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CHAPTER FUNCTIONS 1.1 FUNCTIONS AND THEIR GRAPHS domain (, ); range [1, ) domain [0, ); range (, 1] domain [2, ); y in range and y x 10 y can be any nonnegative real number range [0, ) domain (, 0] [3, ); y in range and y x 3x y can be any nonnegative real number range [0, ) domain (, 3) (3, ); y in range and y 3t 3t , 3t now if t t 3t 0, or if t y can be any nonzero real number range (, 0) (0, ) domain (, 4) ( 4, 4) (4, ); y in range and y 4 t 16 t 16 16 t 16 , t 16 now if t 4 t 16 , or if t t 16 t 16 t 16 0, or if y can be any nonzero real number range (, 18 ] (0, ) (a) Not the graph of a function of x since it fails the vertical line test (b) Is the graph of a function of x since any vertical line intersects the graph at most once (a) Not the graph of a function of x since it fails the vertical line test (b) Not the graph of a function of x since it fails the vertical line test base x; (height) 2x x height x; area is a( x) (base)(height) 12 ( x ) x x ; perimeter is p ( x) x x x 3x 10 s side length s s d s d ; and area is a s a 12 d 2 11 Let D diagonal length of a face of the cube and the length of an edge Then D d and 2 D 2 3 d d The surface area is 6 d3 2d and the volume is 3 d3 12 The coordinates of P are x, x so the slope of the line joining P to the origin is m xx Thus, x, x m2 x 3/2 d 3 ( x 0) , m1 25 13 x y y 12 x 45 ; L ( x 0)2 ( y 0)2 x ( 12 x 54 ) x 14 x 45 x 16 x2 25 x 16 20 x 20 x 25 16 20 x 20 x 25 14 y x y x; L ( x 4)2 ( y 0)2 ( y 4)2 y ( y 1)2 y y4 y2 y2 y4 y2 1 Copyright 2014 Pearson Education, Inc Chapter Functions 15 The domain is (, ) 16 The domain is (, ) 17 The domain is (, ) 18 The domain is (, 0] 19 The domain is (, 0) (0, ) 20 The domain is (, 0) (0, ) 21 The domain is (, 5) (5, 3] [3, 5) (5, ) 22 The range is [2, 3) 23 Neither graph passes the vertical line test (a) (b) Copyright 2014 Pearson Education, Inc Section 1.1 Functions and Their Graphs 24 Neither graph passes the vertical line test (a) (b) x y 1 y 1 x x y 1 or or x y 1 y 1 x 25 26 x y x , x 27 F ( x) x x, x x y 0 , x 28 G ( x) x x, x 29 (a) Line through (0, 0) and (1, 1): y x; Line through (1, 1) and (2, 0): y x x, x f ( x) x 2, x 2, 0, (b) f ( x) 2, 0, 0 1 2 3 x 1 x2 x3 x4 30 (a) Line through (0, 2) and (2, 0): y x Line through (2, 1) and (5, 0): m x 2, x f ( x) x , x 1 52 31 13 , so y 13 ( x 2) 13 x 53 Copyright 2014 Pearson Education, Inc Chapter Functions (b) Line through (1, 0) and (0, 3): m Line through (0, 3) and (2, 1) : m 3 x 3, x f ( x) 2 x 3, x 3 3, so y 3x ( 1) 1 4 2, so y 2 x 20 31 (a) Line through (1, 1) and (0, 0): y x Line through (0, 1) and (1, 1): y Line through (1, 1) and (3, 0): m 1 31 21 12 , so y 12 ( x 1) 12 x 32 x 1 x f ( x) x 1 1 x x (b) Line through (2, 1) and (0, 0): y 12 x Line through (0, 2) and (1, 0): y 2 x Line through (1, 1) and (3, 1): y 1 32 (a) Line through 1x 2 x f ( x) 2 x x 1 1 x T2 , 0 and (T, 1): m T 1 (T0/2) T2 , so y T2 x T2 T2 x 0, x T2 f ( x) T T x 1, x T A, x T A, T x T (b) f ( x) 3T A, T x A, 32T x 2T 33 (a) x for x [0, 1) (b) x for x (1, 0] 34 x x only when x is an integer 35 For any real number x, n x n 1, where n is an integer Now: n x n (n 1) x n By definition: x n and x n x n So x x for all real x 36 To find f(x) you delete the decimal or fractional portion of x, leaving only the integer part Copyright 2014 Pearson Education, Inc Section 1.1 Functions and