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Solution Manual for Biocalculus: Calculus, Probability, and Statistics for the Life Sciences, 1st Edition

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    DIAGNOSTIC TESTS Test A Algebra 1. (a) ( − 3) 4 = ( − 3)( − 3)( − 3)( − 3) = 81 (b) − 3 4 = − (3)(3)(3)(3) = − 81 (c) 3 − 4 = 1 3 4 = 1 81 (d) 5 23 5 21 = 5 23 − 21 = 5 2 = 25 (e)  2 3  − 2 =  3 2  2 = 9 4 (f ) 16 − 3  4 = 1 16 3  4 = 1  4 √ 16  3 = 1 2 3 = 1 8 2. (a) Note that √ 200 = √ 100 · 2 = 10 √ 2 and √ 32 = √ 16 · 2 = 4 √ 2 . Thus √ 200 − √ 32 = 10 √ 2 − 4 √ 2 = 6 √ 2 . (b) (3  3  3 )(4  2 ) 2 = 3  3  3 16  2  4 = 48  5  7 (c)  3  3  2  3  2  − 1  2  − 2 =   2  − 1  2 3  3  2  3  2 = (  2  − 1  2 ) 2 (3  3  2  3 ) 2 =  4  − 1 9  3  6 =  4 9  3  6  =  9  7 3. (a) 3(  + 6) + 4(2  − 5) = 3  + 18 + 8  − 20 = 11  − 2 (b) (  + 3)(4  − 5) = 4  2 − 5  + 12  − 15 = 4  2 + 7  − 15 (c)  √  + √    √  − √   =  √   2 − √  √  + √  √  −  √   2 =  −  Or: Use the formula for the difference of two squares to see that  √  + √   √  − √   =  √   2 −  √   2 =  −  . (d) (2  + 3) 2 = (2  + 3)(2  + 3) = 4  2 + 6  + 6  + 9 = 4  2 + 12  + 9 . Note: A quicker way to expand this binomial is to use the formula (  +  ) 2 =  2 + 2  +  2 with  = 2  and  = 3 : (2  + 3) 2 = (2  ) 2 + 2(2  )(3) + 3 2 = 4  2 + 12  + 9 (e) See Reference Page 1 for the binomial formula (  +  ) 3 =  3 + 3  2  + 3  2 +  3 . Using it, we get (  + 2) 3 =  3 + 3  2 (2) + 3  (2 2 ) + 2 3 =  3 + 6  2 + 12  + 8 . 4. (a) Using the difference of two squares formula,  2 −  2 = (  +  )(  −  ) , we have 4  2 − 25 = (2  ) 2 − 5 2 = (2  + 5)(2  − 5) . (b) Factoring by trial and error, we get 2  2 + 5  − 12 = (2  − 3)(  + 4) . (c) Using factoring by grouping and the difference of two squares formula, we have  3 − 3  2 − 4  + 12 =  2 (  − 3) − 4(  − 3) = (  2 − 4)(  − 3) = (  − 2)(  + 2)(  − 3) . (d)  4 + 27  =  (  3 + 27) =  (  + 3)(  2 − 3  + 9) This last expression was obtained using the sum of two cubes formula,  3 +  3 = (  +  )(  2 −  +  2 ) with  =  and  = 3 . [See Reference Page 1 in the textbook.] (e) The smallest exponent on  is − 1 2 , so we will factor out  − 1  2 . 3  3  2 − 9  1  2 + 6  − 1  2 = 3  − 1  2 (  2 − 3  + 2) = 3  − 1  2 (  − 1)(  − 2) (f )  3  − 4  =  (  2 − 4) =  (  − 2)(  + 2) uplic ible 1

