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Solution Manual For Calculus, 6th Edition

Struggling with textbook exercises? Solution Manual For Calculus, 6th Edition breaks down solutions in a way that�s easy to understand.

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    Section 1.1 C01S01.001: If f ( x ) = 1 x , then: (a) f ( − a ) = 1 − a = − 1 a ; (b) f ( a − 1 ) = 1 a − 1 = a ; (c) f ( √ a ) = 1 √ a = 1 a 1 / 2 = a − 1 / 2 ; (d) f ( a 2 ) = 1 a 2 = a − 2 . C01S01.002: If f ( x ) = x 2 + 5, then: (a) f ( − a ) = ( − a ) 2 + 5 = a 2 + 5; (b) f ( a − 1 ) = ( a − 1 ) 2 + 5 = a − 2 + 5 = 1 a 2 + 5 = 1 + 5 a 2 a 2 ; (c) f ( √ a ) = ( √ a ) 2 + 5 = a + 5; (d) f ( a 2 ) = ( a 2 ) 2 + 5 = a 4 + 5. C01S01.003: If f ( x ) = 1 x 2 + 5 , then: (a) f ( − a ) = 1 ( − a ) 2 + 5 = 1 a 2 + 5 ; (b) f ( a − 1 ) = 1 ( a − 1 ) 2 + 5 = 1 a − 2 + 5 = 1 · a 2 a − 2 · a 2 + 5 · a 2 = a 2 1 + 5 a 2 ; (c) f ( √ a ) = 1 ( √ a ) 2 + 5 = 1 a + 5 ; (d) f ( a 2 ) = 1 ( a 2 ) 2 + 5 = 1 a 4 + 5 . C01S01.004: If f ( x ) = √ 1 + x 2 + x 4 , then: (a) f ( − a ) = √ 1 + ( − a ) 2 + ( − a ) 4 = √ 1 + a 2 + a 4 ; (b) f ( a − 1 ) = √ 1 + ( a − 1 ) 2 + ( a − 1 ) 4 = √ 1 + a − 2 + a − 4 = √ ( a 4 ) · (1 + a − 2 + a − 4 ) a 4 = √ a 4 + a 2 + 1 a 4 = √ a 4 + a 2 + 1 √ a 4 = √ a 4 + a 2 + 1 a 2 ; (c) f ( √ a ) = √ 1 + ( √ a ) 2 + ( √ a ) 4 = √ 1 + a + a 2 ; (d) f ( a 2 ) = √ 1 + ( a 2 ) 2 + ( a 4 ) 2 = √ 1 + a 4 + a 8 . C01S01.005: If g ( x ) = 3 x + 4 and g ( a ) = 5, then 3 a + 4 = 5, so 3 a = 1; therefore a = 1 3 . C01S01.006: If g ( x ) = 1 2 x − 1 and g ( a ) = 5, then: 1

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    1 2 a − 1 = 5; 1 = 5 · (2 a − 1); 1 = 10 a − 5; 10 a = 6; a = 3 5 . C01S01.007: If g ( x ) = √ x 2 + 16 and g ( a ) = 5, then: √ a 2 + 16 = 5; a 2 + 16 = 25; a 2 = 9; a = 3 or a = − 3 . C01S01.008: If g ( x ) = x 3 − 3 and g ( a ) = 5, then a 3 − 3 = 5, so a 3 = 8. Hence a = 2. C01S01.009: If g ( x ) = 3 √ x + 25 = ( x + 25) 1 / 3 and g ( a ) = 5, then ( a + 25) 1 / 3 = 5; a + 25 = 5 3 = 125; a = 100 . C01S01.010: If g ( x ) = 2 x 2 − x + 4 and g ( a ) = 5, then: 2 a 2 − a + 4 = 5; 2 a 2 − a − 1 = 0; (2 a + 1)( a − 1) = 0; 2 a + 1 = 0 or a − 1 = 0; a = − 1 2 or a = 1 . C01S01.011: If f ( x ) = 3 x − 2, then f ( a + h ) − f ( a ) = [3( a + h ) − 2] − [3 a − 2] = 3 a + 3 h − 2 − 3 a + 2 = 3 h. C01S01.012: If f ( x ) = 1 − 2 x , then 2

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    f ( a + h ) − f ( a ) = [1 − 2( a + h )] − [1 − 2 a ] = 1 − 2 a − 2 h − 1 + 2 a = − 2 h. C01S01.013: If f ( x ) = x 2 , then f ( a + h ) − f ( a ) = ( a + h ) 2 − a 2 = a 2 + 2 ah + h 2 − a 2 = 2 ah + h 2 = h · (2 a + h ) . C01S01.014: If f ( x ) = x 2 + 2 x , then f ( a + h ) − f ( a ) = [( a + h ) 2 + 2( a + h )] − [ a 2 + 2 a ] = a 2 + 2 ah + h 2 + 2 a + 2 h − a 2 − 2 a = 2 ah + h 2 + 2 h = h · (2 a + h + 2) . C01S01.015: If f ( x ) = 1 x , then f ( a + h ) − f ( a ) = 1 a + h − 1 a = a a ( a + h ) − a + h a ( a + h ) = a − ( a + h ) a ( a + h ) = − h a ( a + h ) . C01S01.016: If f ( x ) = 2 x + 1 , then f ( a + h ) − f ( a ) = 2 a + h + 1 − 2 a + 1 = 2( a + 1) ( a + h + 1)( a + 1) − 2( a + h + 1) ( a + h + 1)( a + 1) = 2 a + 2 ( a + h + 1)( a + 1) − 2 a + 2 h + 2 ( a + h + 1)( a + 1) = (2 a + 2) − (2 a + 2 h + 2) ( a + h + 1)( a + 1) = 2 a + 2 − 2 a − 2 h − 2 ( a + h + 1)( a + 1) = − 2 h ( a + h + 1)( a + 1) . C01S01.017: If x > 0 then f ( x ) = x | x | = x x = 1 . If x < 0 then f ( x ) = x | x | = x − x = − 1 . We are given f (0) = 0, so the range of f is {− 1 , 0 , 1 } . That is, the range of f is the set consisting of the three real numbers − 1, 0, and 1. C01S01.018: Given f ( x ) = [[3 x ]], we see that 3