Their Graphs 37 Symmetric about the origin Dec: x Inc: nowhere 38 Symmetric about the y-axis Dec: x Inc: x 39 Symmetric about the origin Dec: nowhere Inc: x 0x 40 Symmetric about the y-axis Dec: x Inc: x 41 Symmetric about the y-axis Dec: x Inc: x 42 No symmetry Dec: x Inc: nowhere Copyright 2014 Pearson Education, Inc Chapter Functions 43 Symmetric about the origin Dec: nowhere Inc: x 44 No symmetry Dec: x Inc: nowhere 45 No symmetry Dec: x Inc: nowhere 46 Symmetric about the y-axis Dec: x Inc: x 47 Since a horizontal line not through the origin is symmetric with respect to the y-axis, but not with respect to the origin, the function is even 48 f ( x) x 5 15 and f ( x) ( x)5 x ( x )5 f ( x) Thus the function is odd x5 49 Since f ( x) x ( x)2 f ( x) The function is even 50 Since [ f ( x) x x ] [ f ( x) ( x)2 x] and [ f ( x) x x] [ f ( x ) ( x )2 x] the function is neither even nor odd 51 Since g ( x) x3 x, g ( x) x3 x ( x3 x) g ( x) So the function is odd 52 g ( x) x x ( x)4 3( x)2 g ( x), thus the function is even 53 g ( x) x2 54 g ( x) x ; x2 55 h(t ) ( x )2 g ( x) Thus the function is even g ( x ) ; h( t ) t 1 x x2 g ( x) So the function is odd ; h (t ) t 1 t Since h(t ) h(t ) and h(t ) h(t ), the function is neither even nor odd Copyright 2014 Pearson Education, Inc Section 1.1 Functions and Their Graphs 56 Since |t | |( t )3 |, h(t ) h( t ) and the function is even 57 h(t ) 2t 1, h(t ) 2t So h(t ) h(t ) h(t ) 2t 1, so h(t ) h(t ) The function is neither even nor odd 58 h(t ) 2| t | and h(t ) 2| t | 2| t | So h(t ) h(t ) and the function is even 59 s kt 25 k (75) k 13 s 13 t ; 60 13 t t 180 60 K c v 12960 c(18)2 c 40 K 40v ; K 40(10) 4000 joules 61 r k s 6 k k 24 r 24 ; 10 s 24 s s 12 k k 14700 P 14700 ; 23.4 14700 V 62 P Vk 14.7 1000 V V 24500 39 628.2 in 63 V f ( x ) x (14 x )(22 x ) x 72 x 308 x; x AB 64 (a) Let h height of the triangle Since the triangle is isosceles, AB 2 22 AB So, h2 12 h B is at (0, 1) slope of AB 1 The equation of AB is y f ( x) x 1; x [0, 1] (b) A( x) xy x( x 1) 2 x x; x [0, 1] 65 (a) Graph h because it is an even function and rises less rapidly than does Graph g (b) Graph f because it is an odd function (c) Graph g because it is an even function and rises more rapidly than does Graph h 66 (a) Graph f because it is linear (b) Graph g because it contains (0, 1) (c) Graph h because it is a nonlinear odd function 67 (a) From the graph, (b) x x 4x x (2, 0) (4, ) 4x 2x 4x x 0: x2 x 4x 0 2x x since x is positive; x x2 x ( x 4)( x 2) 2x 0 ( x 4)( x 2) 2x 0 x 0: 2x 4x 0 2x x 2 since x is negative; sign of ( x 4)( x 2) Solution interval: (2, 0) (4, ) Copyright 2014 Pearson Education, Inc Chapter Functions 68 (a) From the graph, (b) Case x 1: x ( , 5) ( 1, 1) x 1 3( x 1) x 1 x 1 x 1 x 1 3x x x 5 Thus, x (, 5) solves the inequality Case 1 x 1: x 1 3( x 1) x 1 x 1 2 3x x x 5 which is true if x 1 Thus, x (1, 1) solves the inequality Case x : x 3 x 2 3x x x 5 which is never true if x, so no solution here In conclusion, x ( , 5) (1, 1) 69 A curve symmetric about the x-axis will not pass the vertical line test because the points (x, y) and ( x, y ) lie on the same vertical line The graph of the function y f ( x ) is the x-axis, a horizontal line for which there is a single y-value, 0, for any x 70 price 40 x, quantity 300 25x R ( x) (40 x)(300 25 x) 71 x x h2 x h 2h ; cost 5(2 x) 10h C (h) 10 10h 5h 2h 22 72 (a) Note that mi 