Solution manual for calculus and its applications 13th edition by goldstein

Chapter Functions 0.1 Functions and Their Graphs 16 h( s ) = s (1 + s ) ⎛1⎞ h⎜ ⎟ = = ⎝ ⎠ 1+ ( h(a + 1) = 17 [–1, 0) 12 ⎡⎣ 2, ∞ ) f ( x) = x − 3x f (0) = − 3(0) = f (5) = − 3(5) = 25 − 15 = 10 f (3) = − 3(3) = − = f (−7) = (−7) − 3(−7) = 49 + 21 = 70 f ( x) = x + x − x − f ( x) = x − x f (a + 2) = (a + 2) − 2(a + 2) = (a + 4a + 4) − 2a − = a + 2a 10 [–1, 8) (−∞,3) 18 f ( x) = x + x + f (a − 1) = (a − 1) + 4(a − 1) + = (a − 2a + 1) + (4a − 4) + = a + 2a f (a − 2) = (a − 2) + 4(a − 2) + = (a − 4a + 4) + (4a − 8) + = a2 −1 19 a f (1) = 13 + 12 − − = f (−1) = (−1) + (−1) − (−1) − = b ⎛1⎞ ⎛1⎞ ⎛1⎞ ⎛1⎞ f ⎜ ⎟ = ⎜ ⎟ + ⎜ ⎟ − ⎜ ⎟ −1 = − ⎝2⎠ ⎝2⎠ ⎝2⎠ ⎝2⎠ f (a) = a + a − a − 15 g (t ) = t − 3t + t g (2) = − 3(2) + = − 12 + = −2 3 ⎛2⎞ ⎛ 2⎞ ⎛2⎞ ⎛2⎞ g ⎜ ⎟ = ⎜ ⎟ − 3⎜ ⎟ + ⎜ ⎟ ⎝3⎠ ⎝ 3⎠ ⎝3⎠ ⎝3⎠ 12 10 = − + =− ≈ −.37037 27 27 f (0) represents the number of laptops sold in 2010 f (6) = 150 + 2(6) + = 150 + 12 + 36 = 198 100 x , x≥0 b+x b = 20, x = 60 100(60) R(60) = = 75 20 + 60 The solution produces a 75% response 20 R( x) = a ⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞ g ⎜− ⎟ = ⎜− ⎟ − 3⎜− ⎟ + ⎜− ⎟ ⎝ 2⎠ ⎝ 2⎠ ⎝ 2⎠ ⎝ 2⎠ 11 =− − − =− 8 a +1 a +1 = + (a + 1) a + f (a + 1) = (a + 1) − 2(a + 1) = (a + 2a + 1) − 2a − = a − 3⎞ ⎛ ⎜ −1, ⎟ ⎝ 2⎠ [2, 3) 14 ( 2) 13 = − 32 − 32 ⎛ 3⎞ = =3 h ⎜− ⎟ = ⎝ ⎠ 1+ − − 12 11 ) b If R(50) = 60, then 100(50) 60 = b + 50 60b + 3000 = 5000 100 b= This particular frog has a positive constant of 33.3 g (a ) = a − 3a + a Copyright © 2014 Pearson Education Inc Chapter Functions 21 22 8x ( x − 1)( x − 2) all real numbers such that x ≠ 1, or (−∞, −1) ∪ (−1, 2) ∪ (2, ∞ ) f ( x) = t all real numbers such that t > or (0, ∞ ) f (t ) = 23 g ( x) = 3− x all real numbers such that x < or (−∞, −3) x( x + 2) all real numbers such that x ≠ 0, –2 or (−∞, −2) ∪ (−2, 0) ∪ (0, ∞ ) 24 g ( x) = 25 26 27 f ( x) = 3x − 39 [−1, 3] 41 (−∞, − 1] ∪ [5, 9] 43 f (1) ≈ 03; f (5) ≈ 037 44 f (6) ≈ 03 45 [0, 05] 47 1⎞ ⎛ f ( x) = ⎜ x − ⎟ ( x + 2) ⎝ 2⎠ 2x + 1− x The denominator is greater than for all x < 1, and the numerator is defined for all x ≥ − 1 ⎡ ⎞ Thus, the domain is − ≤ x < 1, or ⎢ − , 1⎟ ⎣ ⎠ 29 function 30 not a function 31 not a function 32 not a function 33 not a function 34 function 35 f (0) = 1; f (7 ) = −1 36 f (2) = 3; f (−1) = 37 positive = ⎛1 2⎞ So ⎜ , ⎟ is