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    f ( x ) = 0 if 0  x < 1 3 , f ( x ) = 1 if 1 3  x < 2 3 , f (2) = 2 if 2 3  x < 1; moreover, f ( x ) = − 3 if − 1  x < − 2 3 , f ( x ) = − 2 if − 2 3  x < − 1 3 , f ( x ) = − 1 if − 1 3  x < 0 . In general, if m is any integer, then f ( x ) = 3 m if m  x < m + 1 3 , f ( x ) = 3 m + 1 if m + 1 3  x < m + 2 3 , f ( x ) = 3 m + 2 if m + 2 3  x < m + 1 . Because every integer is equal to 3 m or to 3 m + 1 or to 3 m + 2 for some integer m , we see that the range of f includes the set Z of all integers. Because f can assume no values other than integers, we can conclude that the range of f is exactly Z . C01S01.019: Given f ( x ) = ( − 1) [[ x ]] , we first note that the values of the exponent [[ x ]] consist of all the integers and no other numbers. So all that matters about the exponent is whether it is an even integer or an odd integer, for if even then f ( x ) = 1 and if odd then f ( x ) = − 1. No other values of f ( x ) are possible, so the range of f is the set consisting of the two numbers − 1 and 1. C01S01.020: If 0 < x  1, then f ( x ) = 34. If 1 < x  2 then f ( x ) = 34 + 21 = 55. If 2 < x  3 then f ( x ) = 34 + 2 · 21 = 76. We continue in this way and conclude with the observation that if 11 < x < 12 then f ( x ) = 34 + 11 · 21 = 265. So the range of f is the set { 34 , 55 , 76 , 97 , 118 , 139 , 160 , 181 , 202 , 223 , 244 , 265 } . C01S01.021: Given f ( x ) = 10 − x 2 , note that for every real number x , x 2 is defined, and for every such real number x 2 , 10 − x 2 is also defined. Therefore the domain of f is the set R of all real numbers. C01S01.022: Given f ( x ) = x 3 + 5, we note that for each real number x , x 3 is defined; moreover, for each such real number x 3 , x 3 + 5 is also defined. Thus the domain of f is the set R of all real numbers. C01S01.023: Given f ( t ) = √ t 2 , we observe that for every real number t , t 2 is defined and nonnegative, and hence that √ t 2 is defined as well. Therefore the domain of f is the set R of all real numbers. C01S01.024: Given g ( t ) = ( √ t ) 2 , we observe that √ t is defined exactly when t  0. In this case, ( √ t ) 2 is also defined, and hence the domain of g is the set [0 , + ∞ ) of all nonnegative real numbers. C01S01.025: Given f ( x ) = √ 3 x − 5, we note that 3 x − 5 is defined for all real numbers x , but that its square root will be defined when and only when 3 x − 5 is nonnegative; that is, when 3 x − 5  0, so that x  5 3 . So the domain of f consists of all those real numbers x in the interval [ 5 3 , + ∞ ) . 4

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    C01S01.026: Given g ( t ) = 3 √ t + 4 = ( t + 4) 1 / 3 , we note that t + 4 is defined for every real number t and the cube root of t + 4 is defined for every possible resulting value of t + 4. Therefore the domain of g is the set R of all real numbers. C01S01.027: Given f ( t ) = √ 1 − 2 t , we observe that 1 − 2 t is defined for every real number t , but that its square root is defined only when 1 − 2 t is nonnegative. We solve the inequality 1 − 2 t  0 to find that f ( t ) is defined exactly when t  1 2 . Hence the domain of f is the interval ( −∞ , 1 2 ] . C01S01.028: Given g ( x ) = 1 ( x + 2) 2 , we see that ( x + 2) 2 is defined for every real number x , but that g ( x ), its reciprocal, will be defined only when ( x + 2) 2  = 0; that is, when x + 2  = 0. So the domain of g consists of those real numbers x  = − 2. C01S01.029: Given f ( x ) = 2 3 − x , we see that 3 − x is defined for all real values of x , but that f ( x ), double its reciprocal, is defined only when 3 − x  = 0. So the domain of f consists of those real numbers x  = 3. C01S01.030: Given g ( t ) = √ 2 3 − t , it is necessary that 3 − t be both nonzero (so that its reciprocal is defined) and nonnegative (so that the square root is defined). Thus 3 − t > 0, and therefore the domain of g consists of those real numbers t < 3. C01S01.031: Given f ( x ) = √ x 2 + 9, observe that for each real number x , x 2 + 9 is defined and, moreover, is positive. So its square root is defined for every real number x . Hence the domain of f is the set R of all real numbers. C01S01.032: Given h ( z ) = 1 √ 4 − z 2 , we note that 4 − z 2 is defined for every real number z , but that its square root will be defined only if 4 − z 2  0. Moreover, the square root cannot be zero, else its reciprocal will be undefined, so we need to solve the inequality 4 − z 2 > 0; that is, z 2 < 4. The solution is − 2 < z < 2, so the domain of h is the open interval ( − 2 , 2). C01S01.033: Given f ( x ) = √ 4 − √ x , note first that we require x  0 in order that √ x be defined. In addition, we require 4 − √ x  0 so that its square root will be defined as well. So we solve [simultaneously] x  0 and √ x  4 to find that 0  x  16. So the domain of f is the closed interval [0 , 16]. C01S01.034: Given f ( x ) = √ x + 1 x − 1 , 5