10,560 ft, so there are 8002 x feet of river cable at $180 per foot and (10,560 x) feet of land cable at $100 per foot The cost is C ( x) 180 8002 x 100(10,560 - x) (b) C (0) $1, 200, 000 C (500) $1,175,812 C (1000) $1,186,512 C (1500) $1, 212, 000 C (2000) $1, 243, 732 C (2500) $1, 278, 479 C (3000) $1,314,870 Values beyond this are all larger It would appear that the least expensive location is less than 2000 feet from the point P 1.2 COMBINING FUNCTIONS; SHIFTING AND SCALING GRAPHS D f : x , Dg : x D f g D fg : x R f : y , Rg : y 0, R f g : y 1, R fg : y D f : x x 1, Dg : x x Therefore D f g D fg : x R f Rg : y 0, R f g : y 2, R fg : y D f : x , Dg : x , D f /g : x , Dg /f : x , R f : y 2, Rg : y 1, R f /g : y 2, Rg /f : 12 y D f : x , Dg : x 0, D f /g : x 0, Dg /f : x 0; R f : y 1, Rg : y 1, R f /g : y 1, Rg /f : y Copyright 2014 Pearson Education, Inc Section 1.2 Combining Functions; Shifting and Scaling Graphs (a) (d) ( x 5) x 10 x 22 (g) x 10 (b) 22 (e) (h) ( x 3)2 x x (c) x (f ) 2 (a) 13 (b) (c) (d) x (e) (g) x (h) (f ) 1 x 1 1 x2 x 1 x 1 x2 1 x x 1 x 1 ( f g h)( x) f ( g (h( x))) f ( g (4 x)) f (3(4 x )) f (12 x) (12 3x) 13 3x ( f g h)( x) f ( g ( h( x))) f ( g ( x )) f (2( x ) 1) f (2 x 1) 3(2 x 1) x 1x f ( f g h)( x) f ( g (h( x))) f g x 5x f x4 x x4 x x x 2 f f x 2 2x 3 x 2x 2 3 x 2x 3 x 3x 10 ( f g h)( x) f ( g (h( x))) f g 2 x 11 (a) ( f g )( x) (d) ( j j )( x) (b) ( j g )( x) (e) ( g h f )( x) (c) ( g g )( x) (f ) (h j f )( x) 12 (a) ( f j )( x) (d) ( f f )( x) (b) ( g h)( x) (e) ( j g f )( x) (c) (hh)( x) (f ) ( g f h)( x) f (x) ( f g )( x) (a) x x x7 (b) x 3x 13 g(x) x2 (d) x x 1 x x 1 (e) x 1 1x x x x (f ) x 2x 3( x 2) x x 5 (c) x x x 1 x 1 x 1 x (xx 1) x 14 (a) ( f g )( x) |g ( x)| (b) ( f g )( x) x 1 g ( x) x x g ( x) g (1x) x x x x g (1x) x 1 g (1x) , so g ( x) x (c) Since ( f g )( x) g ( x) | x |, g ( x) x (d) Since ( f g )( x) f x | x |, f ( x) x (Note that the domain of the composite is [0, ).) Copyright 2014 Pearson Education, Inc 10 Chapter Functions The completed table is shown Note that the absolute value sign in part (d) is optional g (x) f (x) ( f g )(x) x 1 | x| x 1 x 1 x 1 x x x 1 x2 x | x| x | x| x 15 (a) f ( g (1)) f (1) (d) g ( g (2)) g (0) 16 (a) (b) (c) (d) (e) (b) g ( f (0)) g (2) (e) g ( f (2)) g (1) 1 (c) f ( f (1)) f (0) 2 (f) f ( g (1)) f (1) f ( g (0)) f (1) (1) 3, where g (0) 1 g ( f (3)) g (1) (1) 1, where f (3) 1 g ( g (1)) g (1) 0, where g (1) (1) f ( f (2)) f (0) 2, where f (2) g ( f (0)) g (2) 1, where f (0) f 12 12 52 , where g 12 12 12 (f ) f g 12 1 x 17 (a) ( f g )( x) f ( g ( x)) 1x x ( g f )( x) g ( f ( x)) x 1 (b) Domain ( f g ): (, 1] (0, ), domain ( g f ): (1, ) (c) Range ( f g ): (1, ), range ( g f ): (0, ) 18 (a) ( f g )( x ) f ( g ( x)) x x ( g f )( x ) g ( f ( x)) | x | (b) Domain ( f g ): [0, ), domain ( g f ): (, ) (c) Range ( f g ): (0, ), range ( g f ): (, 1] 19 ( f g )( x) x f ( g ( x)) x g ( x) x g ( x) ( g ( x) 2) x x g ( x) x g ( x) x g ( x) 2 x g ( x) 2xx x2x g ( x) 20 ( f g )( x ) x f ( g ( x )) x 2( g ( x))3 x ( g ( x))3 21 (a) y ( x 7)2 (b) y ( x 4) 22 (a) y x (b) y x x6 g ( x) x6 23 (a) Position (b) Position (c) Position (d) Position 24 (a) y ( x 1) (b) y ( x 2)2 (c) y ( x 4)2 (d) y ( x 2) Copyright 2014 Pearson Education, Inc 28 Chapter Functions 17 [ 5, 1] by [ 5, 5] 18 [ 5, 1] by [ 2, 4] 19 [ 4, 4] by [0, 3] 20 [ 5, 5] by [ 2, 2] 21 [ 10, 10] by [ 6, 6] 22 [ 5, 5] by [ 