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Chapter 0 Functions © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 1 of 78 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 © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 2 of 78 § 0.1 Functions and Their Graphs © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 3 of 78 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 © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 4 of 78 Rational & Irrational Numbers Definition Rational Number: A number that may be written as a finite or infinite repeating decimal, in other words, a number that can be written in the form m/n such that m, n are integers Irrational Number: A number that has an infinite decimal representation whose digits form no repeating pattern © 2010 Pearson Education Inc. Example 2 0.285714 7 3 1.73205 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 5 of 78 The Number Line The Number Line A geometric representation of the real numbers is shown below. -6 © 2010 Pearson Education Inc. -5 -4 -3 -2 -1 2 7 0 3 1 2 3 4 5 6 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 6 of 78 Open & Closed Intervals Definition Open Interval: The set of numbers that lie between two given endpoints, not including the endpoints themselves Closed Interval: The set of numbers that lie between two given endpoints, including the endpoints themselves © 2010 Pearson Education Inc. Example -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 2 3 4 5 6 4, x4 -6 -5 -4 -3 -2 -1 0 1 [-1, 4] 1 x 4 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 7 of 78 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 Rx 100 x b x 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) Rx 100 x b x © 2010 Pearson Education Inc. This is the given function. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 8 of 78 Functions in Application CONTINUED R60 100 60 20 60 Replace b with 20 and x with 60. R60 6000 80 Simplify the numerator and denominator. R60 75 Divide. Therefore, when b = 20 and x = 60, R (x) = 75%. (b) Rx 100 x b x This is the given function. R50 100 50 b 50 Replace x with 50. 60 100 50 b 50 Replace R(50) with 60. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 9 of 78 Functions in Application CONTINUED 60 b 50 60 5000 b 50 Simplify the numerator. 5000 b 50 b 50 Multiply both sides by b + 50 and cancel. 60b 3000 5000 Distribute on the left side. 60b 2000 Subtract 3000 from both sides. b 33.3 Divide both sides by 60. Therefore, when R (50) = 60, b = 33.3. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 10 of 78 Functions EXAMPLE If f x x 2 4 x 3 , find f (a - 2). SOLUTION f x x 2 4 x 3 This is the given function. f a 2 a 2 4a 2 3 2 Replace each occurrence of x with a – 2. f a 2 a 2 4a 4 4a 2 3 Evaluate (a – 2)2 = a2 – 4a + 4. f a 2 a 2 4a 4 4a 8 3 Remove parentheses and distribute. f a 2 a 2 1 Combine like terms. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 11 of 78 Domain Definition Domain of a Function: The set of acceptable values for the variable x. Example The domain of the function f x is © 2010 Pearson Education Inc. 1 3 x 3 x 0 3 x Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 12 of 78 Graphs of Functions Definition Example Graph of a Function: The set of all points (x, f (x)) where x is the domain of f (x). Generally, this forms a curve in the xyplane. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 13 of 78 The Vertical Line Test Definition Example Vertical Line Test: A curve in the xy-plane is the graph of a function if and only if each vertical line cuts or touches the curve at no more than one point. Although the red line intersects the graph no more than once (not at all in this case), there does exist a line (the yellow line) that intersects the graph more than once. Therefore, this is not the graph of a function. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 14 of 78 Graphing Calculators Graphing Using a Graphing Calculator Step Display 1) Enter the expression for the function. 2) Enter the specifications for the viewing window. 3) Display the graph. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 15 of 78 Graphs of Equations EXAMPLE 1 2 Is the point (3, 12) on the graph of the function f x x x 2 ? SOLUTION 1 f x x x 2 2 This is the given function. 1 f 3 3 3 2 2 Replace x with 3. 1 12 3 3 2 2 Replace f (3) with 12. 12 2.55 Simplify. 12 12.5 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. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 16 of 78 § 0.2 Some Important Functions © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 17 of 78 Section Outline Linear Equations Applications of Linear Functions Piece-Wise Functions Quadratic Functions Polynomial Functions Rational Functions Power Functions Absolute Value Function © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 18 of 78 Linear Equations Equation Example y = mx + b (This is a linear function) x=a (This is not the graph of a function) © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 19 of 78 Linear Equations CONTINUED Equation Example y=b © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 20 of 78 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 f x K / V x 1/ V , where K and V are constants. (a) If f (x) = 0.2x + 50, find K and V so that f (x) may be written in the form, f x K / V x 1/ V. (b) Find the x-intercept and y-intercept of the line f x K / V x 1/ V 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 50V = 1 V = 0.02 Explained above. Multiply both sides by V. Divide both sides by 50. 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. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 21 of 78 Applications of Linear Functions CONTINUED 0.2 = K/V 0.2V = K 0.2(0.02) = K 0.004 = K Explained above. Multiply both sides by V. Replace V with 0.02. 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. f x K / V x 1/ V 0 K / V x 1/ V 1/ V K / V x © 2010 Pearson Education Inc. This is the original function. Replace f (x) with 0. Solve for x by first subtracting 1/V from both sides. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 22 of 78 Applications of Linear Functions CONTINUED 1/ V V / K x 1 / K x 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. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 23 of 78 Piece-Wise Functions EXAMPLE 1 x for x 3 . 2 for x 3 Sketch the graph of the following function f x SOLUTION We graph the function f (x) = 1 + x only for those values of x that are less than or equal to 3. 6 4 2 0 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 -2 -4 -6 Notice that for all values of x greater than 3, there is no line. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 24 of 78 Piece-Wise Functions CONTINUED Now we graph the function f (x) = 4 only for those values of x that are greater than 3. 6 5 4 3 2 1 0 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 Notice that for all values of x less than or equal to 3, there is no line. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 25 of 78 Piece-Wise Functions CONTINUED Now we graph both functions on the same set of axes. -6 © 2010 Pearson Education Inc. -5 -4 -3 -2 6 5 4 3 2 1 0 -1 -1 0 -2 -3 -4 -5 -6 1 2 3 4 5 6 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 26 of 78 Quadratic Functions Definition Example Quadratic Function: A function of the form f x ax 2 bx c where a, b, and c are constants and a 0. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 27 of 78 Polynomial Functions Definition Example Polynomial Function: A function of the form f x an x n an 1 x n 1 a0 f x 17 x3 x 2 5 where n is a nonnegative integer and a0, a1, ...an are given numbers. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 28 of 78 Rational Functions Definition Rational Function: A function expressed as the quotient of two polynomials. © 2010 Pearson Education Inc. Example 3x x 4 g x 2 5x x 1 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 29 of 78 Power Functions Definition Example Power Function: A function of the form f x x 5.2 f x x r . © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 30 of 78 Absolute Value Function Definition Example Absolute Value Function: The function defined for all numbers x by f x x f x x , such that |x| is understood to be x if x is positive and –x if x is negative © 2010 Pearson Education Inc. f 1 2 1 2 1 2 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 31 of 78 § 0.3 The Algebra of Functions © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 32 of 78 Section Outline Adding Functions Subtracting Functions Multiplying Functions Dividing Functions Composition of Functions © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 33 of 78 Adding Functions EXAMPLE Given f x 2 1 and g x , express f (x) + g (x) as a rational function. x3 x2 SOLUTION f (x) + g (x) = 2 1 x 3 x 2 Replace f (x) and g (x) with the given functions. x2 2 1 x3 x 2 x 3 x 2 x 3 Multiply to get common denominators. 2x 4 x 3 x 2x 3 x 2x 3 Evaluate. 2x 4 x 3 x 2x 3 Add. 3x 1 x 2x 3 Simplify the numerator. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 34 of 78 Adding Functions CONTINUED 3x 1 x 2 3x 2 x 6 Evaluate the denominator. 3x 1 x x6 Simplify the denominator. 2 © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 35 of 78 Subtracting Functions EXAMPLE Given f x 2 1 and g x , express f (x) - g (x) as a rational function. x3 x2 SOLUTION f (x) - g (x) = 2 1 x 3 x 2 Replace f (x) and g (x) with the given functions. x2 2 1 x 3 x 2 x 3 x 2 x 3 Multiply to get common denominators. 2x 4 x 3 x 2x 3 x 2x 3 Evaluate. 2 x 4 x 3 x 2x 3 Subtract. x7 x 2x 3 Simplify the numerator. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 36 of 78 Subtracting Functions CONTINUED x7 x 2 3x 2 x 6 Evaluate the denominator. x7 x x6 Simplify the denominator. 