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    2 ¤ DIAGNOSTIC TESTS 5. (a)  2 + 3  + 2  2 −  − 2 = (  + 1)(  + 2) (  + 1)(  − 2) =  + 2  − 2 (b) 2  2 −  − 1  2 − 9 ·  + 3 2  + 1 = (2  + 1)(  − 1) (  − 3)(  + 3) ·  + 3 2  + 1 =  − 1  − 3 (c)  2  2 − 4 −  + 1  + 2 =  2 (  − 2)(  + 2) −  + 1  + 2 =  2 (  − 2)(  + 2) −  + 1  + 2 ·  − 2  − 2 =  2 − (  + 1)(  − 2) (  − 2)(  + 2) =  2 − (  2 −  − 2) (  + 2)(  − 2) =  + 2 (  + 2)(  − 2) = 1  − 2 (d)   −   1  − 1  =   −   1  − 1  ·   =  2 −  2  −  = (  −  )(  +  ) − (  −  ) =  +  − 1 = − (  +  ) 6. (a) √ 10 √ 5 − 2 = √ 10 √ 5 − 2 · √ 5 + 2 √ 5 + 2 = √ 50 + 2 √ 10  √ 5  2 − 2 2 = 5 √ 2 + 2 √ 10 5 − 4 = 5 √ 2 + 2 √ 10 (b) √ 4 +  − 2  = √ 4 +  − 2  · √ 4 +  + 2 √ 4 +  + 2 = 4 +  − 4   √ 4 +  + 2  =    √ 4 +  + 2  = 1 √ 4 +  + 2 7. (a)  2 +  + 1 =   2 +  + 1 4  + 1 − 1 4 =   + 1 2  2 + 3 4 (b) 2  2 − 12  + 11 = 2(  2 − 6  ) + 11 = 2(  2 − 6  + 9 − 9) + 11 = 2(  2 − 6  + 9) − 18 + 11 = 2(  − 3) 2 − 7 8. (a)  + 5 = 14 − 1 2  ⇔  + 1 2  = 14 − 5 ⇔ 3 2  = 9 ⇔  = 2 3 · 9 ⇔  = 6 (b) 2   + 1 = 2  − 1  ⇒ 2  2 = (2  − 1)(  + 1) ⇔ 2  2 = 2  2 +  − 1 ⇔  = 1 (c)  2 −  − 12 = 0 ⇔ (  + 3)(  − 4) = 0 ⇔  + 3 = 0 or  − 4 = 0 ⇔  = − 3 or  = 4 (d) By the quadratic formula, 2  2 + 4  + 1 = 0 ⇔  = − 4 ±  4 2 − 4(2)(1) 2(2) = − 4 ± √ 8 4 = − 4 ± 2 √ 2 4 = 2  − 2 ± √ 2  4 = − 2 ± √ 2 2 = − 1 ± 1 2 √ 2 . (e)  4 − 3  2 + 2 = 0 ⇔ (  2 − 1)(  2 − 2) = 0 ⇔  2 − 1 = 0 or  2 − 2 = 0 ⇔  2 = 1 or  2 = 2 ⇔  = ± 1 or  = ± √ 2 (f ) 3 |  − 4 | = 10 ⇔ |  − 4 | = 10 3 ⇔  − 4 = − 10 3 or  − 4 = 10 3 ⇔  = 2 3 or  = 22 3 (g) Multiplying through 2  (4 −  ) − 1  2 − 3 √ 4 −  = 0 by (4 −  ) 1  2 gives 2  − 3(4 −  ) = 0 ⇔ 2  − 12 + 3  = 0 ⇔ 5  − 12 = 0 ⇔ 5  = 12 ⇔  = 12 5 . 9. (a) − 4  5 − 3  ≤ 17 ⇔ − 9  − 3  ≤ 12 ⇔ 3   ≥ − 4 or − 4 ≤   3 . In interval notation, the answer is [ − 4  3) . (b)  2  2  + 8 ⇔  2 − 2  − 8  0 ⇔ (  + 2)(  − 4)  0 . Now, (  + 2)(  − 4) will change sign at the critical values  = − 2 and  = 4 . Thus the possible intervals of solution are ( −∞  − 2) , ( − 2  4) , and (4  ∞ ) . By choosing a single test value from each interval, we see that ( − 2  4) is the only interval that satis fi es the inequality. post , in