2, 2] 23 [ 6, 10] by [ 6, 6] 24 [ 3, 5] by [ 2, 10] 25 [0.03, 0.03] by [1.25, 1.25] 26 [0.1, 0.1] by [3, 3] Copyright 2014 Pearson Education, Inc Section 1.4 Graphing with Software 27 [300, 300] by [1.25, 1.25] 28 [50, 50] by [0.1, 0.1] 29 [0.25, 0.25] by[0.3, 0.3] 30 [0.15, 0.15] by [0.02, 0.05] 31 x x y y y x x The lower half is produced by graphing y x x 32 y 16 x y 16 x The upper branch is produced by graphing y 16 x 33 34 Copyright 2014 Pearson Education, Inc 29 30 Chapter Functions 35 36 38 37 200 150 100 50 60 64 68 72 76 1970 1980 1990 2000 2010 2020 80 40 39 (in thousands) 300 26 225 22 R 18 T 150 14 75 10 1972 1980 1988 1996 2004 2012 2000 2002 2004 2006 2008 42 41 600 450 0.5 300 1955 1935 1975 1995 2015 150 0.5 Copyright 2014 Pearson Education, Inc 10 Chapter Practice Exercises 31 CHAPTER PRACTICE EXERCISES The area is A r and the circumference is C 2 r Thus, r 2C A 2C 2 C4 4S The volume is V 43 r r 34V Substitution into the formula 2/3 for surface area gives S 4 r 4 34V 1/2 The surface area is S 4 r r The coordinates of a point on the parabola are (x, x2) The angle of inclination joining this point to the origin satisfies the equation tan xx x Thus the point has coordinates ( x, x ) (tan , tan ) tan rise run h 500 h 500 tan ft Symmetric about the origin Symmetric about the y-axis Neither Symmetric about the y-axis y ( x) ( x)2 x y ( x) Even 10 y ( x) ( x)5 ( x )3 ( x) x5 x3 x y ( x) Odd 11 y ( x) cos( x) cos x y ( x) Even 12 y ( x) sec( x) tan( x) 13 y ( x) x 4 x 3 2 x sin x cos x x4 x 2x sin x cos x x4 x3 x sec x tan x y ( x) Odd y ( x) Odd 14 y ( x) ( x) sin( x) ( x) sin x ( x sin x) y ( x) Odd Copyright 2014 Pearson Education, Inc 32 Chapter Functions 15 y ( x) x cos( x) x cos x Neither even nor odd 16 y ( x) ( x) cos( x) x cos x y ( x) Odd 17 Since f and g are odd f ( x) f ( x) and g ( x) g ( x) (a) ( f g )( x) f ( x) g ( x) [ f ( x)] [ g ( x )] f ( x) g ( x) ( f g )( x) f g is even (b) f ( x) f ( x) f ( x) f ( x) [ f ( x)] [ f x ] [ f ( x)] f ( x) f ( x) f ( x ) f ( x) f is odd (c) f (sin( x)) f (sin( x)) f (sin( x)) f (sin( x)) is odd (d) g (sec( x)) g (sec( x)) g (sec( x)) is even (e) | g ( x )| | g ( x)| | g ( x) | | g | is even 18 Let f (a x ) f (a x) and define g ( x) f ( x a) Then g ( x) f (( x) a ) f (a x) f (a x) f ( x a) g ( x) g ( x) f ( x a) is even 19 (a) The function is defined for all values of x, so the domain is (, ) (b) Since | x | attains all nonnegative values, the range is [2, ) 20 (a) Since the square root requires x 0, the domain is (,1] (b) Since x attains all nonnegative values, the range is [2, ) 21 (a) Since the square root requires 16 x 0, the domain is [4, 4] (b) For values of x in the domain, 16 x 16, so 16 x The range is [0, 4] 22 (a) The function is defined for all values of x, so the domain is (, ) (b) Since 32 x attains all positive values, the range is (1, ) 23 (a) The function is defined for all values of x, so the domain is (, ) (b) Since 2e x attains all positive values, the range is (3, ) 24 (a) The function is equivalent to y tan x, so we require x k2 for odd integers k The domain is given by x k4 for odd integers k (b) Since the tangent function attains all values, the range is (, ) 25 (a) The function is defined for all values of x, so the domain is (, ) (b) The sine function attains values from –1 to 1, so 2 2sin (3x ) and hence 3 sin (3x ) The range is [3, 1] 26 (a) The function is