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    we require that x  = 1 so that the fraction is defined. In addition we require that the fraction be nonnegative so that its square root will be defined. These conditions imply that both numerator and denominator be positive or that both be negative; moreover, the numerator may also be zero. But if the denominator is positive then the [larger] numerator will be positive as well; if the numerator is nonpositive then the [smaller] denominator will be negative. So the domain of f consists of those real numbers for which either x − 1 > 0 or x + 1  0; that is, either x > 1 or x  − 1. So the domain of f is the union of the two intervals ( −∞ , − 1] and (1 , + ∞ ). Alternatively, it consists of those real numbers x not in the interval ( − 1 , 1]. C01S01.035: Given: g ( t ) = t | t | . This fraction will be defined whenever its denominator is nonzero, thus for all real numbers t  = 0. So the domain of g consists of the nonzero real numbers; that is, the union of the two intervals ( −∞ , 0) and (0 , + ∞ ). C01S01.036: If a square has edge length x , then its area A is given by A = x 2 and its perimeter P is given by P = 4 x . To express A in terms of P : x = 1 4 P ; A = x 2 = ( 1 4 P ) 2 = 1 16 P 2 . Thus to express A as a function of P , we write A ( P ) = 1 16 P 2 , 0  P < + ∞ . (It will be convenient later in the course to allow the possibility that P , x , and A are zero. If this produces an answer that fails to meet real-world criteria for a solution, then that possibility can simply be eliminated when the answer to the problem is stated.) C01S01.037: If a circle has radius r , then its circumference C is given by C = 2 πr and its area A by A = πr 2 . To express C in terms of A , we first express r in terms of A , then substitute in the formula for C : A = πr 2 ; r = √ A π ; C = 2 πr = 2 π √ A π = 2 √ π 2 A π = 2 √ πA. Therefore to express C as a function of A , we write C ( A ) = 2 √ πA, 0  A < + ∞ . It is also permissible simply to write C ( A ) = 2 √ πA without mentioning the domain, because the “default” domain is correct. In the first displayed equation we do not write r = ± √ A/π because we know that r is never negative. C01S01.38: If r denotes the radius of the sphere, then its volume is given by V = 4 3 πr 3 and its surface area by S = 4 πr 2 . Hence 6

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    r = 1 2 √ S π ; V = 4 3 πr 3 = 4 3 π · 1 8 ( S π ) 3 / 2 = 1 6 π ( S π ) 3 / 2 . Answer: V ( S ) = 1 6 π ( S π ) 3 / 2 , 0  S < + ∞ . C01S01.039: To avoid decimals, we note that a change of 5 ◦ C is the same as a change of 9 ◦ F, so when the temperature is 10 ◦ C it is 32 + 18 = 50 ◦ F; when the temperature is 20 ◦ C then it is 32 + 2 · 18 = 68 ◦ F. In general we get the Fahrenheit temperature F by adding 32 to the product of 1 10 C and 18, where C is the Celsius temperature. That is, F = 32 + 9 5 C, and therefore C = 5 9 ( F − 32). Answer: C ( F ) = 5 9 ( F − 32) , F > − 459 . 67 . C01S01.040: Suppose that a rectangle has base length x and perimeter 100. Let h denote the height of such a rectangle. Then 2 x + 2 h = 100, so that h = 50 − x . Because x  0 and h  0, we see that 0  x  50. The area A of the rectangle is xh , so that A ( x ) = x (50 − x ) , 0  x  50 . C01S01.041: Let y denote the height of such a rectangle. The rectangle is inscribed in a circle of diameter 4, so the bottom side x and the left side y are the two legs of a right triangle with hypotenuse 4. Consequently x 2 + y 2 = 16, so y = √ 16 − x 2 (not − √ 16 − x 2 because y  0). Because x  0 and y  0, we see that 0  x  4. The rectangle has area A = xy , so A ( x ) = x √ 16 − x 2 , 0  x  4 . C01S042.042: We take the problem to mean that current production is 200 barrels per day per well, that if one new well is drilled then the 21 wells will produce 195 barrels per day per well; in general, that if x new wells are drilled then the 20 + x wells will produce 200 − 5 x barrels per day per well. So total production would be p = (20 + x )(200 − 5 x ) barrels per day. But because 200 − 5 x  0, we see that x  40. Because x  0 as well (you don’t “undrill” wells), here’s the answer: p ( x ) = 4000 + 100 x − 5 x 2 , 0  x  40 , x an integer . C01S01.043: The square base of the box measures x by x centimeters; let y denote its height (in centime- ters). Because the volume of the box is 324 cm 3 , we see that x 2 y = 324. The base of the box costs 2 x 2 cents, each of its four sides costs xy cents, and its top costs x 2 cents. So the total cost of the box is C = 2 x 2 + 4 xy + x 2 = 3 x 2 + 4 xy. (1) 7

