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Applied Calculus © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 1 of 78 Course Contents Functions Derivatives Applications of Derivative Techniques of Differentiation Logarithmic Functions and Applications The Definite Integrals The Trigonometric Functions © 2010 Pearson Education Inc. … Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 2 of 78 Chapter 0 Functions © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 3 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 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 0.2857142857 14... 7 0.285714 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 Example Infinite Interval: The set of numbers that lie between a given endpoint and the infinity Closed Interval: The set of numbers that lie between two given endpoints, including the endpoints themselves Open Interval: The set of numbers that lie between two given endpoints, not including the endpoints themselves © 2010 Pearson Education Inc. -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 2 3 4 5 6 3 4 5 6 4, x4 -6 -5 -4 -3 -2 -1 0 1 [−1, 4] 1 x 4 -6 -5 -4 -3 -2 -1 0 1 2 (−1, 4) 1 x 4 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 7 of 78 Functions • A function f is a rule that assigns to each value of a real variable x exactly one value of another real variable y. • The variable x is called the independent variable and the variable y is called the dependent variable. • We usually write y = f (x) to express the fact that y is a function of x. Here f (x) is the name of the function. x y = f (x) y EXAMPLES: f x x 2 4 1 g x x 3 h x x 2 1 k t e 2t sin 2 t © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 8 of 78 Functions in Application EXAMPLE 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 R x 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 R = 60% response. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 9 of 78 Functions in Application SOLUTION R60 100 60 75 20 60 Replace b with 20 and x with 60. Therefore, when b = 20 and x = 60, R (x) = 75%. 100 x b x 100 50 R50 b 50 100 50 60 b 50 b 50 60 5000 b 50 b 50 60b 3000 5000 (b) R x This is the given function. Replace x with 50. Replace R(50) with 60. Multiply both sides by b + 50. Distribute on the left side. 60b 2000 Subtract 3000 from both sides. b 33.3 Divide both sides by 60. Therefore, b = 33.3 when R (50) = 60. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 10 of 78 Functions EXAMPLE If f (x) = x2 + 4x + 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 Replace each x with a – 2. f a 2 a2 4a 4 4a 2 3 Evaluate (a – 2)2 = a2 – 4a + 4. f a 2 a2 4a 4 4a 8 3 Remove parentheses and distribute. 2 f a 2 a 2 1 © 2010 Pearson Education Inc. Combine like terms. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 11 of 78 Domain of a Function Definition Domain of a Function: The set of acceptable values for the variable x. Example The domain of the function f x x 2 is x20 x 2 [- 2 ,) © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 12 of 78 Domain of a Function Definition Example Domain of a Function: The The domain of the function 2x set of acceptable values for f x 2 x 4 the variable x. is x2 4 0 x 2 x R \ 2 © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 13 of 78 Domain of a Function Definition Example The domain of the function Domain of a Function: The set of acceptable x2 f x values for the variable x. 3 x is 3 x 0 x 3 ( ,3) © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 14 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 15 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 16 of 78 Graphs of Equations EXAMPLE 1 f x x x 2 ? Is the point (3, 12) on the graph of the function 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 17 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 18 of 78 Linear Equations CONTINUED Equation Example y=b © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 19 of 78 Piece-Wise Functions EXAMPLE 1 x for x 3 Sketch the graph of the following function f x . 2 for x 3 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 20 of 78 Piece-Wise Functions CONTINUED Now we graph the function f (x) = 2 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 21 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 22 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 23 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 24 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 25 of 78 Power Functions Definition Power Function: A function of the form f x x r . © 2010 Pearson Education Inc. Example f x x5.2 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 26 of 78 Absolute Value Function Definition Example Absolute Value Function: The function defined for all f x x numbers x by f x x , f 1 2 1 2 1 2 such that |x| is understood to be x if x is positive and –x if x is negative © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 27 of 78 Adding Functions EXAMPLE Given f x 2 and x3 g x 1 , express f (x) + g(x) as a rational function. x2 SOLUTION f (x) + g(x) = 2 1 x3 x 2 x2 2 1 x3 x 2 x3 x 2 x3 2x 4 x 3 x 2x 3 x 2x 3 2x 4 x 3 x 2x 3 3x 1 x2 x 6 © 2010 Pearson Education Inc. 3x 1 x 2x 3 Replace f (x) and g(x) with the given functions. Multiply to get common denominators. Evaluate. Add and simplify the numerator. Evaluate the denominator. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 28 of 78 Subtracting Functions EXAMPLE f x Given 2 and x3 g x 1 , express f (x) − g(x) as a rational function. x2 SOLUTION f (x) − g(x) = 2 1 x3 x 2 x2 2 1 x3 x 2 x3 x 2 x3 2x 4 x 3 x 2x 3 x 2x 3 2 x 4 x 3 x 2x 3 x7 x7 2 x 2x 3 x x 6 © 2010 Pearson Education Inc. Replace f (x) and g(x) with the given functions. Multiply to get common denominators. Evaluate. Subtract. Simplify the numerator and denominator. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 29 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 x3 x 2 Replace f (x) and g(x) with the given functions. 2 1 x 3x 2 Multiply the numerators and denominators. 2 x2 x 6 © 2010 Pearson Education Inc. Evaluate. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 30 of 78 Dividing Functions EXAMPLE Given f x 2 1 and g x , express x3 x2 f x as a rational function. g x SOLUTION f x g x 2 x3 1 x2 2 x2 x3 1 2 x 2 x 31 2x 4 x3 © 2010 Pearson Education Inc. 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. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 31 of 78 Composition of Functions EXAMPLE Table 1 shows a conversion table for men’s hat sizes for three countries. The function 1 g x 8 x 1 converts from British sizes to French sizes, and the function f x x 8 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. 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 32 of 78 Composition of Functions CONTINUED 1 1 8x 1 8 8 1 x 8 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) British French f (x) 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 33 of 78 Composition of Functions EXAMPLE Given f x 2 1 g x and , find f (g (x)). x3 x2 SOLUTION f (g (x)) = 2 g x 3 2 1 3 x2 2 x2 1 3 x 2 x2 2x 4 3x 5 © 2010 Pearson Education Inc. Replace x by g(x) in the function f (x) Substitute. Multiply the numerators and denominators by x + 2. Simplify. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 34 of 78 Zeros of Functions Definition Zero of a Function: For a function f (x), all values of x such that f (x) = 0. © 2010 Pearson Education Inc. Example f x x 2 1 0 x2 1 x 1 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 35 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 36 of 78 Graphs of Intersecting Functions EXAMPLE Find the points of intersection of the pair of curves. y x2 10 x 9; y x 9 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 37 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. x2 10 x 9 x 9 Now we solve the equation for x using the quadratic formula. x2 10 x 9 x 9 This is the equation to solve. x2 11x 9 9 Subtract x from both sides. x2 11x 18 0 Add 9 to both sides. Here, a = 1, b = −11, and c =18. x 11 x 11 21 11 121 72 2 © 2010 Pearson Education Inc. 2 4118 Use the quadratic formula. Simplify. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 38 of 78 Graphs of Intersecting Functions CONTINUED x 11 49 2 Simplify. 11 7 2 11 7 11 7 x , 2 2 x Simplify. 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 x 9 y 99 y x 9 y 29 y0 y 7 © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 39 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 40 of 78 Factoring EXAMPLE Factor the following quadratic polynomial. 6 x 2 x3 SOLUTION 6 x 2 x3 2 x3 x 2 2 2 x 3 x2 This is the given polynomial. Factor 2x out of each term. 2 Rewrite 3 as 3 . Now I can use the factorization pattern: a2 – b2 = (a – b)(a + b). 2x 3 x © 2010 Pearson Education Inc. 3x Rewrite 2 3 x 2 as 3x 3x . Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 41 of 78 Factoring EXAMPLE Solve the equation for x. 1 SOLUTION 5 6 2 x x 5 6 x 2 1 2 x 2 x x 1 5 2 6 2 x 2 x x x x 2 5x 6 x2 x 2 5x 6 0 x 1x 6 0 x 1 0 x 1 x60 x6 © 2010 Pearson Education Inc. 5 6 x x2 This is the given equation. Multiply everything by the LCD: x2. Distribute. Multiply. Subtract 5x + 6 from both sides. Factor. Set each factor equal to zero. Solve. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 42 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 43 of 78 Exponents Definition m n b b b m n n 1 m Example b n 1 m n m b bn © 2010 Pearson Education Inc. 3 4 5 5 m 1 b n m 3 4 4 5 4 1 5 3 4 3 5 54 1 3 3 1 5 4 3 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 44 of 78 Exponents Definition b b b r s 1 b r r b © 2010 Pearson Education Inc. r s Example 1 3 2 3 6 6 6 1 2 1 2 3 3 3 3 6 61 6 1 1 4 1 4 2 42 1 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 45 of 78 Exponents Definition Example br r s b bs b r s © 2010 Pearson Education Inc. b rs 7 7 4 3 1 3 7 4 1 3 3 3 3 7 71 7 5 8 45 4 1 9 95 8 98 92 9 3 4 5 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 46 of 78 Exponents Definition Example 125 27 1/ 3 ab r a r br r a a r b b © 2010 Pearson Education Inc. r 1251 / 3 271 / 3 3 125 3 27 5 3 15 4 10 10 4 2 16 4 5 5 4 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 47 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 This is the given expression. x 5 x 27 2/ 3 2/ 3 3 27 2/ 3 x 5 2/ 3 1/ 3 x 27 2/ 3 x10 / 3 1/ 3 3 abr a r b r x 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 48 of 78 Applications of Exponents CONTINUED 3 2 x10 / 3 1/ 3 3 27 3 x 10 / 3 9x x1 / 3 32 9 9x br br s s b 9 x9 / 3 Subtract. 9x3 Divide. 10 / 31 / 3 © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 49 of 78 Geometric Problems EXAMPLE 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 50 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 51 of 78 Surface Area EXAMPLE Assign letters to the dimensions of the geometric box and then determine an expression representing the volume and the surface area of the box. SOLUTION First we assign letters to represent the dimensions of the box. z x y Therefore, an expression that represents the volume is: V = xyz. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 52 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: S = yz + yz + xz + xz + xy = 2yz + 2xz + xy. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 53 of 78