defined for all values of x, so the domain is (, ) (b) The function is equivalent to y x , which attains all nonnegative values The range is [0, ) 27 (a) The logarithm requires x 0, so the domain is (3, ) (b) The logarithm attains all real values, so the range is (, ) 28 (a) The function is defined for all values of x, so the domain is (, ) (b) The cube root attains all real values, so the range is (, ) Copyright 2014 Pearson Education, Inc Chapter Practice Exercises 29 (a) (b) (c) (d) Increasing because volume increases as radius increases Neither, since the greatest integer function is composed of horizontal (constant) line segments Decreasing because as the height increases, the atmospheric pressure decreases Increasing because the kinetic (motion) energy increases as the particles velocity increases 30 (a) Increasing on [2, ) (c) Increasing on (, ) (b) Increasing on [1, ) (d) Increasing on 12 , 31 (a) The function is defined for 4 x 4, so the domain is [4, 4] (b) The function is equivalent to y | x |, x 4, which attains values from to for x in the domain The range is [0, 2] 32 (a) The function is defined for 2 x 2, so the domain is [2, 2] (b) The range is [1, 1] 1 33 First piece: Line through (0, 1) and (1, 0) m 11 1 y x x Second piece: Line through (1, 1) and (2, 0) m 1 x, x f ( x) x, x 1 1 1 1 y ( x 1) x x 50 52 y 52 x 20 05 0) m 25 52 34 First piece: Line through (0, 0) and (2, 5) m Second piece: Line through (2, 5) and (4, y 52 ( x 2) 52 x 10 10 52x x, x (Note: x can be included on either piece.) f ( x) 5x 10 , x 35 (a) ( f g )(1) f ( g (1)) f f (1) 11 (b) ( g f )(2) g ( f (2)) g 12 11 or 25 2.5 ( f f )( x) f ( f ( x)) f 1x 1/1x x, x (c) (d) ( g g )( x) g ( g ( x)) g 36 (a) ( f g )(1) f ( g (1)) f x2 1 2 x2 x2 1 x 1 1 f (0) (b) ( g f )(2) f ( g (2)) g (2 2) g (0) (c) ( f f )( x) f ( f ( x)) f (2 x) (2 x) x (d) ( g g )( x) g ( g ( x)) g 37 (a) ( f g )( x) f ( g ( x)) f x 1 3 x 1 1 x2 2 ( g f )( x) g ( f ( x)) g (2 x ) (b) Domain of f g : [2, ) Domain of g f : [2, 2] x2 x, x 2 x2 x2 (c) Range of f g : (, 2] Range of g f : [0, 2] Copyright 2014 Pearson Education, Inc 33 34 Chapter Functions 1 x ( g f )( x) g ( f ( x)) g x 38 (a) ( f g )( x) f ( g ( x)) f x x x (b) Domain of f g : (, 1] Domain of g f : [0, 1] 39 (c) Range of f g : [0, ) Range of g f : [0, 1] y ( f f )( x) y f ( x) 40 41 42 The graph of f ( x) f1 (| x |) is the same as the graph of f1 ( x) to the right of the y-axis The graph of f ( x) to the left of the y-axis is the reflection of y f1 ( x), x across the y-axis It does not change the graph Copyright 2014 Pearson Education, Inc Chapter Practice Exercises 35 44 43 Whenever g1 ( x) is positive, the graph of y g ( x) | g1 ( x)| is the same as the graph of y g1 ( x) When g1 ( x) is negative, the graph of y g ( x) is the reflection of the graph of y g1 ( x ) across the x-axis Whenever g1 ( x) is positive, the graph of y g ( x) g1 ( x) is the same as the graph of y g1 ( x) When g1 ( x) is negative, the graph of y g ( x) is the reflection of the graph of y g1 ( x ) across the x-axis 45 46 The graph of f ( x) f1 (| x |) is the same as the graph of f1 ( x) to the right of the y-axis The graph of f ( x) to the left of the y-axis is the reflection of y f1 ( x), x across the y-axis Whenever g1 ( x) is positive, the graph of y g ( x) | g1 ( x)| is the same as graph of y g1 ( x) When g1 ( x) is negative, the graph of y g ( x) is the reflection