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    TEST B ANALYTIC GEOMETRY ¤ 3 (c) The inequality  (  − 1)(  + 2)  0 has critical values of − 2  0  and 1 . The corresponding possible intervals of solution are ( −∞  − 2) , ( − 2  0) , (0  1) and (1  ∞ ) . By choosing a single test value from each interval, we see that both intervals ( − 2  0) and (1  ∞ ) satisfy the inequality. Thus, the solution is the union of these two intervals: ( − 2  0) ∪ (1  ∞ ) . (d) |  − 4 |  3 ⇔ − 3   − 4  3 ⇔ 1    7 . In interval notation, the answer is (1  7) . (e) 2  − 3  + 1 ≤ 1 ⇔ 2  − 3  + 1 − 1 ≤ 0 ⇔ 2  − 3  + 1 −  + 1  + 1 ≤ 0 ⇔ 2  − 3 −  − 1  + 1 ≤ 0 ⇔  − 4  + 1 ≤ 0 . Now, the expression  − 4  + 1 may change signs at the critical values  = − 1 and  = 4 , so the possible intervals of solution are ( −∞  − 1) , ( − 1  4] , and [4  ∞ ) . By choosing a single test value from each interval, we see that ( − 1  4] is the only interval that satis fi es the inequality. 10. (a) False. In order for the statement to be true, it must hold for all real numbers, so, to show that the statement is false, pick  = 1 and  = 2 and observe that (1 + 2) 2 6 = 1 2 + 2 2 . In general, (  +  ) 2 =  2 + 2  +  2 . (b) True as long as  and  are nonnegative real numbers. To see this, think in terms of the laws of exponents: √  = (  ) 1  2 =  1  2  1  2 = √  √  . (c) False. To see this, let  = 1 and  = 2 , then √ 1 2 + 2 2 6 = 1 + 2 . (d) False. To see this, let  = 1 and  = 2 , then 1 + 1(2) 2 6 = 1 + 1 . (e) False. To see this, let  = 2 and  = 3 , then 1 2 − 3 6 = 1 2 − 1 3 . (f ) True since 1   −  ·   = 1  −  , as long as  6 = 0 and  −  6 = 0 . Test B Analytic Geometry 1. (a) Using the point (2  − 5) and  = − 3 in the point-slope equation of a line,  −  1 =  (  −  1 ) , we get  − ( − 5) = − 3(  − 2) ⇒  + 5 = − 3  + 6 ⇒  = − 3  + 1 . (b) A line parallel to the  -axis must be horizontal and thus have a slope of 0 . Since the line passes through the point (2  − 5) , the  -coordinate of every point on the line is − 5 , so the equation is  = − 5 . (c) A line parallel to the  -axis is vertical with unde fi ned slope. So the  -coordinate of every point on the line is 2 and so the equation is  = 2 . (d) Note that 2  − 4  = 3 ⇒ − 4  = − 2  + 3 ⇒  = 1 2  − 3 4 . Thus the slope of the given line is  = 1 2 . Hence, the slope of the line we’re looking for is also 1 2 (since the line we’re looking for is required to be parallel to the given line). So the equation of the line is  − ( − 5) = 1 2 (  − 2) ⇒  + 5 = 1 2  − 1 ⇒  = 1 2  − 6 . 2. First we’ll fi nd the distance between the two given points in order to obtain the radius,  , of the circle:  =  [3 − ( − 1)] 2 + ( − 2 − 4) 2 =  4 2 + ( − 6) 2 = √ 52 . Next use the standard equation of a circle, (  −  ) 2 + (  −  ) 2 =  2 , where (   ) is the center, to get (  + 1) 2 + (  − 4) 2 = 52 . or du cessi

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    4 ¤ DIAGNOSTIC TESTS 3. We must rewrite the equation in standard form in order to identify the center and radius. Note that  2 +  2 − 6  + 10  + 9 = 0 ⇒  2 − 6  + 9 +  2 + 10  = 0 . For the left-hand side of the latter equation, we factor the fi rst three terms and complete the square on the last two terms as follows:  2 − 6  + 9 +  2 + 10  = 0 ⇒ (  − 3) 2 +  2 + 10  + 25 = 25 ⇒ (  − 3) 2 + (  + 5) 2 = 25 . Thus, the center of the circle is (3  − 5) and the radius is 5 . 4. (a)  ( − 7  4) and  (5  − 12) ⇒   = − 12 − 4 5 − ( − 7) = − 16 12 = − 4 3 (b)  − 4 = − 4 3 [  − ( − 7)] ⇒  − 4 = − 4 3  − 28 3 ⇒ 3  − 12 = − 4  − 28 ⇒ 4  + 3  + 16 = 0 . Putting  = 0 , we get 4  + 16 = 0 , so the  -intercept is − 4 , and substituting 0 for  results in a  -intercept of − 16 3 . (c) The midpoint is obtained by averaging the corresponding coordinates of both points:  − 7+5 2  4+( − 12) 2  = ( − 1  − 4) . (d)  =  [5 − ( − 7)] 2 + ( − 12 − 4) 2 =  12 2 + ( − 16) 2 = √ 144 + 256 = √ 400 = 20 (e) The perpendicular bisector is the line that intersects the line segment  at a right angle through its midpoint. Thus the perpendicular bisector passes through ( − 1  − 4) and has slope 3 4 [the slope is obtained by taking the negative reciprocal of the answer from part (a)]. So the perpendicular bisector is given by  + 4 = 3 4 [  − ( − 1)] or 3  − 4  = 13 . (f ) The center of the required circle is the midpoint of  , and the radius is half the length of  , which is 10 . Thus, the equation is (  + 1) 2 + (  + 4) 2 = 100 . 5. (a) Graph the corresponding horizontal lines (given by the equations  = − 1 and  = 3 ) as solid lines. The inequality  ≥ − 1 describes the points (   ) that lie on or above the line  = − 1 . The inequality  ≤ 3 describes the points (   ) that lie on or below the line  = 3 . So the pair of inequalities − 1 ≤  ≤ 3 describes the points that lie on or between the lines  = − 1 and  = 3 . (b) Note that the given inequalities can be written as − 4    4 and − 2    2 , respectively. So the region lies between the vertical lines  = − 4 and  = 4 and between the horizontal lines  = − 2 and  = 2 . As shown in the graph, the region common to both graphs is a rectangle (minus its edges) centered at the origin. (c) We fi rst graph  = 1 − 1 2  as a dotted line. Since   1 − 1 2  , the points in the region lie below this line. post , in