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    Because x > 0 and y > 0 (the box has positive volume), but because y can be arbitrarily close to zero (as well as x ), we see also that 0 < x < + ∞ . We use the equation x 2 y = 324 to eliminate y from Eq. (1) and thereby find that C ( x ) = 3 x 2 + 1296 x , 0 < x < + ∞ . C01S01.044: If the rectangle is rotated around its side S of length x to produce a cylinder, then x will also be the height of the cylinder. Let y denote the length of the two sides perpendicular to S ; then y will be the radius of the cylinder; moreover, the perimeter of the original rectangle is 2 x + 2 y = 36. Hence y = 18 − x . Note also that x  0 and that x  18 (because y  0). The volume of the cylinder is V = πy 2 x , and so V ( x ) = πx (18 − x ) 2 , 0  x  18 . C01S01.045: Let h denote the height of the cylinder. Its radius is r , so its volume is πr 2 h = 1000. The total surface area of the cylinder is A = 2 πr 2 + 2 πrh (look inside the front cover of the book); h = 1000 πr 2 , so A = 2 πr 2 + 2 πr · 1000 πr 2 = 2 πr 2 + 2000 r . Now r cannot be negative; r cannot be zero, else πr 2 h  = 1000. But r can be arbitrarily small positive as well as arbitrarily large positive (by making h sufficiently close to zero). Answer: A ( r ) = 2 πr 2 + 2000 r , 0 < r < + ∞ . C01S01.046: Let y denote the height of the box (in centimeters). Then 2 x 2 + 4 xy = 600 , so that y = 600 − 2 x 2 4 x . (1) The volume of the box is V = x 2 y = (600 − 2 x 2 ) · x 2 4 x = 1 4 (600 x − 2 x 3 ) = 1 2 (300 x − x 3 ) by Eq. (1). Also x > 0 by Eq. (1), but the maximum value of x is attained when Eq. (1) forces y to be zero, at which point x = √ 300 = 10 √ 3. Answer: V ( x ) = 300 x − x 3 2 , 0 < x  10 √ 3 . C01S01.047: The base of the box will be a square measuring 50 − 2 x in. on each side, so the open-topped box will have that square as its base and four rectangular sides each measuring 50 − 2 x by x (the height of the box). Clearly 0  x and 2 x  50. So the volume of the box will be V ( x ) = x (50 − 2 x ) 2 , 0  x  25 . 8

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    C01S01.048: Recall that A ( x ) = x (50 − x ), 0  x  50. Here is a table of a few values of the function A at some special numbers in its domain: x 0 5 10 15 20 25 30 35 40 45 50 A 0 225 400 525 600 625 600 525 400 225 0 It appears that when x = 25 (so the rectangle is a square), the rectangle has maximum area 625. C01S01.049: Recall that the total daily production of the oil field is p ( x ) = (20 + x )(200 − 5 x ) if x new wells are drilled (where x is an integer satisfying 0  x  40). Here is a table of all of the values of the production function p : x 0 1 2 3 4 5 6 7 p 4000 4095 4180 4255 4320 4375 4420 4455 x 8 9 10 11 12 13 14 15 p 4480 4495 4500 4495 4480 4455 4420 4375 x 16 17 18 19 20 21 22 23 p 4320 4255 4180 4095 4000 3895 3780 3655 x 24 25 26 27 28 29 30 31 p 3520 3375 3220 3055 2880 2695 2500 2295 x 32 33 34 35 36 37 38 39 p 2080 1855 1620 1375 1120 855 580 295 and, finally, p (40) = 0. Answer: Drill ten new wells. C01S01.050: The surface area A of the box of Example 8 was A ( x ) = 2 x 2 + 500 x , 0 < x < ∞ . The restrictions x  1 and y  1 imply that 1  x  √ 125. A small number of values of A , rounded to three places, are given in the following table. x 1 2 3 4 5 6 7 8 9 10 11 A 502 258 185 157 150 155 169 191 218 250 287 It appears that A is minimized when x = y = 5. C01S01.051: If x is an integer, then Ceiling ( x ) = x and − Floor ( − x ) = − ( − x ) = x . If x is not an integer, then choose the integer n so that n < x < n + 1. Then Ceiling ( x ) = n + 1, − ( n + 1) < − x < − n , and − Floor ( − x ) = − [ − ( n + 1)] = n + 1 . In both cases we see that Ceiling ( x ) = − Floor ( − x ). C01S01.052: The range of Round ( x ) is the set Z of all integers. If k is a nonzero constant, then as x varies through all real number values, so does kx . Hence the range of Round ( kx ) is Z if k  = 0. If k = 0 then the range of Round ( kx ) consists of the single number zero. 9

