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Chapter 0 Functions

Chapter 0 Functions. Chapter Outline. Functions and Their Graphs Some Important Functions The Algebra of Functions Zeros of Functions – The Quadratic Formula and Factoring Exponents and Power Functions Functions and Graphs in Applications. § 0.1. Functions and Their Graphs.

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Chapter 0 Functions

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  1. Chapter 0Functions

  2. Chapter Outline • Functions and Their Graphs • Some Important Functions • The Algebra of Functions • Zeros of Functions – The Quadratic Formula and Factoring • Exponents and Power Functions • Functions and Graphs in Applications

  3. § 0.1 Functions and Their Graphs

  4. Section Outline • Rational and Irrational Numbers • The Number Line • Open and Closed Intervals • Applications of Functions • Domain of a Function • Graphs of Functions • The Vertical Line Test • Graphing Calculators • Graphs of Equations

  5. Rational & Irrational Numbers

  6. The Number Line -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

  7. Open & Closed Intervals -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

  8. Functions in Application EXAMPLE (Response to a Muscle) When a solution of acetylcholine is introduced into the heart muscle of a frog, it diminishes the force with which the muscle contracts. The data from experiments of the biologist A. J. Clark are closely approximated by a function of the form where x is the concentration of acetylcholine (in appropriate units), b is a positive constant that depends on the particular frog, and R(x) is the response of the muscle to the acetylcholine, expressed as a percentage of the maximum possible effect of the drug. (a) Suppose that b = 20. Find the response of the muscle when x = 60. (b) Determine the value of b if R(50) = 60 – that is, if a concentration of x = 50 units produces a 60% response. SOLUTION (a) This is the given function.

  9. Functions in Application CONTINUED Replace b with 20 and x with 60. Simplify the numerator and denominator. Divide. Therefore, when b = 20 and x = 60, R(x) = 75%. (b) This is the given function. Replace x with 50. Replace R(50) with 60.

  10. Functions in Application CONTINUED Simplify the numerator. Multiply both sides by b + 50 and cancel. Distribute on the left side. Subtract 3000 from both sides. Divide both sides by 60. Therefore, when R(50) = 60, b = 33.3.

  11. Functions EXAMPLE If , find f (a - 2). SOLUTION This is the given function. Replace each occurrence of x with a – 2. Evaluate (a – 2)2 = a2 – 4a + 4. Remove parentheses and distribute. Combine like terms.

  12. Domain

  13. Graphs of Functions

  14. The Vertical Line Test

  15. Graphing Calculators

  16. Graphs of Equations EXAMPLE Is the point (3, 12) on the graph of the function ? SOLUTION This is the given function. Replace x with 3. Replace f (3) with 12. Simplify. false Multiply. Since replacing x with 3 and f(x) with 12 did not yield a true statement in the original function, we conclude that the point (3, 12) is not on the graph of the function.

  17. § 0.2 Some Important Functions

  18. Section Outline • Linear Equations • Applications of Linear Functions • Piece-Wise Functions • Quadratic Functions • Polynomial Functions • Rational Functions • Power Functions • Absolute Value Function

  19. Linear Equations

  20. Linear Equations CONTINUED

  21. Applications of Linear Functions EXAMPLE • (Enzyme Kinetics) In biochemistry, such as in the study of enzyme kinetics, one encounters a linear function of the form , where K and V are constants. • If f (x) = 0.2x + 50, find K and V so that f (x) may be written in the form, . • Find the x-intercept and y-intercept of the line in terms of K and V. SOLUTION (a) Since the number 50 in the equation f (x) = 0.2x + 50 is in place of the term 1/V (from the original function), we know the following. 50 = 1/V Explained above. 50V = 1 Multiply both sides by V. Divide both sides by 50. V = 0.02 Now that we know what V is, we can determine K. Since the number 0.2 in the equation f (x) = 0.2x + 50 is in place of K/V (from the original function), we know the following.

  22. Applications of Linear Functions CONTINUED 0.2 = K/V Explained above. 0.2V = K Multiply both sides by V. Replace V with 0.02. 0.2(0.02) = K 0.004 = K Multiply. Therefore, in the equation f (x) = 0.2x + 50, K = 0.004 and V = 0.02. (b) To find the x-intercept of the original function, replace f (x) with 0. This is the original function. Replace f (x) with 0. Solve for x by first subtracting 1/V from both sides.

  23. Applications of Linear Functions CONTINUED Multiply both sides by V/K. Simplify. Therefore, the x-intercept is -1/K. To find the y-intercept of the original function, we recognize that this equation is in the form y = mx + b. Therefore we know that 1/V is the y-intercept.