of the graph of y g1 ( x ) across the x-axis 47 48 The graph of f ( x) f1 (| x |) is the same as the graph of f1 ( x) to the right of the y-axis The graph of f ( x) to the left of the y-axis is the reflection of y f1 ( x), x across the y-axis 49 (a) y g ( x 3) 12 (c) y g ( x) (e) y g ( x) The graph of f ( x) f1 (| x |) is the same as the graph of f1 ( x) to the right of the y-axis The graph of f ( x) to the left of the y-axis is the reflection of y f1 ( x), x across the y-axis (b) y g x 23 (d) y g ( x) (f ) y g (5 x) Copyright 2014 Pearson Education, Inc 36 Chapter Functions 50 (a) (c) (d) (e) (f ) (b) Horizontally compress the graph of f by a factor of Shift the graph of f right units Horizontally compress the graph of f by a factor of and then reflect the graph about the y-axis Horizontally compress the graph of f by a factor of and then shift the graph left 12 unit Horizontally stretch the graph of f by a factor of and then shift the graph down units Vertically stretch the graph of f by a factor of 3, then reflect the graph about the x-axis, and finally shift the graph up 14 unit 51 Reflection of the graph of y x about the x-axis followed by a horizontal compression by a factor of then a shift left units 52 Reflect the graph of y x about the x-axis, followed by a vertical compression of the graph by a factor of 3, then shift the graph up unit 53 Vertical compression of the graph of y 12 by a x factor of 2, then shift the graph up unit 54 Reflect the graph of y x1/3about the y-axis, then compress the graph horizontally by a factor of Copyright 2014 Pearson Education, Inc Chapter Practice Exercises 56 55 period 4 period 57 58 period period 59 60 period 2 period 2 61 (a) sin B sin 3 b c b2 b 2sin 3 23 By the theorem of Pythagoras, a b2 c a c b (b) sin B sin 3 b c 2c c Thus, a c b2 sin 3 (2) 62 (a) sin A ac a c sin A (b) tan A ab a b tan A 63 (a) tan B ba a tanb B (b) sin A ac c sina A 64 (a) sin A ac c2 b2 (b) sin A ac c Copyright 2014 Pearson Education, Inc 43 37 38 Chapter Functions 65 Let h height of vertical pole, and let b and c denote the distances of points B and C from the base of the pole, measured along the flat ground, respectively Then, tan 50 hc , tan 35 bh , and b c 10 Thus, h c tan 50and h b tan 35 (c 10) tan 35 c tan 50 (c 10) tan 35 c(tan 50 tan 35) 10 tan 35 c 10 tan 35 h c tan 50 tan 50 tan 35 tan 35 tan 50 16.98 m 10 tan 50 tan 35 66 Let h height of balloon above ground From the figure at the right, tan 40 ah , tan 70 bh , and a b Thus, h b tan 70 h (2 a) tan 70 and h a tan 40 (2 a) tan 70 a tan 40 a (tan 40 tan 70) tan 70 tan 70 a tan 40 h a tan 40 tan 70 tan 70 tan 40 1.3 km tan 40 tan 70 67 (a) (b) The period appears to be 4 (c) f ( x 4 ) sin( x 4 ) cos x 4 sin( x 2 ) cos x 2 sin x cos 2x since the period of sine and cosine is 2 Thus, f (x) has period 4 68 (a) (b) D ( ,0) (0, ); R [ 1, 1] But then f 21 kp sin (1/(21)) kp Choose k so large that 21 kp 1 1/(2 ) kp (c) f is not periodic For suppose f has period p Then f 21 kp f 21 sin 2 for all integers k which is a contradiction Thus f has no period, as claimed Copyright 2014 Pearson Education, Inc Chapter Additional and Advanced Exercises 39 CHAPTER ADDITIONAL AND ADVANCED EXERCISES There are (infinitely) many such function pairs For example, f ( x ) 3x and g ( x) x satisfy f ( g ( x)) f (4 x) 3(4 x) 12 x 4(3x) g (3x) g ( f ( x)) Yes, there are many such function pairs