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    TEST C FUNCTIONS ¤ 5 (d) We fi rst graph the parabola  =  2 − 1 using a solid curve. Since  ≥  2 − 1 , the points in the region lie on or above the parabola. (e) We graph the circle  2 +  2 = 4 using a dotted curve. Since   2 +  2  2 , the region consists of points whose distance from the origin is less than 2, that is, the points that lie inside the circle. (f ) The equation 9  2 + 16  2 = 144 is an ellipse centered at (0  0) . We put it in standard form by dividing by 144 and get  2 16 +  2 9 = 1 . The  -intercepts are located at a distance of √ 16 = 4 from the center while the  -intercepts are a distance of √ 9 = 3 from the center (see the graph). Test C Functions 1. (a) Locate − 1 on the  -axis and then go down to the point on the graph with an  -coordinate of − 1 . The corresponding  -coordinate is the value of the function at  = − 1 , which is − 2 . So,  ( − 1) = − 2 . (b) Using the same technique as in part (a), we get  (2) ≈ 2  8 . (c) Locate 2 on the  -axis and then go left and right to fi nd all points on the graph with a  -coordinate of 2 . The corresponding  -coordinates are the  -values we are searching for. So  = − 3 and  = 1 . (d) Using the same technique as in part (c), we get  ≈ − 2  5 and  ≈ 0  3 . (e) The domain is all the  -values for which the graph exists, and the range is all the  -values for which the graph exists. Thus, the domain is [ − 3  3] , and the range is [ − 2  3] . 2. Note that  (2 +  ) = (2 +  ) 3 and  (2) = 2 3 = 8 . So the difference quotient becomes  (2 +  ) −  (2)  = (2 +  ) 3 − 8  = 8 + 12  + 6  2 +  3 − 8  = 12  + 6  2 +  3  =  (12 + 6  +  2 )  = 12 + 6  +  2 . 3. (a) Set the denominator equal to 0 and solve to fi nd restrictions on the domain:  2 +  − 2 = 0 ⇒ (  − 1)(  + 2) = 0 ⇒  = 1 or  = − 2 . Thus, the domain is all real numbers except 1 or − 2 or, in interval notation, ( −∞  − 2) ∪ ( − 2  1) ∪ (1  ∞ ) . (b) Note that the denominator is always greater than or equal to 1 , and the numerator is de fi ned for all real numbers. Thus, the domain is ( −∞  ∞ ) . (c) Note that the function  is the sum of two root functions. So  is de fi ned on the intersection of the domains of these two root functions. The domain of a square root function is found by setting its radicand greater than or equal to 0 . Now, or du cessi

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    6 ¤ DIAGNOSTIC TESTS 4 −  ≥ 0 ⇒  ≤ 4 and  2 − 1 ≥ 0 ⇒ (  − 1)(  + 1) ≥ 0 ⇒  ≤ − 1 or  ≥ 1 . Thus, the domain of  is ( −∞  − 1] ∪ [1  4] . 4. (a) Re fl ect the graph of  about the  -axis. (b) Stretch the graph of  vertically by a factor of 2 , then shift 1 unit downward. (c) Shift the graph of  right 3 units, then up 2 units. 5. (a) Make a table and then connect the points with a smooth curve:  − 2 − 1 0 1 2  − 8 − 1 0 1 8 (b) Shift the graph from part (a) left 1 unit. (c) Shift the graph from part (a) right 2 units and up 3 units. (d) First plot  =  2 . Next, to get the graph of  (  ) = 4 −  2 , re fl ect  about the x -axis and then shift it upward 4 units. (e) Make a table and then connect the points with a smooth curve:  0 1 4 9  0 1 2 3 (f ) Stretch the graph from part (e) vertically by a factor of two. post , in

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    TEST D TRIGONOMETRY ¤ 7 (g) First plot  = 2  . Next, get the graph of  = − 2  by re fl ecting the graph of  = 2  about the x -axis. (h) Note that  = 1 +  − 1 = 1 + 1  . So fi rst plot  = 1  and then shift it upward 1 unit. 6. (a)  ( − 2) = 1 − ( − 2) 2 = − 3 and  (1) = 2(1) + 1 = 3 (b) For  ≤ 0 plot  (  ) = 1 −  2 and, on the same plane, for   0 plot the graph of  (  ) = 2  + 1 . 7. (a) (  ◦  )(  ) =  (  (  )) =  (2  − 3) = (2  − 3) 2 + 2(2  − 3) − 1 = 4  2 − 12  + 9 + 4  − 6 − 1 = 4  2 − 8  + 2 (b) (  ◦  )(  ) =  (  (  )) =  (  2 + 2  − 1) = 2(  2 + 2  − 1) − 3 = 2  2 + 4  − 2 − 3 = 2  2 + 4  − 5 (c) (  ◦  ◦  )(  ) =  (  (  (  ))) =  (  (2  − 3)) =  (2(2  − 3) − 3) =  (4  − 9) = 2(4  − 9) − 3 = 8  − 18 − 3 = 8  − 21 Test D Trigonometry 1. (a) 300 ◦ = 300 ◦   180 ◦  = 300  180 = 5  3 (b) − 18 ◦ = − 18 ◦   180 ◦  = − 18  180 = −  10 2. (a) 5  6 = 5  6  180   ◦ = 150 ◦ (b) 2 = 2  180   ◦ =  360   ◦ ≈ 114  6 ◦ 3. We will use the arc length formula,  =  , where  is arc length,  is the radius of the circle, and  is the measure of the central angle in radians. First, note that 30 ◦ = 30 ◦   180 ◦  =  6 . So  = (12)   6  = 2  cm. 4. (a) tan(  3) = √ 3  You can read the value from a right triangle with sides 1, 2, and √ 3 .  (b) Note that 7  6 can be thought of as an angle in the third quadrant with reference angle  6 . Thus, sin(7  6) = − 1 2 , since the sine function is negative in the third quadrant. (c) Note that 5  3 can be thought of as an angle in the fourth quadrant with reference angle  3 . Thus, sec(5  3) = 1 cos(5  3) = 1 1  2 = 2 , since the cosine function is positive in the fourth quadrant. or du cessi

