2.1 Increasing, Decreasing, and Piecewise Functions; Applications


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1 2.1 Increasing, Decreasing, and Piecewise Functions; Applications Graph functions, looking for intervals on which the function is increasing, decreasing, or constant, and estimate relative maxima and minima. Given an application, find a function that models the application; find the domain of the function and function values, and then graph the function. Graph functions defined piecewise.
2 Increasing, Decreasing, and Constant Functions On a given interval, if the graph of a function rises from left to right, it is said to be increasing on that interval. If the graph drops from left to right, it is said to be decreasing. If the function values stay the same from left to right, the function is said to be constant. Slide 2.12
3 Definitions A function f is said to be increasing on an open interval I, if for all a and b in that interval, a < b implies f(a) < f(b). Slide 2.13
4 Definitions continued A function f is said to be decreasing on an open interval I, if for all a and b in that interval, a < b implies f(a) > f(b). Slide 2.14
5 Definitions continued A function f is said to be constant on an open interval I, if for all a and b in that interval, f(a) = f(b). Slide 2.15
6 Relative Maximum and Minimum Values Suppose that f is a function for which f(c) exists for some c in the domain of f. Then: f(c) is a relative maximum if there exists an open interval I containing c such that f(c) > f(x), for all x in I where x c; and f(c) is a relative minimum if there exists an open interval I containing c such that f(c) < f(x), for all x in I where x c. Slide 2.16
7 Relative Maximum and Minimum Values y Relative maximum f Relative minimum c 1 c 2 c 3 x Slide 2.17
8 Applications of Functions Many realworld situations can be modeled by functions. Example A man plans to enclose a rectangular area using 80 yards of fencing. If the area is w yards wide, express the enclosed area as a function of w. Solution We want area as a function of w. Since the area is rectangular, we have A = lw. We know that the perimeter, 2 lengths and 2 widths, is 80 yds, so we have 40 yds for one length and one width. If the width is w, then the length, l, can be given by l = 40 w. Now A(w) = (40 w)w = 40w w 2. Slide 2.18
9 Functions Defined Piecewise Some functions are defined piecewise using different output formulas for different parts of the domain. For the function defined as: find f(3), f(1), and f(5). x 2, for x 0, f ( x) 4, for 0 x 2, x 1, for x 2, Since 3 0, use f(x) = x 2 : f( 3) = ( 3) 2 = 9. Since 0 < 1 2, use f(x) = 4: f(1) = 4. Since 5 > 2 use f(x) = x 1: f(5) = 5 1 = 4. Slide 2.19
10 Functions Defined Piecewise Graph the function defined as: 3 for x 0 f x x x x 1 for x ( ) 3 for 0 2 a) We graph f(x) = 3 only for inputs x less than or equal to 0. f(x) = 3, for x 0 x f ( x) 1for x 2 2 f(x) = 3 + x 2, for 0 < x 2 b) We graph f(x) = 3 + x 2 only for inputs x greater than 0 and x less than or equal to 2. c) We graph f(x) = 1 2 only for inputs x greater than 2. Slide
11 Greatest Integer Function = the greatest integer less than or equal to x. The greatest integer function pairs the input with the greatest integer less than or equal to that input Slide
12 2.2 The Algebra of Functions Find the sum, the difference, the product, and the quotient of two functions, and determine the domains of the resulting functions. Find the difference quotient for a function.
13 Sums, Differences, Products, and Quotients of Functions If f and g are functions and x is in the domain of each function, then ( f g)( x) f ( x) g( x) ( f g)( x) f ( x) g( x) ( fg)( x) f ( x) g( x) ( f / g)( x) f ( x) / g( x), provided g( x) 0 Slide
14 Example Given that f(x) = x + 2 and g(x) = 2x + 5, find each of the following. a) (f + g)(x) b) (f + g)(5) Solution: a) ( f g)( x) f ( x) g( x) x 2 2x 5 3x 7 b) We can find (f + g)(5) provided 5 is in the domain of each function. This is true. f(5) = = 7 g(5) = 2(5) + 5 = 15 (f + g)(5) = f(5) + g(5) = = 22 or (f + g)(5) = 3(5) + 7 = 22 Slide
15 Another Example Given that f(x) = x and g(x) = x 3, find each of the following. a) The domain of f + g, f g, fg, and f/g b) (f g)(x) c) (f/g)(x) Solution: a) The domain of f is the set of all real numbers. The domain of g is also the set of all real numbers. The domains of f + g, f g, and fg are the set of numbers in the intersection of the domains that is, the set of numbers in both domains, or all real numbers. For f/g, we must exclude 3, since g(3) = 0. Slide
16 Another Example continued b) (f g)(x) = f(x) g(x) = (x 2 + 2) (x 3) = x 2 x + 5 c) (f/g)(x) = ( f / g)( x) f( x) gx ( ) 2 x 2 x 3 Remember to add the stipulation that x 3, since 3 is not in the domain of (f/g)(x). Slide
17 Difference Quotient The ratio below is called the difference quotient, or average rate of change. f ( x h) f ( x) h Slide
18 Example For the function f given by f(x) = 5x 1, find the difference quotient f ( x h) f ( x). h Solution: We first find f(x + h): f ( x h) 5( x h) 1 5x 5h 1 Slide
19 Example continued f ( x h) f ( x) h 5x 5h 1 (5x 1) h 5h h 5 Slide
20 Another Example For the function f given by f(x) = x 2 + 2x 3, find the difference quotient. Solution: We first find f(x + h): f ( x h) 2 ( x h) 2( x h) x xh h x h Slide
21 Example continued f ( x h) f ( x) h x 2xh h 2x 2h 3 ( x 2x 3) h x 2xh h 2x 2h 3 x 2x 3 h 2 2xh h 2h h h(2x h 2) 2x h 2 h Slide
22 2.3 The Composition of Functions Find the composition of two functions and the domain of the composition. Decompose a function as a composition of two functions.