For example, if g ( x) (2 x 3)3 and f ( x) x1/3, then ( f g )( x) f ( g ( x)) f ((2 x 3)3 ) ((2 x 3)3 )1/3 x 3 If f is odd and defined at x, then f ( x) f ( x) Thus g ( x) f ( x) f ( x) whereas g ( x) ( f ( x) 2) f ( x ) Then g cannot be odd because g ( x) g ( x) f ( x) 2 f ( x) 0, which is a contradiction Also, g ( x) is not even unless f ( x ) for all x On the other hand, if f is even, then g ( x) f ( x) is also even: g ( x) f ( x) f ( x) g ( x) If g is odd and g(0) is defined, then g (0) g ( 0) g (0) Therefore, g (0) g (0) For (x, y) in the 1st quadrant, | x | | y | x x y x y For (x, y) in the 2nd quadrant, | x | | y | x x y x y x In the 3rd quadrant, | x | | y | x x y x y 2 x In the 4th quadrant, | x | | y | x x ( y ) x y 1 The graph is given at the right We use reasoning similar to Exercise (1) 1st quadrant: y | y | x | x | y x y x (2) 2nd quadrant: y | y | x | x | y x ( x) y (3) 3rd quadrant: y | y | x | x | y ( y ) x ( x) all points in the 3rd quadrant satisfy the equation (4) 4th quadrant: y | y | x | x | y ( y ) x x Combining these results we have the graph given at the right: sin x cos x (a) sin x cos x sin x cos x (1 cos x) (1 cos x) 1 cos x cos x sin x (b) Using the definition of the tangent function and the double angle formulas, we have tan 2x cos x x x 2 cos2 2x cos2 2 2x 11 cos cos x sin 2 The angles labeled in the accompanying figure are equal since both angles subtend arc CD Similarly, the two angles labeled are equal since they both subtend arc AB Thus, triangles AED and BEC are similar which ac 2a cos b implies b a c (a c)(a c) b(2a cos b) a c 2ab cos b c a b 2ab cos Copyright 2014 Pearson Education, Inc x sincos x 40 Chapter Functions As in the proof of the law of sines of Section 1.3, Exercise 61, ah bc sin A ab sin C ac sin B the area of ABC 12 (base)(height) 12 ah 12 bc sin A 12 ab sin C 12 ac sin B 10 As in Section 1.3, Exercise 61, (Area of ABC )2 14 (base)2 (height)2 14 a h 14 a b sin C a2 b2 c2 Thus, 14 a 2b (1 cos C ) By the law of cosines, c a b 2ab cos C cos C ab 2 2 a b2 c 2 a 2b ( a b c ) (area of ABC )2 14 a 2b (1 cos C ) 14 a b 1 ab a 2b 4a 2b ( a b c ) [(2ab ( a b c )) (2ab ( a b c ))] 16 16 16 2 [(( a b) c)(( a b) c)(c ( a b))(c ( a b))] [((a b) c )(c (a b)2 )] 16 a b c a b c a b c a b c abc s ( s a )( s b)( s c), where s 2 2 Therefore, the area of ABC equals s ( s a)( s b)( s c) 11 If f is even and odd, then f ( x) f ( x) and f ( x) f ( x) f ( x) f ( x ) for all x in the domain of f Thus f ( x) f ( x) 12 (a) As suggested, let E ( x) f ( x) f ( x) E ( x) function Define O ( x) f ( x) E ( x) f ( x) f ( x) f ( x) f ( x) f ( x) f ( x ) f ( ( x )) f ( x) f ( x) O ( x ) O f ( x) f ( x) f ( x) f ( x) Then E ( x) E is an even O( x) f ( x) f ( x ) is an odd function f ( x) E ( x ) O( x) is the sum of an even and an odd function (b) Part (a) shows that f ( x) E ( x) O ( x) is the sum of an even and an odd function If also f ( x) E1 ( x) O1 ( x), where E1 is even and O1 is odd, then f ( x) f ( x) ( E1 ( x) O1 ( x)) ( E ( x) O( x)) Thus, E ( x) E1 ( x) O1 ( x) O ( x) for all x in the domain of f (which is the same as the domain of E E1 and O O1) Now ( E E1 )( x) E ( x) E1 ( x) E ( x) E1 ( x) (since E and E1 are even) ( E E1 )( x) E E1 is even Likewise, (O1 O )( x) O1 ( x) O( x) O1 ( x) (O ( x)) (since O and O1 are odd) (O1 ( x) O( x)) (O1 O ) ( x) O1 O is odd Therefore, E E1 and O1 O are both even and odd so they must