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    8 ¤ DIAGNOSTIC TESTS 5. sin  =  24 ⇒  = 24 sin  and cos  =  24 ⇒  = 24 cos  6. sin  = 1 3 and sin 2  + cos 2  = 1 ⇒ cos  =  1 − 1 9 = 2 √ 2 3 . Also, cos  = 4 5 ⇒ sin  =  1 − 16 25 = 3 5 . So, using the sum identity for the sine, we have sin(  +  ) = sin  cos  + cos  sin  = 1 3 · 4 5 + 2 √ 2 3 · 3 5 = 4 + 6 √ 2 15 = 1 15  4 + 6 √ 2  7. (a) tan  sin  + cos  = sin  cos  sin  + cos  = sin 2  cos  + cos 2  cos  = 1 cos  = sec  (b) 2 tan  1 + tan 2  = 2 sin  (cos  ) sec 2  = 2 sin  cos  cos 2  = 2 sin  cos  = sin 2  8. sin 2  = sin  ⇔ 2 sin  cos  = sin  ⇔ 2 sin  cos  − sin  = 0 ⇔ sin  (2 cos  − 1) = 0 ⇔ sin  = 0 or cos  = 1 2 ⇒  = 0 ,  3 ,  , 5  3 , 2  . 9. We fi rst graph  = sin 2  (by compressing the graph of sin  by a factor of 2) and then shift it upward 1 unit. post , in

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    1 FUNCTIONS AND SEQUENCES 1.1 Four Ways to Represent a Function 1. The functions  (  ) =  + √ 2 −  and  (  ) =  + √ 2 −  give exactly the same output values for every input value, so  and  are equal. 2.  (  ) =  2 −   − 1 =  (  − 1)  − 1 =  for  − 1 6 = 0 , so  and  [where  (  ) =  ] are not equal because  (1) is unde fi ned and  (1) = 1 . 3. (a) The point (1  3) is on the graph of  , so  (1) = 3 . (b) When  = − 1 ,  is about − 0  2 , so  ( − 1) ≈ − 0  2 . (c)  (  ) = 1 is equivalent to  = 1  When  = 1 , we have  = 0 and  = 3 . (d) A reasonable estimate for  when  = 0 is  = − 0  8 . (e) The domain of  consists of all  -values on the graph of  . For this function, the domain is − 2 ≤  ≤ 4 , or [ − 2  4] . The range of  consists of all  -values on the graph of  . For this function, the range is − 1 ≤  ≤ 3 , or [ − 1  3] . (f) As  increases from − 2 to 1 ,  increases from − 1 to 3 . Thus,  is increasing on the interval [ − 2  1] . 4. (a) The point ( − 4  − 2) is on the graph of  , so  ( − 4) = − 2 . The point (3  4) is on the graph of  , so  (3) = 4 . (b) We are looking for the values of  for which the  -values are equal. The  -values for  and  are equal at the points ( − 2  1) and (2  2) , so the desired values of  are − 2 and 2 . (c)  (  ) = − 1 is equivalent to  = − 1 . When  = − 1 , we have  = − 3 and  = 4 . (d) As  increases from 0 to 4 ,  decreases from 3 to − 1 . Thus,  is decreasing on the interval [0  4] . (e) The domain of  consists of all  -values on the graph of  . For this function, the domain is − 4 ≤  ≤ 4 , or [ − 4  4] . The range of  consists of all  -values on the graph of  . For this function, the range is − 2 ≤  ≤ 3 , or [ − 2  3] . (f) The domain of  is [ − 4  3] and the range is [0  5  4] . 5. No, the curve is not the graph of a function because a vertical line intersects the curve more than once. Hence, the curve fails the Vertical Line Test. 6. Yes, the curve is the graph of a function because it passes the Vertical Line Test. The domain is [ − 2  2] and the range is [ − 1  2] . 7. Yes, the curve is the graph of a function because it passes the Vertical Line Test. The domain is [ − 3  2] and the range is [ − 3  − 2) ∪ [ − 1  3] . 8. No, the curve is not the graph of a function since for  = 0 , ± 1 , and ± 2 , there are in fi nitely many points on the curve. 9. (a) The graph shows that the global average temperature in 1950 was  (1950) ≈ 13  8 ◦ C  (b) By drawing the horizontal line  = 14  2 to the curve and then drawing the vertical line down to the horizontal axis, we see that  ≈ 1992  (c) The temperature was smallest in 1910 and largest in 2006  (d) The range is {  | 13  5 ≤  ≤ 14  5 } = [13  5  14  5] uplic ible 9