23 Composition of Functions Definition: The composite function f g, the composition of f and g, is defined as ( f g)( x) f ( g( x)), where x is in the domain of g and g( x) is in the domain of f. Slide
24 Example Given that f(x) = 3x 1 and g(x) = x 2 + x 3, find: a) ( f g)( x) b) ( g f )( x) a) 2 ( f g)( x) f ( g( x)) f ( x x 3) 2 3( x x 3) 1 2 3x 3x x 3x 10 Slide
25 Example Given that f(x) = 3x 1 and g(x) = x 2 + x 3, find: a) ( f g)( x) b) ( g f )( x) b) ( g f )( x) g( f ( x)) g( 3x 1) 2 ( x ) ( x ) 3 2 9x 6x 1 3x x 3x 3 Slide
26 Example Given that f(x) = 3x 1 and g(x) = x 2 + x 3, find: a) ( f g)(2) b) ( g f)(2) a) ( f g)(2) f ( g(2)) f 2 ( 2 2 3) f ( 3) 3( 3) 1 8 Slide
27 Example Given that f(x) = 3x 1 and g(x) = x 2 + x 3, find: a) ( f g)(2) b) ( g f)(2) b) ( g f )(2) g( f (2)) g( 3(2) 1) 2 ( 5) ( 5) 3 27 Slide
28 Example 4 Given f ( x) x and g( x) 2x 3, find the domain of ( f g)( x). Solution: f(x) is not defined for negative radicands. Since the inputs of f g are the outputs of g, the domain of f g consists of all the values in the domain of g for which g(x) is nonnegative. gx ( ) 0 2x 3 0 x 3/ 2 The domain is { x x 3/ 2}, or [ 3/ 2, ). Slide
29 Decomposing a Function as a Composition In calculus, one needs to recognize how a function can be expressed as the composition of two functions. This can be thought of as decomposing the function. Slide
30 Example If h(x) = (3x 1) 4, find f(x) and g(x) such that h( x) ( f g)( x). Solution: The function h(x) raises (3x 1) to the fourth power. Two functions that can be used for the composition are: h( x) ( f g)( x) f ( g( x)) f (3x 1) 3x 1 4 f(x) = x 4 and g(x) = 3x 1. Slide
31 2.4 Symmetry and Transformations Determine whether a graph is symmetric with respect to the xaxis, the yaxis, and the origin. Determine whether a function is even, odd, or neither even nor odd. Given the graph of a function, graph its transformation under translations, reflections, stretchings, and shrinkings.