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    C01S01.053: By the result of Problem 52, the range of Round (10 x ) is the set of all integers, so the range of g ( x ) = 1 10 Round (10 x ) is the set of all integral multiple of 1 10 . C01S01.054: What works for π will work for every real number; let Round 2( x ) = 1 100 Round (100 x ). To be certain that this is correct (we will verify it only for positive numbers), write the [positive] real number x in the form x = k + t 10 + h 100 + m 1000 + r, where k is a nonnnegative integer, t (the “tenths” digit) is a nonnegative integer between 0 and 9, h (the “hundredths” digit) is a nonnegative integer between 0 and 9, as is m , and 0  r < 0 . 001. Then Round 2( x ) = 1 100 Floor (100 x + 0 . 5) = 1 100 Floor (100 k + 10 t + h + 1 10 ( m + 5) + 100 r ) . If 0  m  4, the last expression becomes 1 100 (100 k + 10 t + h ) = k + t 10 + h 100 , which is the correct two-digit rounding of x . If 5  m  9, it becomes 1 100 (100 k + 10 t + h + 1) = k + t 10 + h + 1 100 , also the correct two-digit rounding of x in this case. C01S01.055: Let Round 4( x ) = 1 10000 Round (10000 x ). To verify that Round 4 has the desired property for [say] positive values of x , write such a number x in the form x = k + d 1 10 + d 2 100 + d 3 1000 + d 4 10000 + d 5 100 , 000 + r, where k is a nonnegative integer, each d i is an integer between 0 and 9, and 0  r < 0 . 00001. Application of Round 4 to x then produces 1 10000 Floor (10000 k + 1000 d 1 + 100 d 2 + 10 d 3 + d 4 + 1 10 ( d 5 + 5) + 10000 r ) . Then consideration of the two cases 0  d 5  4 and 5  d 5  9 will show that Round 4 produces the correct four-place rounding of x in both cases. C01S01.056: Let Chop 4( x ) = 1 10000 Floor (10000 x ). Suppose that x > 0. Write x in the form x = k + d 1 10 + d 2 100 + d 3 1000 + d 4 10000 + r, where k is a nonnegative integer, each of the d i is an integer between 0 and 9, and 0  r < 0 . 0001. Then Chop 4( x ) produces 1 10000 Floor (10000 k + 1000 d 1 + 100 d 2 + 10 d 3 + d 4 + 10000 r ) = 1 10000 (10000 k + 1000 d 1 + 100 d 2 + 10 d 3 + d 4 ) 10

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    because 10000 r < 1. It follows that Chop 4 has the desired effect. C01S01.057: x 0 . 0 0 . 2 0 . 4 0 . 6 0 . 8 1 . 0 y 1 . 0 0 . 44 − 0 . 04 − 0 . 44 − 0 . 76 − 1 . 0 The sign change occurs between x = 0 . 2 and x = 0 . 4. x 0 . 20 0 . 25 0 . 30 0 . 35 0 . 40 y 0 . 44 0 . 3125 0 . 19 0 . 0725 − 0 . 04 The sign change occurs between x = 0 . 35 and x = 0 . 40. x 0 . 35 0 . 36 0 . 37 0 . 38 0 . 39 0 . 40 y 0 . 0725 0 . 0496 0 . 0269 0 . 0044 − 0 . 0179 − 0 . 04 From this point on, the data for y will be rounded. x 0 . 380 0 . 382 0 . 384 0 . 386 0 . 388 0 . 390 y 0 . 0044 − 0 . 0001 − 0 . 0045 − 0 . 0090 − 0 . 0135 − 0 . 0179 Answer (rounded to two places): 0 . 38. The quadratic formula yields the two roots 1 2 ( 3 ± √ 5 ) ; the smaller of these is approximately 0.381966011250105151795. Problems 58 through 66 are worked in the same way as Problem 57. C01S01.058: The sign change intervals are [2 , 3], [2 . 6 , 2 . 8], [2 . 60 , 2 . 64], and [2 . 616 , 2 . 624]. Answer: 1 2 ( 3 + √ 5 ) ≈ 2 . 62. C01S01.059: The sign change intervals are [1 , 2], [1 . 2 , 1 . 4], [1 . 20 , 1 . 24], [1 . 232 , 1 . 240], and [1 . 2352 , 1 . 2368]. Answer: − 1 + √ 5 ≈ 1 . 24. C01S01.060: The sign change intervals are [ − 4 , − 3], [ − 3 . 4 , − 3 . 2], [ − 3 . 24 , − 3 . 20], [ − 3 . 240 , − 3 . 232], and [ − 3 . 2368 , − 3 . 2352]. Answer: − 1 − √ 5 ≈ − 3 . 24. C01S01.061: The sign change intervals are [0 , 1], [0 . 6 , 0 . 8], [0 . 68 , 0 . 72], [0 . 712 , 0 . 720], and [0 . 7184 , 0 . 7200]. Answer: 1 4 ( 7 − √ 17 ) ≈ 0 . 72. C01S01.062: The sign change intervals are [2 , 3], [2 . 6 , 2 . 8], [2 . 76 , 2 . 80], [2 . 776 , 2 . 784], and [2 . 7792 , 2 . 7808]. Answer: 1 4 ( 7 + √ 17 ) ≈ 2 . 78. C01S01.063: The sign change intervals are [3 , 4], [3 . 2 , 3 . 4], [3 . 20 , 3 . 24], [3 . 208 , 3 . 216], and [3 . 2080 , 3 . 2096]. Answer: 1 2 ( 11 − √ 21 ) ≈ 3 . 21. C01S01.064: The sign change intervals are [7 , 8], [7 . 6 , 7 . 8], [7 . 76 , 7 . 80], [7 . 784 , 7 . 792], and [7 . 7904 , 7 . 7920]. Answer: 1 2 ( 11 + √ 21 ) ≈ 7 . 79. C01S01.065: The sign change intervals are [1 , 2], [1 . 6 , 1 . 8], [1 . 60 , 1 . 64], [1 . 608 , 1 . 616], [1 . 6144 , 1 . 6160], and [1 . 61568 , 1 . 61600]. Answer: 1 6 ( − 23 + √ 1069 ) ≈ 1 . 62. C01S01.066: The sign change intervals are [ − 10 , − 9], [ − 9 . 4 , − 9 . 2], [ − 9 . 32 , − 9 . 28], [ − 9 . 288 , − 9 . 280], and [ − 9 . 2832 , − 9 . 2816]. Answer: 1 6 ( − 23 − √ 1069 ) ≈ − 9 . 28. 11