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    10 ¤ CHAPTER 1 FUNCTIONS AND SEQUENCES 10. (a) The range is { Width | 0  Width ≤ 1  6 } = (0  1  6] (b) The graph shows an overall decline in global temperatures from 1500 to 1700, followed by an overall rise in temperatures. The fl uctuations in temperature in the mid and late 19th century are re fl ective of the cooling effects caused by several large volcanic eruptions. 11. If we draw the horizontal line pH = 4  0  we can see that the pH curve is less than 4.0 between 12:23 AM and 12:52 AM . Therefore, a clinical acid re fl ux episode occurred approximately between 12:23 AM and 12:52 AM at which time the esophageal pH was less than 4  0  12. The graphs indicate that tadpoles raised in densely populated regions take longer to put on weight. This is sensible since more crowding leads to fewer resources available for each tadpole. 13. (a) At 30 ◦ S and 20 ◦ N, we expect approximately 100 and 134 ant species respectively. (b) By drawing the horizontal line at a species richness of 100, we see there are two points of intersection with the curve, each having latitude values of roughly 30 ◦ N and 30 ◦ S. (c) The function is even since its graph is symmetric with respect to the y-axis. 14. Example 1: A car is driven at 60 mi  h for 2 hours. The distance  traveled by the car is a function of the time  . The domain of the function is {  | 0 ≤  ≤ 2 } , where  is measured in hours. The range of the function is {  | 0 ≤  ≤ 120 } , where  is measured in miles. Example 2: At a certain university, the number of students  on campus at any time on a particular day is a function of the time  after midnight. The domain of the function is {  | 0 ≤  ≤ 24 } , where  is measured in hours. The range of the function is {  | 0 ≤  ≤  } , where  is an integer and  is the largest number of students on campus at once. Example 3: A certain employee is paid $8  00 per hour and works a maximum of 30 hours per week. The number of hours worked is rounded down to the nearest quarter of an hour. This employee’s gross weekly pay  is a function of the number of hours worked  . The domain of the function is [0  30] and the range of the function is { 0  2  00  4  00      238  00  240  00 } . 240 pay hours 0.25 0.50 0.75 0 29.50 29.75 30 2 4 238 236 15. The person’s weight increased to about 160 pounds at age 20 and stayed fairly steady for 10 years. The person’s weight dropped to about 120 pounds for the next 5 years, then increased rapidly to about 170 pounds. The next 30 years saw a gradual increase to 190 pounds. Possible reasons for the drop in weight at 30 years of age: diet, exercise, health problems. 16. Initially, the person’s forward moving heel contacts the ground resulting in a ground reaction force in the opposite or negative direction. In moving from heel-strike to toe-off, the foot transitions from a forward push to a backward push. Hence, the ground reaction force switches from a negative value to a positive value, becoming zero at some point in between. post , in

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    SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION ¤ 11 17. The water will cool down almost to freezing as the ice melts. Then, when the ice has melted, the water will slowly warm up to room temperature. 18. Runner A won the race, reaching the fi nish line at 100 meters in about 15 seconds, followed by runner B with a time of about 19 seconds, and then by runner C who fi nished in around 23 seconds. B initially led the race, followed by C, and then A. C then passed B to lead for a while. Then A passed fi rst B, and then passed C to take the lead and fi nish fi rst. Finally, B passed C to fi nish in second place. All three runners completed the race. 19. Initially, the bacteria population size remains constant during which nutrients are consumed in preparation for reproduction. In the second phase, the population size increases rapidly as the bacteria replicate. The population size plateaus in phase three at which point the "carrying capacity" has been reached and the available resources and space cannot support a larger population. Finally, the bacteria die due to starvation and waste toxicity and the population declines. 20. The summer solstice (the longest day of the year) is around June 21, and the winter solstice (the shortest day) is around December 22. (Exchange the dates for the southern hemisphere.) 21. Of course, this graph depends strongly on the geographical location! 22. The temperature of the pie would increase rapidly, level off to oven temperature, decrease rapidly, and then level off to room temperature. 23. As the price increases, the amount sold decreases. 24. The value of the car decreases fairly rapidly initially, then somewhat less rapidly. or du cessi