32 Symmetry Algebraic Tests of Symmetry xaxis: If replacing y with y produces an equivalent equation, then the graph is symmetric with respect to the xaxis. yaxis: If replacing x with x produces an equivalent equation, then the graph is symmetric with respect to the yaxis. Origin: If replacing x with x and y with y produces an equivalent equation, then the graph is symmetric with respect to the origin. Slide
33 Example Test x = y for symmetry with respect to the xaxis, the yaxis, and the origin. xaxis: We replace y with y: x x 2 ( y) 2 y 2 2 The resulting equation is equivalent to the original so the graph is symmetric with respect to the xaxis. Slide
34 Example continued Test x = y for symmetry with respect to the xaxis, the yaxis, and the origin. yaxis: We replace x with x: x y x y The resulting equation is not equivalent to the original equation, so the graph is not symmetric with respect to the yaxis. Slide
35 Example continued Origin: We replace x with x and y with y: x x x y 2 2 ( y) 2 y 2 The resulting equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin. 2 2 Slide
36 Even and Odd Functions If the graph of a function f is symmetric with respect to the yaxis, we say that it is an even function. That is, for each x in the domain of f, f(x) = f( x). If the graph of a function f is symmetric with respect to the origin, we say that it is an odd function. That is, for each x in the domain of f, f( x) = f(x). Slide
37 Example Determine whether the function is even, odd, or neither. h( x) x 4x h( x) ( x) 4( x) x 4 2 4x 4 2 y = x 4 4x 2 We see that h(x) = h( x). Thus, h is even. Slide
38 Example Determine whether the function is even, odd, or neither. h( x) x 4x h( x) ( x 4 x ) x 4x 4 2 y = x 4 4x 2 We see that h( x) h(x). Thus, h is not odd. Slide
39 Vertical Translation Vertical Translation For b > 0, the graph of y = f(x) + b is the graph of y = f(x) shifted up b units; y = 3x 2 +2 y = 3x 2 the graph of y = f(x) b is the graph of y = f(x) shifted down b units. y = 3x 2 3 Slide
40 Horizontal Translation Horizontal Translation For d > 0, the graph of y = f(x d) is the graph of y = f(x) shifted right d units; y = (3x 2) 2 the graph of y = f(x + d) is the graph of y = f(x) shifted left d units. y = (3x + 2) 2 y = 3x 2 Slide
41 Reflections The graph of y = f(x) is the reflection of the graph of y = f(x) across the xaxis. The graph of y = f( x) is the reflection of the graph of y = f(x) across the yaxis. If a point (x, y) is on the graph of y = f(x), then (x, y) is on the graph of y = f(x), and ( x, y) is on the graph of y = f( x). Slide
42 Example Reflection of the graph y = 3x 3 4x 2 across the xaxis. y = 3x 3 4x 2 y = 3x 3 + 4x 2 Slide
43 Example Reflection of the graph y = x 3 2x 2 across the yaxis. y = x3 + 2x 2 y = x 3 2x 2 Slide
44 Vertical Stretching and Shrinking The graph of y = af(x) can be obtained from the graph of y = f(x) by stretching vertically for a > 1, or shrinking vertically for 0 < a < 1. For a < 0, the graph is also reflected across the x axis. (The ycoordinates of the graph of y = af(x) can be obtained by multiplying the ycoordinates of y = f(x) by a.) Slide
45 Examples Stretch y = x 3 x vertically. Slide
46 Examples Shrink y = x 3 x vertically. Slide
47 Examples Stretch and reflect y = x 3 x across the x axis Slide
48 Horizontal Stretching or Shrinking The graph of y = f(cx) can be obtained from the graph of y = f(x) by shrinking horizontally for c > 1, or stretching horizontally for 0 < c < 1. For c < 0, the graph is also reflected across the y axis. (The xcoordinates of the graph of y = f(cx) can be obtained by dividing the xcoordinates of the graph of y = f(x) by c.) Slide
49 Examples Shrink y = x 3 x horizontally. Slide
50 Examples Stretch y = x 3 x horizontally. Slide
51 Examples Stretch horizontally and reflect y = x 3 x. Slide
52 2.5 Variation and Applications Find equations of direct, inverse, and combined variation given values of the variables. Solve applied problems involving variation.