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    Section 1.2 C01S02.001: The slope of L is m = (3 − 0) / (2 − 0) = 3 2 , so L has equation y − 0 = 3 2 ( x − 0); that is, 2 y = 3 x. C01S02.002: Because L is vertical and (7 , 0) lies on L , every point on L has Cartesian coordinates (7 , y ) for some number y (and every such point lies on L ). Hence an equation of L is x = 7. C01S02.003: Because L is horizontal, it has slope zero. Hence an equation of L is y − ( − 5) = 0 · ( x − 3); that is, y = − 5 . C01S02.004: Because (2 , 0) and (0 , − 3) lie on L , it has slope (0 + 3) / (2 − 0) = 3 2 . Hence an equation of L is y − 0 = 3 2 ( x − 2); that is, y = 3 2 x − 3 . C01S02.005: The slope of L is (3 − ( − 3)) / (5 − 2) = 2, so an equation of L is y − 3 = 2( x − 5); that is, y = 2 x − 7 . C01S02.006: An equation of L is y − ( − 4) = 1 2 ( x − ( − 1)); that is, 2 y + 7 = x. C01S02.007: The slope of L is tan(135 ◦ ) = − 1, so L has equation y − 2 = − 1 · ( x − 4); that is, x + y = 6 . C01S02.008: Equation: y − 7 = 6( x − 0); that is, y = 6 x + 7. C01S02.009: The second line’s equation can be written in the form y = − 2 x + 10 to show that it has slope − 2. Because L is parallel to the second line, L also has slope − 2 and thus equation y − 5 = − 2( x − 1). C01S02.010: The equation of the second line can be rewritten as y = − 1 2 x + 17 2 to show that it has slope − 1 2 . Because L is perpendicular to the second line, L has slope 2 and thus equation y − 4 = 2( x + 2). C01S02.011: x 2 − 4 x + 4 + y 2 = 4: ( x − 2) 2 + ( y − 0) 2 = 2 2 . Center (2 , 0), radius 2. C01S02.012: x 2 + y 2 + 6 y + 9 = 9: ( x − 0) 2 + ( y + 3) 2 = 3 2 . Center (0 , − 3), radius 3. C01S02.013: x 2 + 2 x + 1 + y 2 + 2 y + 1 = 4: ( x + 1) 2 + ( y + 1) 2 = 2 2 . Center ( − 1 , − 1), radius 2. C01S02.014: x 2 + 10 x + 25 + y 2 − 20 y + 100 = 25: ( x + 5) 2 + ( y − 10) 2 = 5 2 . Center ( − 5 , 10), radius 5. C01S02.015: x 2 + y 2 + x − y = 1 2 : x 2 + x + 1 4 + y 2 − y + 1 4 = 1; ( x + 1 2 ) 2 + ( y − 1 2 ) 2 = 1. Center: ( − 1 2 , 1 2 ), radius 1. C01S02.016: x 2 + y 2 − 2 3 x − 4 3 y = 11 9 : x 2 − 2 3 x + 1 9 + y 2 − 4 3 y + 4 9 = 16 9 ; ( x − 1 3 ) 2 + ( y − 2 3 ) 2 = ( 4 3 ) 2 . Center ( 1 3 , 2 3 ), radius 4 3 . 1