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    12 ¤ CHAPTER 1 FUNCTIONS AND SEQUENCES 25. (a) 1980 1990 2000 2010 20 40 60 80 100 120 x y 0 Year Count (b) 1980 1990 2000 2010 20 40 60 80 100 120 x y 0 Year Count We see from the graph that there were approximately 92,000 birds in 1997. 26. (a) 0 1 2 3 4 0.2 t C 0.4 (b) Alcohol concentration increases rapidly within the fi rst hour of consumption and then slowly decreases over the following three hours. 27.  (  ) = 3  2 −  + 2   (2) = 3(2) 2 − 2 + 2 = 12 − 2 + 2 = 12   ( − 2) = 3( − 2) 2 − ( − 2) + 2 = 12 + 2 + 2 = 16   (  ) = 3  2 −  + 2   ( −  ) = 3( −  ) 2 − ( −  ) + 2 = 3  2 +  + 2   (  + 1) = 3(  + 1) 2 − (  + 1) + 2 = 3(  2 + 2  + 1) −  − 1 + 2 = 3  2 + 6  + 3 −  + 1 = 3  2 + 5  + 4  2  (  ) = 2 ·  (  ) = 2(3  2 −  + 2) = 6  2 − 2  + 4   (2  ) = 3(2  ) 2 − (2  ) + 2 = 3(4  2 ) − 2  + 2 = 12  2 − 2  + 2   (  2 ) = 3(  2 ) 2 − (  2 ) + 2 = 3(  4 ) −  2 + 2 = 3  4 −  2 + 2  [  (  )] 2 =  3  2 −  + 2  2 =  3  2 −  + 2  3  2 −  + 2  = 9  4 − 3  3 + 6  2 − 3  3 +  2 − 2  + 6  2 − 2  + 4 = 9  4 − 6  3 + 13  2 − 4  + 4   (  +  ) = 3(  +  ) 2 − (  +  ) + 2 = 3(  2 + 2  +  2 ) −  −  + 2 = 3  2 + 6  + 3  2 −  −  + 2  28. A spherical balloon with radius  + 1 has volume  (  + 1) = 4 3  (  + 1) 3 = 4 3  (  3 + 3  2 + 3  + 1) . We wish to fi nd the amount of air needed to in fl ate the balloon from a radius of  to  + 1 . Hence, we need to fi nd the difference  (  + 1) −  (  ) = 4 3  (  3 + 3  2 + 3  + 1) − 4 3  3 = 4 3  (3  2 + 3  + 1) . 29.  (  ) = 4 + 3  −  2 , so  (3 +  ) = 4 + 3(3 +  ) − (3 +  ) 2 = 4 + 9 + 3  − (9 + 6  +  2 ) = 4 − 3  −  2 , and  (3 +  ) −  (3)  = (4 − 3  −  2 ) − 4  =  ( − 3 −  )  = − 3 −  . post , in

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    SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION ¤ 13 30.  (  ) =  3 , so  (  +  ) = (  +  ) 3 =  3 + 3  2  + 3  2 +  3 , and  (  +  ) −  (  )  = (  3 + 3  2  + 3  2 +  3 ) −  3  =  (3  2 + 3  +  2 )  = 3  2 + 3  +  2 . 31.  (  ) −  (  )  −  = 1  − 1   −  =  −    −  =  −   (  −  ) = − 1(  −  )  (  −  ) = − 1  32.  (  ) −  (1)  − 1 =  + 3  + 1 − 2  − 1 =  + 3 − 2(  + 1)  + 1  − 1 =  + 3 − 2  − 2 (  + 1)(  − 1) = −  + 1 (  + 1)(  − 1) = − (  − 1) (  + 1)(  − 1) = − 1  + 1 33.  (  ) = (  + 4)  (  2 − 9) is de fi ned for all  except when 0 =  2 − 9 ⇔ 0 = (  + 3)(  − 3) ⇔  = − 3 or 3 , so the domain is {  ∈ R |  6 = − 3  3 } = ( −∞  − 3) ∪ ( − 3  3) ∪ (3  ∞ ) . 34.  (  ) = (2  3 − 5)  (  2 +  − 6) is de fi ned for all  except when 0 =  2 +  − 6 ⇔ 0 = (  + 3)(  − 2) ⇔  = − 3 or 2 , so the domain is {  ∈ R |  6 = − 3  2 } = ( −∞  − 3) ∪ ( − 3  2) ∪ (2  ∞ ) . 35.  (  ) = 3 √ 2  − 1 is de fi ned for all real numbers. In fact 3   (  ) , where  (  ) is a polynomial, is de fi ned for all real numbers. Thus, the domain is R  or ( −∞  ∞ ) . 36.  (  ) = √ 3 −  − √ 2 +  is de fi ned when 3 −  ≥ 0 ⇔  ≤ 3 and 2 +  ≥ 0 ⇔  ≥ − 2 . Thus, the domain is − 2 ≤  ≤ 3 , or [ − 2  3] . 37.  (  ) = 1  4 √  2 − 5  is de fi ned when  2 − 5   0 ⇔  (  − 5)  0 . Note that  2 − 5  6 = 0 since that would result in division by zero. The expression  (  − 5) is positive if   0 or   5 . (See Appendix A for methods for solving inequalities.) Thus, the domain is ( −∞  0) ∪ (5  ∞ ) . 38.  (  ) =  + 1 1 + 1  + 1 is de fi ned when  + 1 6 = 0 [  6 = − 1 ] and 1 + 1  + 1 6 = 0 . Since 1 + 1  + 1 = 0 ⇒ 1  + 1 = − 1 ⇒ 1 = −  − 1 ⇒  = − 2 , the domain is {  |  6 = − 2 ,  6 = − 1 } = ( −∞  − 2) ∪ ( − 2  − 1) ∪ ( − 1  ∞ ) . 39.  (  ) =  2 − √  is de fi ned when  ≥ 0 and 2 − √  ≥ 0 . Since 2 − √  ≥ 0 ⇒ 2 ≥ √  ⇒ √  ≤ 2 ⇒ 0 ≤  ≤ 4 , the domain is [0  4] . 40.  (  ) = √ 4 −  2 . Now  = √ 4 −  2 ⇒  2 = 4 −  2 ⇔  2 +  2 = 4 , so the graph is the top half of a circle of radius 2 with center at the origin. The domain is   | 4 −  2 ≥ 0  =   | 4 ≥  2  = {  | 2 ≥ |  |} = [ − 2  2] . From the graph, the range is 0 ≤  ≤ 2 , or [0  2] . or du cessi

