Chapter 0 Functions Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 1 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
Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 2 0.1 Functions and Their Graphs Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 3 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 Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 4 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 Example 2 0.285714 7 3 1.73205 Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 5 The Number Line The Number Line A geometric representation of the real numbers is shown below. -6 -5 -4 -3
-2 -1 2 7 3 0 1 2 3 4 5 6 Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 6 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 Example -6 -5 -4 -3 -2 -1 0 1 2 3 4 5
6 1 2 3 4 5 6 4, x4 -6 -5 -4 -3 -2 -1 0 [-1, 4] 1 x 4
Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 7 Functions in Application EXAMPL E (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 100 x R x bx 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. SOLUTIO N (a) 100 x R x bx This is the given function.
Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 8 Functions in Application CONTINU ED 100 60 R 60 20 60 R 60 Replace b with 20 and x with 60. 6000 80 Simplify the numerator and denominator. Divide. R 60 75 Therefore, when b = 20 and x = 60, R (x) = 75%. (b) 100 x R x bx
100 50 R 50 b 50 100 50 60 b 50 This is the given function. Replace x with 50. Replace R(50) with 60. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 9 Functions in Application CONTINU ED 60 5000 b 50 Simplify the numerator. b 50 60 5000 b 50 Multiply both sides by b + 50 and cancel. 60b 3000 5000
Distribute on the left side. b 50 Subtract 3000 from both sides. 60b 2000 Divide both sides by 60. b 33.3 Therefore, when R (50) = 60, b = 33.3. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 10 Functions EXAMPL E If f x x 2 4 x 3 , find f (a - 2). SOLUTIO N f x x 2 4 x 3 This is the given function. f a 2 a 2 4 a 2 3
Replace each occurrence of x with a 2. f a 2 a 2 4a 4 4 a 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. 2 Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 11 Domain Definition Domain of a Function: The set of acceptable values for the variable x. Example The domain of the function f x
is 1 3 x 3 x 0 3x Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 12 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. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 13 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. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 14 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. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 15 Graphs of Equations EXAMPL E Is the point (3, 12) on the graph of the function f x x SOLUTIO N 1 x 2 ? 2 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 2.5 5 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. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 16 0.2 Some Important Functions
Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 17 Section Outline Linear Equations Applications of Linear Functions Piece-Wise Functions Quadratic Functions Polynomial Functions Rational Functions
Power Functions Absolute Value Function Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 18 Linear Equations Equation Example y = mx + b (This is a linear function) x=a (This is not the graph of a function) Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 19 Linear Equations CONTINU ED Equation
Example y=b In this case, f(x) is called a constant function, since it assigns the same number b to every value of x Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 20 Applications of Linear Functions EXAMPL E (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. SOLUTIO N (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 (x) = 0.2x + 50 is in place of K/V (from the original function), we know the following. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 21 f Applications of Linear Functions CONTINU ED 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.
f x K / V x 1 / V 0 K / V x 1 / V 1 / V K / V x 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, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 22 Applications of Linear Functions CONTINU ED 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. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 23
Piece-Wise Functions EXAMPL E 1 x for x 3 . for x 3 2 Sketch the graph of the following function f x SOLUTIO N 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. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 24 Piece-Wise Functions CONTINU ED 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. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc .
Slide 25 Piece-Wise Functions CONTINU ED Now we graph both functions on the same set of axes. -6 -5 -4 -3 -2 6 5 4 3 2 1 0 -1 -1 0 -2 -3 -4 -5 -6 1
2 3 4 Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . 5 6 Slide 26 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. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc .
Slide 27 Polynomial Functions Definition Example Polynomial Function: A function of the form f x an x n an 1 x n 1 a0 where n is a nonnegative integer and a0, a1, ...an are given numbers. f x 17 x 3 x 2 5 Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 28 Rational Functions Definition Rational Function: A function expressed as the quotient of two polynomials. Example
3x x 4 g x 2 5x x 1 Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 29 Power Functions Definition Example Power Function: A function of the form f x x 5.2 f x x r . Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 30 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 f 1 2 1 2 1 2 Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 31 0.3 The Algebra of Functions Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 32 Section Outline Adding Functions
Subtracting Functions Multiplying Functions Dividing Functions Composition of Functions Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 33 Adding Functions EXAMPL E Given f x 2 1 and g x , express f (x) + g (x) as a rational function. x 3 x2
SOLUTIO N f (x) + g (x) = 2 1 x 3 x2 Replace f (x) and g (x) with the given functions. x2 2 1 x 3 x2 x 3 x2 x 3 Multiply to get common denominators. 2x 4 x 3 x 2 x 3 x 2 x 3
Evaluate. 2x 4 x 3 x 2 x 3 Add. 3x 1 x 2 x 3 Simplify the numerator. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 34 Adding Functions CONTINU ED 3x 1 x 2 3x 2 x 6 Evaluate the denominator.
3x 1 x2 x 6 Simplify the denominator. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 35 Subtracting Functions EXAMPL E Given f x 2 1 and g x , express f (x) - g (x) as a rational function. x 3 x2 SOLUTIO N f (x) - g (x) = 2 1 x 3 x2
Replace f (x) and g (x) with the given functions. x2 2 1 x 3 x2 x 3 x2 x 3 Multiply to get common denominators. 2x 4 x 3 x 2 x 3 x 2 x 3 Evaluate. 2 x 4 x 3 x 2 x 3 Subtract.
x 7 x 2 x 3 Simplify the numerator. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 36 Subtracting Functions CONTINU ED x 7 x 2 3x 2 x 6 Evaluate the denominator. x 7 x2 x 6 Simplify the denominator. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 37 Multiplying Functions
EXAMPL E Given f x 2 1 and g x , express f (x)g (x) as a rational function. x 3 x2 SOLUTIO N f (x)g (x) = 2 1 x 3 x2 Replace f (x) and g (x) with the given functions. 2 1 x 3 x 2 Multiply the numerators and denominators.
2 x2 x 6 Evaluate. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 38 Dividing Functions EXAMPL E Given f x 2 1 and g x , express [f (x)]/[g (x)] as a rational function. x 3 x2 SOLUTIO N 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 x 3 Evaluate. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 39
Composition of Functions EXAMPL E (Conversion Scales) Table 1 shows a conversion table for mens hat sizes for three countries. The 1 function g x 8 x 1 converts from British sizes to French sizes, and the function f x x converts from French sizes to U.S. sizes. Determine the function h (x) = f (g (x)) and give 8 its interpretation. SOLUTIO N 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. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 40
Composition of Functions CONTINU ED 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 mens British hat size to a mens U.S. hat size. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 41 0.4 Zeros of Functions The Quadratic Formula and Factoring Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 42 Section Outline Zeros of Functions Quadratic Formula Graphs of Intersecting Lines
Factoring Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 43 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 x 2 1 x 1 Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 44 Quadratic Formula Definition Quadratic Formula: A formula for solving any
quadratic equation of the form ax 2 bx c 0 . The solution is: 2 x b b 4ac . 2a There is no solution if b 2 4ac 0. Example x 2 3x 2 0 a 1; b 3; c 2 2 3 3 41 2 x 21 x 3 17 2 These are the solutions/ zeros of the quadratic function f x x 2 3 x 2.
Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 45 Graphs of Intersecting Functions EXAMPL E Find the points of intersection of the pair of curves. y x 2 10 x 9; y x 9 SOLUTIO N 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 -20 0 5
10 15 -40 Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 46 Graphs of Intersecting Functions CONTINU ED 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 11 x 9 9 Subtract x from both sides. x 2 11 x 18 0 Add 9 to both sides.
We now recognize that, for the quadratic formula, a = 1, b = -11, and c =18. 2 11 11 4118 x 21 Use the quadratic formula. 11 121 72 x 2 Simplify. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 47 Graphs of Intersecting Functions CONTINU ED 11 49 x 2 Simplify. 11 7 x 2
Simplify. 11 7 11 7 x , 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. Lets use y = x 9. x 9 x 2 y x 9 y x 9 y 9 9 y 2 9 y 0 y 7
Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 48 Graphs of Intersecting Functions CONTINU ED 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 Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc .
Slide 49 Factoring EXAMPL E Factor the following quadratic polynomial. 6 x 2 x3 SOLUTIO N 6x 2 x3 This is the given polynomial. 2 x 3 x 2 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 2 Rewrite 3 x 2 as Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . 3 x 3x . Slide 50 Factoring EXAMPL E
Solve the equation for x. SOLUTIO N 5 6 1 2 x x 5 6 1 2 x x This is the given equation. 5 6 x 2 1 2 x 2 x x Multiply everything by the LCD: x2. 5 6 x 2 x 2 2 x 2 x x Distribute. x 2 5 x 6 Multiply.
x 2 5 x 6 0 x 1 x 6 0 Subtract 5x + 6 from both sides. Factor. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 51 Factoring CONTINU ED x 1 0 x 1 x 6 0 x 6 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. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 52 0.5 Exponents and Power Functions
Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 53 Section Outline Exponent Rules Applications of Exponents Compound Interest Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 54 Exponents Definition Example b n bb b
53 5 5 5 n times 1 n n b b 1 3 5 3 5 Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 55 Exponents Definition m n n Example
m n b b b b m n 1 b m n 1 n b m 3 4
m b m 5 3 4 1 5 3 4 4 1 4 Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . 3
3 5 5 5 1 n 4 5 3 1 3 5 4 Slide 56 Exponents Definition r s
b b b 1 b r b r r s 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, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 57 Exponents Definition Example 7 r b
r s b bs r s b 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 9 5 8 9 8 9 2 9 3 Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 58 Exponents Definition ab r a r b r r r a a
br b Example 125 27 1/ 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, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 59 Applications of Exponents EXAMPL E
Use the laws of exponents to simplify the algebraic expression. 5 2/3 27 x 3 SOLUTIO N 5 2/3 27 x 3 x 27 2 / 3 x 5 2 / 3 3 x 27 2 / 3 x 5 2 / 3 x 1/ 3 27 2 / 3 x10 / 3 x1 / 3 3
2 27 x10 / 3 x1/ 3 x This is the given expression. ab r a r b r 1 n b n b r s b b rs m n b n b m n b Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . m
Slide 60 Applications of Exponents CONTINU ED 3 2 x10 / 3 x1/ 3 9 x10 / 3 x1 / 3 9x 10 / 3 1 / 3 3 27 3 3 2 9 br b r s s b 9x 9 / 3 Subtract. 9x 3 Divide.
Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 61 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 (how much money you end 8 A 7001 0.08 up with) 8 A 700 1 .
08 P = the principal amount A 7001.851 invested i = the compound interest rate A 1,295.651 per interest period Therefore, the compound amount n = the number of would be $1,295.65. compounding periods Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 62 Compound Interest - General Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 63 Compound Interest - General EXAMPL E (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. SOLUTIO N
r A P 1 m mt Since interest is not being compounded annually, we must use this formula. r A 500 1 4 4 1 r A 500 1 4 4 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). Simplify. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc .
Slide 64 0.6 Functions and Graphs in Application Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 65 Section Outline Geometric Problems Cost, Revenue, and Profit Surface Area Functions and Graphs Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 66 Geometric Problems
EXAMPL E (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). SOLUTIO N First we will assign letters to represent the dimensions of the corral. y x x x x y Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 67 Geometric Problems CONTINU ED 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 x x x x y y F 4 x 2 y Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 68 Cost Problems EXAMPL E (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. SOLUTIO N 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. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 69 Cost Problems CONTINU ED 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 Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 70 Surface Area EXAMPL E
Assign letters to the dimensions of the geometric box and then determine an expression representing the surface area of the box. SOLUTIO N First we assign letters to represent the dimensions of the box. z y x Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 71 Surface Area CONTINU ED 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. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 72 Cost, Revenue, & Profit EXAMPL E (Cost, Revenue, and Profit) An average sale at a small florist shop is $21, so the shops 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 shops 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? SOLUTIO N (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 Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 73
Cost, Revenue, & Profit CONTINU ED (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. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 74 Cost, Revenue, & Profit CONTINU ED 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. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 75 Functions & Graphs EXAMPL E 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? Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 76 Functions & Graphs CONTINU ED SOLUTIO N 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. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 77 Functions & Graphs CONTINU ED (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. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14e Copyright 2018, 2014, 2010 Pearson Education Inc . Slide 78
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