  24. Piece-Wise Functions EXAMPLE Sketch the graph of the following function . SOLUTION We graph the function f (x) = 1 + x only for those values of x that are less than or equal to 3. Notice that for all values of x greater than 3, there is no line.

  25. Piece-Wise Functions CONTINUED Now we graph the function f (x) = 4 only for those values of x that are greater than 3. Notice that for all values of x less than or equal to 3, there is no line.

  26. Piece-Wise Functions CONTINUED Now we graph both functions on the same set of axes.

  27. Quadratic Functions

  28. Polynomial Functions

  29. Rational Functions

  30. Power Functions

  31. Absolute Value Function

  32. § 0.3 The Algebra of Functions

  33. Section Outline • Adding Functions • Subtracting Functions • Multiplying Functions • Dividing Functions • Composition of Functions

  34. Adding Functions EXAMPLE Given and , express f (x) + g(x) as a rational function. SOLUTION f (x) + g(x) = Replace f (x) and g (x) with the given functions. Multiply to get common denominators. Evaluate. Add. Simplify the numerator.

  35. Adding Functions CONTINUED Evaluate the denominator. Simplify the denominator.

  36. Subtracting Functions EXAMPLE Given and , express f (x) - g(x) as a rational function. SOLUTION f (x) - g(x) = Replace f (x) and g (x) with the given functions. Multiply to get common denominators. Evaluate. Subtract. Simplify the numerator.

  37. Subtracting Functions CONTINUED Evaluate the denominator. Simplify the denominator.

  38. Multiplying Functions EXAMPLE Given and , express f (x)g(x) as a rational function. SOLUTION f (x)g(x) = Replace f (x) and g (x) with the given functions. Multiply the numerators and denominators. Evaluate.

  39. Dividing Functions EXAMPLE Given and , express [f (x)]/[g(x)] as a rational function. SOLUTION f (x)/g(x) = Replace f (x) and g (x) with the given functions. Rewrite as a product (multiply by reciprocal of denominator). Multiply the numerators and denominators. Evaluate.

  40. Composition of Functions EXAMPLE (Conversion Scales) Table 1 shows a conversion table for men’s hat sizes for three countries. The function converts from British sizes to French sizes, and the function converts from French sizes to U.S. sizes. Determine the function h(x) = f (g(x)) and give its interpretation. SOLUTION h(x) = f (g(x)) This is what we will determine. In the function f, replace each occurrence of x with g(x). Replace g(x) with 8x + 1.

  41. Composition of Functions CONTINUED Distribute. Multiply. Therefore, h(x) = f (g(x)) = x + 1/8. Now to determine what this function h(x) means, we must recognize that if we plug a number into the function, we may first evaluate that number plugged into the function g(x). Upon evaluating this, we move on and evaluate that result in the function f (x). This is illustrated as follows. g(x) f (x) British French French U.S. h(x) Therefore, the function h(x) converts a men’s British hat size to a men’s U.S. hat size.

  42. § 0.4 Zeros of Functions – The Quadratic Formula and Factoring

  43. Section Outline • Zeros of Functions • Quadratic Formula • Graphs of Intersecting Lines • Factoring

  44. Zeros of Functions

  45. Quadratic Formula

  46. Graphs of Intersecting Functions EXAMPLE Find the points of intersection of the pair of curves. SOLUTION The graphs of the two equations can be seen to intersect in the following graph. We can use this graph to help us to know whether our final answer is correct.

  47. Graphs of Intersecting Functions CONTINUED To determine the intersection points, set the equations equal to each other, since they both equal the same thing: y. Now we solve the equation for x using the quadratic formula. This is the equation to solve. Subtract x from both sides. Add 9 to both sides. We now recognize that, for the quadratic formula, a = 1, b = -11, and c =18. Use the quadratic formula. Simplify.

  48. Graphs of Intersecting Functions CONTINUED Simplify. Simplify. Rewrite. Simplify. We now find the corresponding y-coordinates for x = 9 and x = 2. We can use either of the original equations. Let’s use y = x – 9.

  49. Graphs of Intersecting Functions CONTINUED Therefore the solutions are (9, 0) and (2, -7). This seems consistent with the two intersection points on the graph. A zoomed in version of the graph follows.

  50. Factoring EXAMPLE Factor the following quadratic polynomial. SOLUTION This is the given polynomial. Factor 2x out of each term. Rewrite 3 as Now I can use the factorization pattern: a2 – b2 = (a – b)(a + b). Rewrite

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