2 © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 37 of 78 Multiplying Functions EXAMPLE Given f x 2 1 and g x , express f (x)g (x) as a rational function. x3 x2 SOLUTION f (x)g (x) = 2 1 x 3 x 2 Replace f (x) and g (x) with the given functions. 2 1 x 3x 2 Multiply the numerators and denominators. 2 x x6 Evaluate. 2 © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 38 of 78 Dividing Functions EXAMPLE Given f x 2 1 and g x , express [f (x)]/[g (x)] as a rational function. x3 x2 SOLUTION f (x)/g (x) = 2 x 3 1 x2 Replace f (x) and g (x) with the given functions. 2 x2 x 3 1 Rewrite as a product (multiply by reciprocal of denominator). 2 x 2 x 31 Multiply the numerators and denominators. 2x 4 x3 Evaluate. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 39 of 78 Composition of Functions EXAMPLE (Conversion Scales) Table 1 shows a conversion table for men’s hat sizes for three countries. The function g x 8x 1 converts from British sizes to French sizes, and the function f x 1 x converts from French sizes to U.S. sizes. Determine the function h (x) = f (g (x)) and give 8 its interpretation. SOLUTION h (x) = f (g (x)) This is what we will determine. 1 g x 8 In the function f, replace each occurrence of x with g (x). 1 8 x 1 8 Replace g (x) with 8x + 1. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 40 of 78 Composition of Functions CONTINUED 1 1 8 x 1 8 8 Distribute. 1 8 Multiply. x 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) British f (x) 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. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 41 of 78 § 0.4 Zeros of Functions – The Quadratic Formula and Factoring © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 42 of 78 Section Outline Zeros of Functions Quadratic Formula Graphs of Intersecting Lines Factoring © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 43 of 78 Zeros of Functions Definition Example Zero of a Function: For a function f (x), all values of x such that f (x) = 0. f x x 2 1 0 x2 1 x 1 © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 44 of 78 Quadratic Formula Definition Quadratic Formula: A formula for solving any quadratic equation of the form ax 2 bx c 0 . The solution is: b b 4ac x . 2a 2 There is no solution if b 2 4ac 0. © 2010 Pearson Education Inc. Example x 2 3x 2 0 a 1; b 3; c 2 x x 3 32 41 2 21 3 17 2 These are the solutions/zeros of the quadratic function f x x 2 3x 2. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 45 of 78 Graphs of Intersecting Functions EXAMPLE Find the points of intersection of the pair of curves. y x 2 10 x 9; y x9 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. 100 80 60 40 20 0 -5 0 5 10 15 -20 -40 © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 46 of 78 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. x 2 10 x 9 x 9 Now we solve the equation for x using the quadratic formula. x 2 10 x 9 x 9 This is the equation to solve. x 2 11x 9 9 Subtract x from both sides. x 2 11x 18 0 Add 9 to both sides. We now recognize that, for the quadratic formula, a = 1, b = -11, and c =18. x x 11 112 4118 21 11 121 72 2 © 2010 Pearson Education Inc. Use the quadratic formula. Simplify. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 47 of 78 Graphs of Intersecting Functions CONTINUED x 11 49 2 Simplify. x 11 7 2 Simplify. x 11 7 11 7 , 2 2 Rewrite. x 9, 2 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. x9 x2 y x9 y x9 y 99 y 29 y0 y 7 © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 48 of 78 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. 10 5 0 0 2 4 6 8 10 -5 -10 -15 -20 © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 49 of 78 Factoring EXAMPLE Factor the following quadratic polynomial. 6 x 2 x3 SOLUTION 6 x 2 x3 2x 3 x2 This is the given polynomial. Factor 2x out of each term. 2 2 x 3 x 2 Rewrite 3 as 2 3 . Now I can use the factorization pattern: a2 – b2 = (a – b)(a + b). 2x 3 x 3x © 2010 Pearson Education Inc. 2 Rewrite 3 x 2 as 3x 3x . Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 50 of 78 Factoring EXAMPLE Solve the equation for x. 1 5 6 x x2 SOLUTION 1 5 6 x x2 5 6 x 2 1 2 x 2 x x x2 5 2 6 2 x 2 x x x x 2 5x 6 x 2 5x 6 0 x 1x 6 0 © 2010 Pearson Education Inc. This is the given equation. Multiply everything by the LCD: x2. Distribute. Multiply. Subtract 5x + 6 from both sides. Factor. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 51 of 78 Factoring CONTINUED x 1 0 x 1 x6 0 x6 Set each factor equal to zero. Solve. Since no denominator from the original equation is zero when x = -1 or when x = 6, these are our solutions. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 52 of 78 § 0.5 Exponents and Power Functions © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 53 of 78 Section Outline Exponent Rules Applications of Exponents Compound Interest © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 54 of 78 Exponents Definition Example b n b b b 53 5 5 5 n times 1 n b b © 2010 Pearson Education Inc. n 1 3 5 3 5 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 55 of 78 Exponents Definition m n b b b m n m n 1 b m n © 2010 Pearson Education Inc. b 1 n b Example m 3 4 5 5 m n 1 b n m 5 3 4 3 4 1 5 3 4 5 1 4 3 5 4 3 1 5 4 3 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 56 of 78 Exponents Definition b b b r b © 2010 Pearson Education Inc. s r 1 r b rs Example 1 3 2 3 6 6 6 4 1 2 1 2 3 3 1 4 1 2 3 3 6 61 6 1 1 4 2 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 57 of 78 Exponents Definition Example 7 br r s b bs b r s © 2010 Pearson Education Inc. b rs 7 9 4 5 4 3 1 3 7 4 1 3 3 3 3 7 71 7 5 8 45 4 1 95 8 98 92 9 3 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 58 of 78 Exponents Definition abr a r b r r ar a r b b © 2010 Pearson Education Inc. Example 125 271/ 3 1251/ 3 271/ 3 3 125 3 27 5 3 15 4 10 4 10 4 2 16 4 5 5 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 59 of 78 Applications of Exponents EXAMPLE Use the laws of exponents to simplify the algebraic expression. 27 x 5 2/3 3 x SOLUTION 27 x 5 2/3 3 x 27 2 / 3 x 5 2 / 3 3 1/ 3 x 27 2 / 3 x10 / 3 1/ 3 3 x abr a r b r x 272 / 3 x5 2 / 3 This is the given expression. 2 27 x10 / 3 x1/ 3 © 2010 Pearson Education Inc. 1 n b n b b r s m n b rs b n bm b n m Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 60 of 78 Applications of Exponents CONTINUED 32 x10 / 3 x1 / 3 9 x10 / 3 x1/ 3 3 27 3 32 9 9x br br s s b 9x 9 / 3 Subtract. 9x 3 Divide. 10 / 31/ 3 © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 61 of 78 Compound Interest - Annual Definition Example Compound Interest Formula: A P1 i n If $700 is invested, compounded annually at 8% for 8 years, this will grow to: A = the compound amount 8 (how much money you end A 7001 0.08 up with) 8 A 7001.08 P = the principal amount A 7001.851 invested A 1,295.651 i = the compound interest rate per interest period Therefore, the compound amount n = the number of would be $1,295.65. compounding periods © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 62 of 78 Compound Interest - General © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 63 of 78 Compound Interest - General EXAMPLE (Quarterly Compound) Assume that a $500 investment earns interest compounded quarterly. Express the value of the investment after one year as a polynomial in the annual rate of interest r. SOLUTION r A P 1 m mt r A 5001 4 r A 5001 4 © 2010 Pearson Education Inc. 4 1 Since interest is not being compounded annually, we must use this formula. Replace P with 500, m with 4 (interest is compounded 4 times each year), and t with 1 (interest is being compounded for 1 year). 4 Simplify. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 64 of 78 § 0.6 Functions and Graphs in Application © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 65 of 78 Section Outline Geometric Problems Cost, Revenue, and Profit Surface Area Functions and Graphs © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 66 of 78 Geometric Problems EXAMPLE (Fencing a Rectangular Corral) Consider a rectangular corral with two partitions, as shown below. Assign letters to the outside dimensions of the corral. Write an equation expressing the fact that the corral has a total area of 2500 square feet. Write an expression for the amount of fencing needed to construct the corral (including both partitions). SOLUTION First we will assign letters to represent the dimensions of the corral. y x x x x y © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 67 of 78 Geometric Problems CONTINUED Now we write an equation expressing the fact that the corral has a total area of 2500 square feet. Since the corral is a rectangle with outside dimensions x and y, the area of the corral is represented by: A xy Now we write an expression for the amount of fencing needed to construct the corral (including both partitions). To determine how much fencing will be needed, we add together the lengths of all the sides of the corral (including the partitions). This is represented by: F xxxx y y F 4x 2 y © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 68 of 78 Cost Problems EXAMPLE (Cost of Fencing) Consider the corral of the last example. Suppose the fencing for the boundary of the corral costs $10 per foot and the fencing for the inner partitions costs $8 per foot. Write an expression for the total cost of the fencing. SOLUTION This is the diagram we drew to represent the corral. y x x x x y Since the boundary of the fence is represented by the red part of the diagram, the length of fencing for this portion of the corral is x + x + y + y = 2x + 2y. Therefore the cost of fencing the boundary of the fence is (2x + 2y)(cost of boundary fencing per foot) = (2x + 2y)(10) = 20x + 20y. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 69 of 78 Cost Problems CONTINUED Since the inner partitions of the fence are represented by the blue part of the diagram, the length of fencing for this portion of the corral is x + x = 2x. Therefore the cost of fencing the inner partitions of the fence is (2x)(cost of inner partition fencing per foot) = (2x)(8) = 16x. Therefore, an expression for the total cost of the fencing is: (cost of boundary fencing) + (cost of inner partition fencing) (20x + 20y) + (16x) 36x + 20y © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 70 of 78 Surface Area EXAMPLE Assign letters to the dimensions of the geometric box and then determine an expression representing the surface area of the box. SOLUTION First we assign letters to represent the dimensions of the box. z y x © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 71 of 78 Surface Area CONTINUED z y x Now we determine an expression for the surface area of the box. Note, the box has 5 sides which we will call Left (L), Right (R), Front (F), Back (B), and Bottom (Bo). We will find the area of each side, one at a time, and then add them all up. L: yz R: yz F: xz B: xz Bo: xy Therefore, an expression that represents the surface area of the box is: yz + yz + xz + xz + xy = 2yz +2xz +xy. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 72 of 78 Cost, Revenue, & Profit EXAMPLE (Cost, Revenue, and Profit) An average sale at a small florist shop is $21, so the shop’s weekly revenue function is R(x) = 21x where x is the number of sales in 1 week. The corresponding weekly cost is C(x) = 9x + 800 dollars. (a) What is the florist shop’s weekly profit function? (b) How much profit is made when sales are at 120 per week? (c) If the profit is $1000 for a week, what is the revenue for the week? SOLUTION (a) Since Profit = Revenue – Cost, the profit function, P(x), would be: P(x) = R(x) – C(x) P(x) = 21x – (9x + 800) P(x) = 21x – 9x - 800 P(x) = 12x - 800 © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 73 of 78 Cost, Revenue, & Profit CONTINUED (b) Since x represents the number of sales in one week, to determine how much profit is made when sales are at 120 per week, we will replace x with 120 in the profit function and then evaluate. P(120) = 12(120) - 800 P(120) = 1,440 - 800 P(120) = 640 Therefore, when sales are at 120 per week, profit is $640 for that week. (c) To determine the revenue for the week when the profit is $1000 for that week, we use an equation that contains profit, namely the profit function: P(x) = 12x - 800 Now we replace P(x) with 1000 and solve for x. 1000 = 12x - 800 1800 = 12x 150 = x Therefore x, the number of units sold in a week, is 150 when profit is $1000. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 74 of 78 Cost, Revenue, & Profit CONTINUED Now, to determine the corresponding revenue, we replace x with 150 in the revenue function. R(x) = 21x R(150) = 21(150) R(150) = 3,150 Therefore, when profit is $1000 in a week, the corresponding revenue is $3,150. NOTE: In order to determine the desired revenue value in part (c), we needed to solve for R(x). But in order to do that, we needed to have a value for x to plug into the R(x) function. In order to acquire that value for x, we needed to use the given information – profit is $1000. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 75 of 78 Functions & Graphs EXAMPLE The function f (r) gives the cost (in cents) of constructing a 100-cubic-inch cylinder of radius r inches. The graph of f (r) is given. (a) What is the cost of constructing a cylinder of radius 6 inches? (b) Interpret the fact that the point (3, 162) is on the graph of the function. (c) Interpret the fact that the point (3, 162) is the lowest point on the graph of the function. What does this say in terms of cost versus radius? © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 76 of 78 Functions & Graphs CONTINUED SOLUTION To determine the cost of constructing a cylinder of radius 6 inches, we look on the graph where r = 6. The corresponding y value will be the cost we are seeking. The red arrow is emphasizing the point in which we are interested. The y value of that point is 270. Therefore, the cost of constructing a cylinder of radius 6 inches is 270 cents or $2.70. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 77 of 78 Functions & Graphs CONTINUED (b) The fact that the point (3, 162) is on the graph tells us that the cost to make 100-cubic-inch cylinders with a radius as small as 3 inches is 162 cents or $1.62. (c) The fact that the point (3, 162) is the lowest point on the graph tells us that the least expensive 100-cubic-inch cylinder that can be made is a 3 inch cylinder at a cost of $1.62. Therefore, the 3 inch cylinder is the most cost-effective one that is offered. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 78 of 78