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    C01S02.017: y = ( x − 3) 2 : Opens upward, vertex at (3 , 0). C01S02.018: y − 16 = − x 2 : Opens downward, vertex at (0 , 16). C01S02.019: y − 3 = ( x + 1) 2 : Opens upward, vertex at ( − 1 , 3). C01S02.020: 2 y = x 2 − 4 x + 4 + 4: y − 2 = 1 2 ( x − 2) 2 . Opens upward, vertex at (2 , 2). C01S02.021: y = 5( x 2 + 4 x + 4) + 3 = 5( x + 2) 2 + 3: Opens upward, vertex at ( − 2 , 3). C01S02.022: y = − ( x 2 − x ) = − ( x 2 − x + 1 4 ) + 1 4 : y − 1 4 = − ( x − 1 2 ) 2 . Opens downward, vertex at ( 1 2 , 1 4 ). C01S02.023: x 2 − 6 x + 9 + y 2 + 8 y + 16 = 25: ( x − 3) 2 + ( y + 4) 2 = 5 5 . Circle, center (3 , − 4), radius 5. C01S02.024: ( x − 1) 2 + ( y + 1) 2 = 0: The graph consists of the single point (1 , − 1). C01S02.025: ( x + 1) 2 + ( y + 3) 2 = − 10: There are no points on the graph. C01S02.026: x 2 + y 2 − x + 3 y + 2 . 5 = 0: x 2 − x + 0 . 25 + y 2 + 3 y + 2 . 25 = 0: ( x − 0 . 5) 2 + ( y + 1 . 5) 2 = 0. The graph consists of the single point (0 . 5 , − 1 . 5). C01S02.027: The graph is the straight line segment connecting the two points ( − 1 , 7) and (1 , − 3) (including those two points). C01S02.028: The graph is the straight line segment connecting the two points (0 , 2) and (2 , − 8), including the first of these two points but not the second. C01S02.029: The graph is the parabola that opens downward, symmetric around the y -axis, with vertex at (0 , 10) and x -intercepts ± √ 10. C01S02.030: The graph of y = 1 + 2 x 2 is a parabola that opens upwards, is symmetric around the y -axis, and has vertex at (0 , 1). C01S02.031: The graph of y = x 3 can be visualized by modifying the familiar graph of the parabola with equation y = x 2 : The former may be obtained by multiplying the y -coordinate of the latter’s point ( x, x 2 ) by x . Thus both have flat spots at the origin. For 0 < x < 1, the graph of y = x 3 is below that of y = x 2 . They cross at (1 , 1), and for x > 1 the graph of y = x 2 is below that of y = x 3 , with the difference becoming arbitrarily large as x increases without bound. If the graph of y = x 3 for x  0 is rotated 180 ◦ around the point (0 , 0), the graph of y = x 3 for x < 0 is the result. C01S02.032: The graph of f ( x ) = x 4 can be visualized by first visualizing the graph of y = x 2 . If the y -coordinate of each point on this graph is replaced with its square ( x 4 ), the result is the graph of f . The effect on the graph of y = x 2 is to multiply the y -coordinate by x 2 , which is between 0 and 1 for 0 < | x | < 1 and which is larger than 1 for | x | > 1. Thus the graph of f superficially resembles that of y = x 2 , but is much closer to the x -axis for | x | < 1 and much farther away for | x | > 1. The two graphs cross at (0 , 0) (where each has a flat spot) and at ( ± 1 , 1), but the graph of f is much steeper at the latter two points. C01S02.033: To graph y = f ( x ) = √ 4 − x 2 , note that y  0 and that y 2 = 4 − x 2 ; that is, x 2 + y 2 = 4. Hence the graph of f is the upper half of the circle with center (0 , 0) and radius 2. C01S02.034: To graph y = f ( x ) = − √ 9 − x 2 , note that y  0 and that y 2 = 9 − x 2 ; that is, that x 2 + y 2 = 9. Hence the graph of f is the lower half of the circle with center (0 , 0) and radius 3. 2

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    -4 -2 2 4 6 -4 -2 2 4 C01S02.035: To graph f ( x ) = √ x 2 − 9, note that there is no graph for − 3 < x < 3, that f ( ± 3) = 0, and that f ( x ) > 0 for x < − 3 and for x > 3. If x is large positive, then √ x 2 − 9 ≈ √ x 2 = x , so the graph of f has x -intercept (3 , 0) and rises as x increases, nearly coinciding with the graph of y = x for x large positive. The case x < − 3 is trickier. In this case, if x is a large negative number, then f ( x ) = √ x 2 − 9 ≈ √ x 2 = − x (Note the minus sign!). So for x  − 3, the graph of f has x -intercept ( − 3 , 0) and, for x large negative, almost coincides with the graph of y = − x . Later we will see that the graph of f becomes arbitrarily steep as x gets closer and closer to ± 3. C01S02.036: As x increases without bound—either positively or negatively— f ( x ) gets arbitrarily close to zero. Moreover, if x is large positive then f ( x ) is negative and close to zero, so the graph of f lies just below the x -axis for such x . Similarly, the graph of f lies just above the x -axis for x large negative. If x is slightly less than 1 but very close to 1, then f ( x ) is the reciprocal of a tiny positive number, hence is a large positive number. So the graph of f just to the left of the vertical line x = 1 almost coincides with the top half of that line. Similarly, just to the right of the line x = 1, then graph of f almost coincides with the bottomhalf of that line. There is no graph where x = 1, so the graph resembles the one in the next figure. The only intercept is the y -intercept (0 , 1). The graph correctly shows that the graph of f is increasing for x < 1 and for x > 1. C01S02.037: Note that f ( x ) is positive and close to zero for x large positive, so that the graph of f is just above the x -axis—and nearly coincides with it—for such x . Similarly, the graph of f is just below the x -axis and nearly coincides with it for x large negative. There is no graph where x = − 2, but if x is slightly greater than − 2 then f ( x ) is the reciprocal of a very small positive number, so f ( x ) is large and nearly coincides with the upper half of the vertical line x = − 2. Similarly, if x is slightly less than − 2, then the graph of f ( x ) is large negative and nearly coincides with the the lower half of the line x = − 3. The graph of f is decreasing for x < − 2 and for x > − 2 and its only intercept is the y -intercept ( 0 , 1 2 ) . C01S02.038: Note that f ( x ) is very small but positive if x is either large positive or large negative. There is no graph for x = 0, but if x is very close to zero, then f ( x ) is the reciprocal of a very small positive number, and hence is large positive. So the graph of f is just above the x -axis and almost coincides with it if | x | is large, whereas the graph of f almost coincides with the positive y -axis for x near zero. There are no intercepts; the graph of f is increasing for x < 0 and is decreasing for x > 0. 3