be zero at each x in the domain of f by Exercise 11 That is, E1 E and O1 O, so the decomposition of f found in part (a) is unique 2 2 13 y ax bx c a x ba x b 4ba c a x 2ba 4ba c 4a (a) If a the graph is a parabola that opens upward Increasing a causes a vertical stretching and a shift of the vertex toward the y-axis and upward If a the graph is a parabola that opens downward Decreasing a causes a vertical stretching and a shift of the vertex toward the y-axis and downward (b) If a the graph is a parabola that opens upward If also b 0, then increasing b causes a shift of the graph downward to the left; if b 0, then decreasing b causes a shift of the graph downward and to the right If a the graph is a parabola that opens downward If b 0, increasing b shifts the graph upward to the right If b 0, decreasing b shifts the graph upward to the left (c) Changing c (for fixed a and b) by c shifts the graph upward c units if c 0, and downward c units if c 14 (a) If a 0, the graph rises to the right of the vertical line x b and falls to the left If a < 0, the graph falls to the right of the line x b and rises to the left If a 0, the graph reduces to the horizontal line y c As | a | increases, the slope at any given point x x0 increases in magnitude and the graph becomes steeper As | a | decreases, the slope at x0 decreases in magnitude and the graph rises or falls more gradually (b) Increasing b shifts the graph to the left; decreasing b shifts it to the right (c) Increasing c shifts the graph upward; decreasing c shifts it downward Copyright 2014 Pearson Education, Inc Chapter Additional and Advanced Exercises 41 15 Each of the triangles pictured has the same base b vt v (1 sec) Moreover, the height of each triangle is the same value h Thus 12 (base)(height) 12 bh A1 A2 A3 … In conclusion, the object sweeps out equal areas in each one second interval 16 (a) Using the midpoint formula, the coordinates of P are y a0 b0 , 2 , Thus the slope a b 2 of OP x ba /2 ba /2 b0 (b) The slope of AB a ba The line segments AB and OP are perpendicular when the product of their ba ba slopes is 1 ba 2 Thus, b a a b (since both are positive) Therefore, AB is perpendicular to OP when a b 17 From the figure we see that 2 and AB AD From trigonometry we have the following: sin We can see that: sin EB EB, cos AE AE , tan CD CD, and tan EB cos AB AB AD AE area ADC ( AE )( EB ) ( AD )2 ( AD ) (CD) area AEB area sector DB 2 sin 12 sin cos 12 (1)2 12 (1)(tan ) 12 sin cos 12 12 cos 18 ( f g )( x) f ( g ( x)) a (cx d ) b acx ad b and ( g f )( x) g ( f ( x )) c (ax b) d acx cb d Thus ( f g )( x ) ( g f )( x) acx ad b acx bc d ad b bc d Note that f (d ) ad b and g (b) cb d , thus ( f g )( x) ( g f )( x) if f (d ) g (b) Copyright 2014 Pearson Education, Inc 42 Chapter Functions Copyright 2014 Pearson Education, Inc ... 1] by [ 2, 4] 19 [ 4, 4] by [0, 3] 20 [ 5, 5] by [ 2, 2] 21 [ 10, 10] by [ 6, 6] 22 [ 5, 5] by [ 2, 2] 23 [ 6, 10] by [ 6, 6] 24 [ 3, 5] by [ 2, 10] 25 [0.03, 0.03] by. .. 4] by [ 8, 8] 13 [ 1, 6] by [ 1, 4] 14 [ 1, 6] by [ 1, 5] 15 [ 3, 3] by [0, 10] 16 [ 1, 2] by [0, 1] Copyright 2014 Pearson Education, Inc 27 28 Chapter Functions 17 [ 5, 1] by. .. 2, 6] by [ 250, 50] [ 1, 5] by [ 5, 30] Copyright 2014 Pearson Education, Inc Section 1.4 Graphing with Software [ 4, 4] by [ 5, 5] 10 [ 2, 2] by [ 2, 8] 11 [ 2, 6] by [
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