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    14 ¤ CHAPTER 1 FUNCTIONS AND SEQUENCES 41.  (  ) = 2 − 0  4  is de fi ned for all real numbers, so the domain is R , or ( −∞  ∞ )  The graph of  is a line with slope − 0  4 and  -intercept 2 . 42.  (  ) =  2 − 2  + 1 = (  − 1) 2 is de fi ned for all real numbers, so the domain is R , or ( −∞  ∞ ) . The graph of  is a parabola with vertex (1  0) . 43.  (  ) = 2  +  2 is de fi ned for all real numbers, so the domain is R , or ( −∞  ∞ ) . The graph of  is a parabola opening upward since the coef fi cient of  2 is positive. To fi nd the  -intercepts, let  = 0 and solve for  . 0 = 2  +  2 =  (2 +  ) ⇒  = 0 or  = − 2 . The  -coordinate of the vertex is halfway between the  -intercepts, that is, at  = − 1 . Since  ( − 1) = 2( − 1) + ( − 1) 2 = − 2 + 1 = − 1 , the vertex is ( − 1  − 1) . 44.  (  ) = 4 −  2 2 −  = (2 +  )(2 −  ) 2 −  , so for  6 = 2 ,  (  ) = 2 +  . The domain is {  |  6 = 2 } . So the graph of  is the same as the graph of the function  (  ) =  + 2 (a line) except for the hole at (2  4) . 45.  (  ) = √  − 5 is de fi ned when  − 5 ≥ 0 or  ≥ 5 , so the domain is [5  ∞ ) . Since  = √  − 5 ⇒  2 =  − 5 ⇒  =  2 + 5 , we see that  is the top half of a parabola. 46.  (  ) = | 2  + 1 | =  2  + 1 − (2  + 1) if 2  + 1 ≥ 0 if 2  + 1  1 =  2  + 1 − 2  − 1 if  ≥ − 1 2 if   − 1 2 The domain is R , or ( −∞  ∞ ) . post , in

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    SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION ¤ 15 47.  (  ) = 3  + |  |  . Since |  | =   if  ≥ 0 −  if   0 , we have  (  ) =      3  +   if   0 3  −   if   0 =      4   if   0 2   if   0 =  4 if   0 2 if   0 Note that  is not de fi ned for  = 0 . The domain is ( −∞  0) ∪ (0  ∞ ) . 48.  (  ) = |  | −  =   −  if  ≥ 0 −  −  if   0 =  0 if  ≥ 0 − 2  if   0 . The domain is R , or ( −∞  ∞ ) . 49.  (  ) =   + 2 if   0 1 −  if  ≥ 0 The domain is R . 50.  (  ) =  3 − 1 2  if  ≤ 2 2  − 5 if   2 The domain is R . 51.  (  ) =   + 2 if  ≤ − 1  2 if   − 1 Note that for  = − 1 , both  + 2 and  2 are equal to 1. The domain is R . 52.  (  ) =       + 9 if   − 3 − 2  if |  | ≤ 3 − 6 if   3 Note that for  = − 3 , both  + 9 and − 2  are equal to 6 ; and for  = 3 , both − 2  and − 6 are equal to − 6 . The domain is R . or du cessi

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