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Chapter 7

Chapter 7. Functions and Graphs. The Algebra of Functions. 7.4. The Sum, Difference, Product, or Quotient of Two Functions Domains and Graphs. The Sum, Difference, Product, or Quotient of Two Functions.

bernard-bray
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Chapter 7

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  1. Chapter 7 Functions and Graphs

  2. The Algebra of Functions 7.4 • The Sum, Difference, Product, or Quotient of Two Functions • Domains and Graphs

  3. The Sum, Difference, Product, or Quotient of Two Functions Suppose that a is in the domain of two functions, f and g. The input a is paired with f(a) by f and with g(a) by g. The outputs can then be added to get f (a) + g(a).

  4. For find the following. a) (f + g)(4) b) (f – g)(x) c) (f /g)(x) d)

  5. For find the following. a) (f + g)(4) b) (f – g)(x) c) (f /g)(x) d) Solution a) Since f (4) = –8 and g(4) = 13, we have ( f + g)(4) = f(4) + g(4) = –8 + 13 = 5.

  6. c) We have, We assume d) Since f(–1) = –3 and g(–1) = –2, we have b) We have,

  7. The Algebra of Functions If f and g are functions and x is in the domain of both functions, then:

  8. Domains and Graphs we must first be able to find f (a) and g(a). This means a must be in the domain of both f and g.

  9. find the domains of

  10. Thus the domain of f + g, f – g, and find the domains of Solution The domain of f is The domain of g is all real numbers.

  11. Thus the domain of f /g is To find the domain of f /g, note that can not be evaluated if x + 1 = 0 or x – 2 = 0.

  12. Formulas, Applications, and Variation 7.5 • Formulas • Direct Variation • Inverse Variation • Joint and Combined Variation

  13. Formulas Formulas occur frequently as mathematical models. Many formulas contain rational expressions, and to solve such formulas for a specified letter, we proceed as when solving rational equations.

  14. In a hydraulic system, a fluid is confined to two connecting chambers. The pressure in each chamber is the same and is given by finding the force exerted (F) divided by the surface area (A). Therefore, we know Solve for A2.

  15. Solution Multiplying both sides by the LCD Simplifying by removing factors Dividing both sides by F1 This formula can be used to calculate A2 whenever A1, F2, and F1 are known.

  16. To Solve a Rational Equation for a Specified Variable 1. Multiply both sides by the LCD to clear fractions, if necessary. 2. Multiply to remove parentheses, if necessary. 3. Get all terms with the specified variable alone on one side. 4. Factor out the specified variable if it is in more than one term. 5. Multiply or divide on both sides to isolate the specified variable.

  17. Variation To extend our study of formulas and functions, we now examine three real-world situations: direct variation, inverse variation, and combined variation.

  18. Direct Variation A mass transit driver earns $17 per hour. In 1 hr, $17 is earned. In 2 hr, $34 is earned. In 3 hr, $51 is earned, and so on. This gives rise to a set of ordered pairs: (1, 17), (2, 34), (3, 51), (4, 68), and so on. Note that the ratio of earnings E to time t is 17/1 in every case. If a situation gives rise to pairs of numbers in which the ratio is constant, we say that there is direct variation. Here earnings varydirectly as the time: We have E/t = 17, so E = 17t, or using function notation, E(t) = 17t.

  19. Direct Variation When a situation is modeled by a linear function of the form f(x) = kx, or y = kx, where k is a nonzero constant, we say that there is direct variation, that y varies directly as x, or that y is proportional to x. The number k is called the variation constant, or constant of proportionality.

  20. Find the variation constant and an equation of variation if y varies directly as x, and y = 15 when x = 3.

  21. Find the variation constant and an equation of variation if y varies directly as x, and y = 15 when x = 3. Solution We know that (3, 15) is a solution of y = kx. Therefore, Substituting Solving for k The variation constant is 5. The equation of variation is y = 5x. The notation y(x) = 5x or f(x) = 5x is also used.

  22. Inverse Variation To see what we mean by inverse variation, suppose it takes one person 8 hours to paint the baseball fields for the local park district. If two people do the job, it will take only 4 hours. If three people paint the fields, it will take only 2 and 2/3 hours, and so on. This gives rise to pairs of numbers, all have the same product: (1, 8), (2, 4), (3, 8/3), (4, 2), and so on.

  23. Inverse Variation Note that the product of each pair of numbers is 8. Whenever a situation gives rise to pairs of numbers for which the product is constant, we say that there is inverse variation. Since pt = 8, the time t, in hours, required for the fields to be painted by p people is given by t = 8/p or, using function notation, t(p) = 8/p.

  24. Inverse Variation When a situation gives rise to a rational function of the form f(x) = k/x, or y = k/x, where k is a nonzero constant, we say that there is inverse variation, that is yvaries inverselyas x, or that y is inversely proportional to x. The number k is called the variation constant, or constant of proportionality.

  25. The time, t, required to empty a tank varies inversely as the rate, r, of pumping. If a pump can empty a tank in 90 minutes at the rate of 1080 kL/min, how long will it take the pump to empty the same tank at the rate of 1500 kL/min?

  26. The time, t, required to empty a tank varies inversely as the rate, r, of pumping. If a pump can empty a tank in 90 minutes at the rate of 1080 kL/min, how long will it take the pump to empty the same tank at the rate of 1500 kL/min? Solution 1. Familiarize. Because of the phrase “ . . . varies inversely as the rate, r, of pumping,” we express the amount of time needed to empty the tank as a function of the rate: t(r) = k/r

  27. Solution(continued) 2. Translate. We use the given information to solve for k. Then we use that result to write the equation of variation. Using function notation Replacing r with 1080 Replacing t(1080) with 90 Solving for k, the variation constant The equation of variation is t(r) = 97,200/r. This is the translation.

  28. Solution(continued) 3. Carry out. To find out how long it would take to pump out the tank at a rate of 1500 mL/min, we calculate t(1500). t = 64.8 when r = 1500 4. Check. We could now recheck each step. Note that, as expected, as the rate goes up, the time it takes goes down. 5. State. If the pump is emptying the tank at a rate of 1500 mL/min, then it will take 64.8 minutes to empty the entire tank.

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