on the graph ⎝2 5⎠ 50 g ( x) = ( x + 4) ( x + 2) ⎛2⎞ g⎜ ⎟= ⎝3⎠ f ( x) = 2x + + x f ( x) = x2 + () () The domain consists of all real numbers greater than or equal to 0, or [ 0, ∞ ) 28 3x − 1 −1 ⎛1⎞ g ⎜ ⎟ = 22 = ⎝3⎠ 1 + 3x + The domain consists of all real numbers, or (−∞, ∞ ) 46 t ≈ 48 f(x) = x(5 + x)(4 – x) f(–2) = –2(5 + (–2))(4 – (–2)) = –36 So (–2, 12) is not on the graph 49 g ( x) = [ −1, 5] ∪ [9, ∞ ] 42 1⎞ 25 ⎛ f (3) = ⎜ − ⎟ (3 + 2) = ⎝ 2⎠ Thus, (3, 12) is not on the graph x2 + x − The denominator is zero for x = −3 and x = 2, so the domain consists of all real numbers such that x ≠ −3, or (−∞, − 3) ∪ (−3, 2) ∪ (2, ∞ ) f ( x) = 40 −1, 5, ( 32 ) +4 +2 = 40 = ⎛ 5⎞ So ⎜ , ⎟ is on the graph ⎝ 3⎠ 51 f ( x) = x f (a + 1) = (a + 1) 52 ⎛5⎞ f ( x) = ⎜ ⎟ − x ⎝x⎠ − (2 + h) (2 + h) − (2 + h) − 4h − h = = (2 + h) 2+h f (2 + h) = 53 ⎧⎪ x for ≤ x < f ( x) = ⎨ ⎪⎩1 + x for ≤ x ≤ f (1) = = 1; f (2) = + = f (3) = + = 38 negative Copyright © 2014 Pearson Education Inc Section 0.1 Functions and Their Graphs 54 60 Entering Y1 = 1/X + will graph the function f ( x) = + In order to graph the function x f ( x) = , you need to include parentheses x +1 in the denominator: Y1 = 1/(X + 1) ⎧1 for ≤ x ≤ ⎪ f ( x) = ⎨ x ⎪ x for < x ⎩ 1 f (1) = = ; f (2) = f (3) = 32 = 61 Entering Y1 = X ^ / will graph the function ⎧π x for x < ⎪ f ( x) = ⎨1 + x for ≤ x ≤ 2.5 ⎪4 x for 2.5 < x ⎩ 55 f ( x) = x3 In order to graph the function y = x , you need to include parentheses in the exponent: Y1 = X ^ (3/4) f (1) = π (1) = π f(2) = + = f(3) = 4(3) = 12 62 Y1 = X^3 − 33X^2 +120X+1500 ⎧ for x < ⎪ 4− x ⎪ 56 f ( x) = ⎨ x for ≤ x < ⎪ ⎪ x − for ≤ x ⎩ f (1) = =1 −1 f(2) = 2(2) = [−8, 30] by [−2000, 2000] 63 Y1 = −X^2+2x+2 f (3) = 32 − = = 57 for 50 ≤ x ≤ 300 ⎧.06 x ⎪ f ( x) = ⎨.02 x + 12 for 300 < x ≤ 600 ⎪.015 x + 15 for 600 < x ⎩ [−2, 4] by [−8, 5] 64 Y1 = (X+1)^(1/2) 58 [0, 10] by [−1, 4] 65 Y1 = 1/(X^2+1) 59 [−4, 4] by [−.5, 1.5] Copyright © 2014 Pearson Education Inc Chapter Functions 0.2 Some Important Functions y = x − x y 1 –1 –1 –3 x = −2 x + f ( x) = x − x y −1 1 y = x = y = x + x y x − y = –1 –2 f ( x) = 3x + x y 1 0 –1 –1 y = − x x−4 y –5 –4 –2 –3 Copyright © 2014 Pearson Education Inc Section 0.2 Some Important Functions x + y = −1 x 12 y −5 –2 −3 x −1 f (0) = − (0) − = −1 The y-intercept is (0, –1) 1 − x − = ⇒ − x = ⇒ x = −2 2 The x-intercept is (–2, 0) f ( x) = − 