and f/gand their domains. • Find (f + g)(4), (f – g)(4), (fg)(4), and (f/g)(4).

  12. Example (a) E.g. 1—Functions & Their Domains • The domain of f is {x | x≠ 2} and the domain of g is {x | x≥ 0}. • The intersection of the domains of f and gis: {x | x≥ 0 and x ≠ 2} = [0, 2)  (2, ∞)

  13. Example (a) E.g. 1—Functions & Their Domains • Thus, we have:

  14. Example (a) E.g. 1—Functions & Their Domains • Also, • Note that, in the domain of f/g, we exclude 0 because g(0) = 0.

  15. Example (b) E.g. 1—Functions & Their Domains • Each of these values exist because x = 4 is in the domain of each function.

  16. Example (b) E.g. 1—Functions & Their Domains

  17. Graphical Addition • The graph of the function f + g can be obtained from the graphs of f and gby graphical addition. • This means that we add corresponding y-coordinates—as illustrated in the next example.

  18. E.g. 2—Using Graphical Addition • The graphs of f and g are shown. • Use graphical addition to graph the function f + g.

  19. E.g. 2—Using Graphical Addition • We obtain the graph of f + g by “graphically adding” the value of f(x) to g(x) as shown here. • This is implemented by copying the line segment PQ on top of PR to obtain the point S on the graph of f + g.

  20. Composition of Functions

  21. Composition of Functions • Now, let’s consider a very important way of combining two functions to get a new function. • Suppose f(x) = and g(x) = x2 + 1. • We may define a function h as:

  22. Composition of Functions • The function h is made up of the functions f and g in an interesting way: • Given a number x, we first apply to it the function g, then apply f to the result.

  23. Composition of Functions • In this case, • f is the rule “take the square root.” • g is the rule “square, then add 1.” • h is the rule “square, then add 1, then take the square root.” • In other words, we get the rule h by applying the rule g and then the rule f.

  24. Composition of Functions • The figure shows a machine diagram for h.

  25. Composition of Functions • In general, given any two functions f and g, we start with a number x in the domain of g and find its image g(x). • If this number g(x) is in the domain of f, we can then calculate the value of f(g(x)).

  26. Composition of Functions • The result is a new function h(x) = f(g(x)) the is obtained by substituting g into f. • It is called the composition(or composite) of f and g and is denoted by f◦g (“f composed with g”).

  27. Composition of Functions—Definition • Given two functions f and g, the composite function f◦g (also called the composition of f and g) is defined by: (f◦g)(x) = f(g(x))

  28. Composition of Functions • The domain of f◦g is the set of all x in the domain of g such that g(x) is in the domain of f. • In other words, (f◦g)(x) is defined whenever both g(x) and f(g(x)) are defined.

  29. Composition of Functions • We can picture f◦g using an arrow diagram.

  30. E.g. 3—Finding the Composition of Functions • Let f(x) = x2 and g(x) = x – 3. • Find the functions f◦g and g◦fand their domains. • Find (f◦g)(5) and (g◦f )(7).

  31. Example (a) E.g. 3—Finding the Composition • We have:and • The domains of both f◦g and g◦f are .

  32. Example (b) E.g. 3—Finding the Composition • We have: (f◦g)(5) = f(g(5)) = f(2) = 22 = 4 (g◦ f )(7) = g(f(7)) = g(49) = 49 – 3 = 46

  33. Composition of Functions • You can see from Example 3 that, in general, f◦g≠g◦f. • Remember that the notation f◦g means that the function g is applied first and then f is applied second.

  34. E.g. 4—Finding the Composition of Functions • If f(x) = and g(x) = , find the following functions and their domains. • f◦ g • g◦ f • f◦ f • g◦ g

  35. Example (a) E.g. 4—Finding the Composition • The domain of f◦g is: {x | 2 – x≥ 0} = {x | x ≤ 2} = (–∞, 2]

  36. Example (b) E.g. 4—Finding the Composition • For to be defined, we must have x ≥ 0. • For to be defined, we must have 2 – ≥ 0, that is ≤ 2, or x ≤ 4.

  37. Example (b) E.g. 4—Finding the Composition • Thus, we have: 0 ≤x ≤ 4 • So, the domain of g◦f is the closed interval [0, 4].

  38. Example (c) E.g. 4—Finding the Composition • The domain of f◦f is [0, ∞).

  39. Example (d) E.g. 4—Finding the Composition • This expression is defined when both 2 – x ≥ 0 and 2 – ≥ 0. • The first inequality means x ≤ 2, and the second is equivalent to ≤ 2, or 2 – x ≤ 4, or x ≥ –2.

  40. Example (d) E.g. 4—Finding the Composition • Thus, –2 ≤x ≤ 2 • So, the domain of g◦g is [–2, 2].

  41. Finding the Composition of Functions • The graphs of f and gof Example 4, as well as f◦g, g◦f, f◦f, and g◦g, are shown.

  42. Finding the Composition of Functions • These graphs indicate that the operation of composition can produce functions quite different from the original functions.

  43. A Composition of Three Functions • It is possible to take the composition of three or more functions. • For instance, the composite function f◦g◦h is found by first applying h, then g, and then f as follows: (f◦ g ◦ h)(x) = f(g(h(x)))

  44. E.g. 5—A Composition of Three Functions • Find f◦ g ◦ h if: • f(x) = x/(x + 1) • g(x) = x10 • h(x) = x + 3

  45. E.g. 5—A Composition of Three Functions

  46. Decomposition • So far, we have used composition to build complicated functions from simpler ones. • However, in calculus, it is useful to be able to “decompose” a complicated function into simpler ones—as shown in the following example.

  47. E.g. 6—Recognizing a Composition of Functions • Given F(x) = , find functions f and g such that F =f◦g. • Since the formula for F says to first add 9 and then take the fourth root, we let:g(x) = x + 9 and f(x) =

  48. E.g. 6—Recognizing a Composition of Functions • Then,

  49. E.g. 7—An Application of Composition of Functions • A ship is traveling at 20 mi/h parallel to a straight shoreline. • It is 5 mi from shore. • It passes a lighthouse at noon.

  50. E.g. 7—An Application of Composition of Functions • (a) Express the distance s between the lighthouse and the ship as a function of d, the distance the ship has traveled since noon. • That is, find f so that s = f(d).

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