53 Direct Variation If a situation gives rise to a linear function f(x) = kx, or y = kx, where k is a positive constant, we say that we have direct variation, or that y varies directly as x, or that y is directly proportional to x. The number k is called the variation constant, or constant of proportionality. Slide
54 Direct Variation The graph of y = kx, k > 0, always goes through the origin and rises from left to right. As x increases, y increases; that is, the function is increasing on the interval (0, ). The constant k is also the slope of the line. y kx, k 0 Slide
55 Direct Variation Example: Find the variation constant and an equation of variation in which y varies directly as x, and y = 42 when x = 3. Solution: We know that (3, 42) is a solution of y = kx. y = kx 42 = k k 14 = k The variation constant 14, is the rate of change of y with respect to x. The equation of variation is y = 14x. Slide
56 Application Example: Wages. A cashier earns an hourly wage. If the cashier worked 18 hours and earned $168.30, how much will the cashier earn if she works 33 hours? Solution: We can express the amount of money earned as a function of the amount of hours worked. I(h) = kh I(18) = k 18 $ = k 18 $9.35 = k The hourly wage is the variation constant. Next, we use the equation to find how much the cashier will earn if she works 33 hours. I(33) = $9.35(33) = $ Slide
57 Inverse Variation If a situation gives rise to a function f(x) = k/x, or y = k/x, where k is a positive constant, we say that we have inverse variation, or that y varies inversely as x, or that y is inversely proportional to x. The number k is called the variation constant, or constant of proportionality. For the graph y = k/x, k 0, as x increases, y decreases; that is, the function is decreasing on the interval (0, ). Slide
58 Inverse Variation For the graph y = k/x, k 0, as x increases, y decreases; that is, the function is decreasing on the interval (0, ). y k x, k 0 Slide
59 Inverse Variation Example: Find the variation constant and an equation of variation in which y varies inversely as x, and y = 22 when x = 0.4. Solution: k y x k (0.4)22 k 8.8 k The variation constant is 8.8. The equation of variation is y = 8.8/x. Slide
60 Application Example: Road Construction. The time t required to do a job varies inversely as the number of people P who work on the job (assuming that they all work at the same rate). If it takes 180 days for 12 workers to complete a job, how long will it take 15 workers to complete the same job? Solution: We can express the amount of time required, in days, as a function of the number of people working. k t varies inversely as P tp ( ) P k t(12) 12 k k This is the variation constant. Slide
61 Application continued The equation of variation is t(p) = 2160/P. Next we compute t(15). tp ( ) t(15) t 2160 P It would take 144 days for 15 people to complete the same job. Slide
62 Combined Variation Other kinds of variation: y varies directly as the nth power of x if there is some n positive constant k such that. y varies inversely as the nth power of x if k there is some positive constant k such that y. y varies jointly as x and z if there is some positive constant k such that y = kxz. y kx x n Slide
63 Example The luminance of a light (E) varies directly with the intensity (I) of the light and inversely with the square distance (D) from the light. At a distance of 10 feet, a light meter reads 3 units for a 50cd lamp. Find the luminance of a 27cd lamp at a distance of 9 feet. I E k D k 50 Solve for k k Substitute the second set of data into the equation. The lamp gives an luminance reading of 2 units. E E Slide
64 2.6 Solving Linear Inequalities Solve linear inequalities. Solve compound inequalities. Solve inequalities with absolute value. Solve applied problems using inequalities. Copyright 2008 Pearson Education, Inc.
65 Inequalities An inequality is a sentence with <, >,, or as its verb. Examples: 5x 7 < 3 + 4x 3(x + 6) 4(x 3) Copyright 2008 Pearson Education, Inc. Slide
66 Principles for Solving Inequalities For any real numbers a, b, and c: The Addition Principle for Inequalities: If a < b is true, then a + c < b + c is true. The Multiplication Principle for Inequalities: If a < b and c > 0 are true, then ac < bc is true. If a < b and c < 0, then ac > bc is true. Similar statements hold for a b. When both sides of an inequality are multiplied or divided by a negative number, we must reverse the inequality sign. Copyright 2008 Pearson Education, Inc. Slide
67 Examples Solve: 4x 6 2x 10 4x 2x 4 2x 4 x 2 Solve: 6( x 3) 7( x 2) 6x 18 7x 14 x 4 x 4 {x x < 2} or (, 2) ) {x x 4} or [ 4, ) [ Copyright 2008 Pearson Education, Inc. Slide
68 Compound Inequalities When two inequalities are joined by the word and or the word or, a compound inequality is formed. Conjunction contains the word and. Example: 7 < 3x + 5 and 3x Disjunction contains the word or. Example: 3x or 3x + 6 > 12 Copyright 2008 Pearson Education, Inc. Slide
69 Examples Solve: 4 3x x x 3 4 x 1 Solve: 4x 5 3 or 4x 5 > 3 4x 5 3 or 4x 5 3 4x 2 4x 8 1 x 2 x 2 ( ] ] ( Copyright 2008 Pearson Education, Inc. Slide
70 Inequalities with Absolute Value Inequalities sometimes contain absolutevalue notation. The following properties are used to solve them. For a > 0 and an algebraic expression X: X < a is equivalent to a < X < a. X > a is equivalent to X < a or X > a. Similar statements hold for X a and X a. Copyright 2008 Pearson Education, Inc. Slide
71 Example Solve: 4x x x x 2 ( ) Copyright 2008 Pearson Education, Inc. Slide
72 Application Johnson Catering charges $100 plus $30 per hour to cater an event. Catherine s Catering charges $50 per hour. For what lengths of time does it cost less to hire Catherine s Catering? 1. Familiarize. Read the problem. 2. Translate. Catherine s is less than Johnson 50x < x Copyright 2008 Pearson Education, Inc. Slide
73 Application continued 3. Carry out. 50x x 20x 100 x 5 4. Check. 50(5)? (5) 250? State. For values of x < 5 hr, Catherine s Catering will cost less. Copyright 2008 Pearson Education, Inc. Slide
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