13 f(x) = The y-intercept is (0, 5) There is no x-intercept 14 f(x) = 14 The y-intercept is (0, 14) There is no x-intercept x = y − x y 1 15 x − y = 0 − 5y = ⇒ y = The x- and y-intercept is (0, 0) 16 + x = y –3 –1 + (0) = y ⇒ y = The y-intercept is (0, 1) + 3x = (0) ⇒ 3x = −2 ⇒ x = − ⎛ The x-intercept is ⎜ − , ⎝ 10 x y + =1 x y 0 –2 17 ⎞ 0⎟ ⎠ ⎛K⎞ f ( x) = ⎜ ⎟ x + ⎝V ⎠ V a f(x) = 2x + 50 K 1 = and = 50 If = 50, We have V V V K = implies then V = Now, V 50 1 K = = 2, so K = ⋅ 50 250 50 b ⎛K ⎞ y= ⎜ ⎟x+ , ⎝V ⎠ V 1 ⎛K⎞ ⎜⎝ ⎟⎠ ⋅ + = , so the V V V ⎛ 1⎞ y-intercept is ⎜ 0, ⎟ ⎝ V⎠ 11 f ( x) = x + f ( ) = (9 ) + = The y-intercept is (0, 3) ⎛K ⎞ Solving ⎜ ⎟ x + = 0, we get ⎝V ⎠ V x + = ⇒ x = −3 ⇒ x = − ⎛ ⎞ The x-intercept is ⎜ − , ⎟ ⎝ ⎠ K 1 x = − ⇒ x = − , so the x-intercept V V K ⎛ ⎞ is ⎜ − , ⎟ ⎝ K ⎠ Copyright © 2014 Pearson Education Inc Chapter Functions ⎛ ⎞ 18 From 17(b), ⎜ − , ⎟ is the x-intercept From ⎝ K ⎠ the experimental data, (–500, 0) is also the 1 x-intercept Thus − = −500, K = K 500 ⎛ 1⎞ Again from 17(b), ⎜ 0, ⎟ is the y-intercept ⎝ V⎠ From the experimental data, (0, 60) is also the 1 y-intercept Thus = 60, V = V 60 19 a b Cost is $(24 + 200(.25)) = $74 28 y = − x + x a = 4, b = –2, c = 29 y = − x a = –1, b = 0, c = 1 x + 3x−π a = , b = 3, c = −π 30 y = 31 f(x) = 25x + 24 0≤x≤1 20 Let x be the volume of gas (in thousands of cubic feet) extracted f(x) = 5000 + 10x 21 Let x be the number of days of hospital confinement f(x) = 700x + 1900 22 x − 40 = 350 ⇒ x = 65 mph 23 50 x , ≤ x ≤ 100 105 − x From example 6, we know that f(70) = 100 The cost to remove 75% of the pollutant is 50 ⋅ 75 f (75) = = 125 105 − 75 The cost of removing an extra 5% is $125 − $100 = $25 million To remove the final 5% the cost is f(100) – f(95) = 1000 – 475 = $525 million This costs 21 times as much as the cost to remove the next 5% after the first 70% is removed for ≤ x ≤ ⎧3x ⎪ f ( x) = ⎨ ⎪⎩ − x for x > x>1 x f ( x ) = 3x x 0 3 f ( x) = − x 2 f ( x) = 24 a b f (85) = 20(85) = $100 million 102 − 85 32 f ( x) = { + x for x ≤ for x > x≤3 x>3 x f ( x) = + x x f ( x) = 4 4 f(100) – f(95) = 1000 – 271.43 ≈ $728.57 million 25 y = 3x − x a = 3, b = –4, c = x − 6x + 2 = x − 2x + 3 a = , b = –2, c = 3 26 y = 27 y = 3x − x + a = –2, b = 3, c = Copyright © 2014 Pearson Education Inc Section 0.2 Some Important Functions 33 f ( x) = { 2x + for x < for x ≥ x≥3 x≥2 x