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    C01S02.039: Note that f ( x ) > 0 for all x other than x = 1, where f is not defined. If | x | is large, then f ( x ) is near zero, so the graph of f almost coincides with the x -axis for such x . If x is very close to 1, then f ( x ) is the reciprocal of a very small positive number, hence f ( x ) is large positive. So for such x , the graph of f ( x ) almost coincides with the upper half of the vertical line x = 1. The only intercept is (0 , 1). C01S02.040: Note first that f ( x ) is undefined at x = 0. To handle the absolute value symbol, we look at two cases: If x > 0, then f ( x ) = 1; if x < 0, then f ( x ) = − 1. So the graph of f consists of the part of the horizontal line y = 1 for which x > 0, together with the part of the horizontal line y = − 1 for which x < 0. C01S02.041: Note that f ( x ) is undefined when 2 x + 3 = 0; that is, when x = − 3 2 . If x is large positive, then f ( x ) is positive and close to zero, so the graph of f is slightly above the x -axis and almost coincides with the x -axis. If x is large negative, then f ( x ) is negative and close to zero, so the graph of f is slightly below the x -axis and almost coincides with the x -axis. If x is slightly greater than − 3 2 then f ( x ) is very large positive, so the graph of f almost coincides with the upper half of the vertical line x = − 3 2 . If x is slightly less than − 3 2 then f ( x ) is very large negative, so the graph of f almost coincides with the lower half of that vertical line. The graph of f is decreasing for x < − 3 2 and also decreasing for x > − 3 2 . The only intercept is at ( 0 , 1 3 ) . C01S02.042: Note that f ( x ) is undefined when 2 x + 3 = 0; that is, when x = − 3 2 . If x is large positive or large negative, then f ( x ) is positive and close to zero, so the graph of f is slightly above the x -axis and almost coincides with the x -axis for | x | large. If x is close to − 3 2 then f ( x ) is very large positive, so the graph of f almost coincides with the upper half of the vertical line x = − 3 2 . The graph of f is increasing for x < − 3 2 and decreasing for x > − 3 2 . The only intercept is at ( 0 , 1 9 ) . C01S02.043: Given y = f ( x ) = √ 1 − x , note that y  0 and that y 2 = 1 − x ; that is, x = 1 − y 2 . So the graph is the part of the parabola x = 1 − y 2 for which y  0. This parabola has horizontal axis of symmetry the y -axis, opens to the left (because the coefficient of y 2 is negative), and has vertex (1 , 0). Therefore the graph of f is the upper half of this parabola. C01S02.044: Note that the interval x < 1 is the domain of f , so there is no graph for x  1. If x is a large negative number, then the denominator is large positive, so that its reciprocal f ( x ) is very small positive. As x gets closer and closer to 1 (while x < 1), the denominator approaches zero, so its reciprocal f ( x ) takes on arbitrarily large positive values. So the graph of f is slightly above the x -axis and almost coincides with that axis for x large negative; the graph of f almost coincides with the upper half of the vertical line x = 1 for x near (and less than) 1. The graph of f is increasing for all x < 1 and (0 , 1) is the only intercept. C01S02.045: Note that f ( x ) is defined only if 2 x + 3 > 0; that is, if x > − 3 2 . Note also that f ( x ) > 0 for all such x . If x is large positive, then f ( x ) is positive but near zero, so the graph of f is just above the x -axis and almost coincides with it. If x is very close to − 3 2 (but larger), then the denominator in f ( x ) is very tiny positive, so the graph of f almost coincides with the upper half of the vertical line x = − 3 2 for such x . The graph of f is decreasing for all x > − 3 2 . C01S02.046: Given: f ( x ) = | 2 x − 2 | . Case 1: x  1. Then 2 x − 2  0, so that f ( x ) = 2 x − 2. Because f (1) = 0, the graph of f for x  1 consists of the part of the straight line through (1 , 0) with slope 2. Case 2: x < 1. Then 2 x − 2 < 0, so that f ( x ) = − 2 x + 2. The line y = − 2 x + 2 passes through (1 , 0), so the graph of f for x < 1 consists of the part of the straight line through (1 , 0) with slope − 2. C01S02.047: Given: f ( x ) = | x | + x . If x  0 then f ( x ) = x + x = 2 x , so if x  0 then the graph of f is the part of the straight line through (0 , 0) with slope 2 for which x  0. If x < 0 then f ( x ) = − x + x = 0, so the rest of the graph of f coincides with the negative x -axis. 4

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