Test Bank for Biocalculus: Calculus for 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)
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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.
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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 .
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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.
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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,
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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.
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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.
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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.
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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]
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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.
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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.
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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 − .
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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] .
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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 ( −∞ ∞ ) .
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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 .
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16 ¤ CHAPTER 1 FUNCTIONS AND SEQUENCES 53. Let the length and width of the rectangle be and . Then the perimeter is 2 + 2 = 20 and the area is = . Solving the fi rst equation for in terms of gives = 20 − 2 2 = 10 − . Thus, ( ) = (10 − ) = 10 − 2 . Since lengths are positive, the domain of is 0 10 . If we further restrict to be larger than , then 5 10 would be the domain. 54. Let the length and width of the rectangle be and . Then the area is = 16 , so that = 16 . The perimeter is = 2 + 2 , so ( ) = 2 + 2(16 ) = 2 + 32 , and the domain of is 0 , since lengths must be positive quantities. If we further restrict to be larger than , then 4 would be the domain. 55. Let the length of a side of the equilateral triangle be . Then by the Pythagorean Theorem, the height of the triangle satis fi es 2 + 1 2 2 = 2 , so that 2 = 2 − 1 4 2 = 3 4 2 and = √ 3 2 . Using the formula for the area of a triangle, = 1 2 ( base )( height ) , we obtain ( ) = 1 2 ( ) √ 3 2 = √ 3 4 2 , with domain 0 . 56. Let the volume of the cube be and the length of an edge be . Then = 3 so = 3 √ , and the surface area is ( ) = 6 2 = 6 3 √ 2 = 6 2 3 , with domain 0 . 57. Let each side of the base of the box have length , and let the height of the box be . Since the volume is 2 , we know that 2 = 2 , so that = 2 2 , and the surface area is = 2 + 4 . Thus, ( ) = 2 + 4 (2 2 ) = 2 + (8 ) , with domain 0 . 58. We can summarize the monthly cost with a piecewise de fi ned function. ( ) = 35 if 0 ≤ ≤ 400 35 + 0 10( − 400) if 400 59. We can summarize the total cost with a piecewise de fi ned function. ( ) = 75 if 0 ≤ 2 150 + 50( − 2) if 2 60. One example is the amount paid for cable or telephone system repair in the home, usually measured to the nearest quarter hour. Another example is the amount paid by a student in tuition fees, if the fees vary according to the number of credits for which the student has registered. 61. The period can be estimated by measuring the peak-to-peak distance on the graph. This is approximately 77 hours. Note that the graph shown is for a single person’s temperature. The period for this species of malaria is, on average, 72 hours. 62. The cycle of increased body temperature followed by a drop in temperature is indicative of a recurrent fever. This is typical of a P. falciparum infection. The period is approximately 48 hours, but the fever is also subsiding. This might be because the person is being treated for infection. 63. is an odd function because its graph is symmetric about the origin. is an even function because its graph is symmetric with respect to the -axis.
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SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION ¤ 17 64. is not an even function since it is not symmetric with respect to the -axis. is not an odd function since it is not symmetric about the origin. Hence, is neither even nor odd. is an even function because its graph is symmetric with respect to the -axis. 65. (a) Because an even function is symmetric with respect to the -axis, and the point (5 3) is on the graph of this even function, the point ( − 5 3) must also be on its graph. (b) Because an odd function is symmetric with respect to the origin, and the point (5 3) is on the graph of this odd function, the point ( − 5 − 3) must also be on its graph. 66. (a) If is even, we get the rest of the graph by re fl ecting about the -axis. (b) If is odd, we get the rest of the graph by rotating 180 ◦ about the origin. 67. ( ) = 2 + 1 . ( − ) = − ( − ) 2 + 1 = − 2 + 1 = − 2 + 1 = − ( ) . So is an odd function. 68. ( ) = 2 4 + 1 . ( − ) = ( − ) 2 ( − ) 4 + 1 = 2 4 + 1 = ( ) . So is an even function. 69. ( ) = + 1 , so ( − ) = − − + 1 = − 1 . Since this is neither ( ) nor − ( ) , the function is neither even nor odd. 70. ( ) = | | . ( − ) = ( − ) | − | = ( − ) | | = − ( | | ) = − ( ) So is an odd function.
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18 ¤ CHAPTER 1 FUNCTIONS AND SEQUENCES 71. ( ) = 1 + 3 2 − 4 . ( − ) = 1+3( − ) 2 − ( − ) 4 = 1+3 2 − 4 = ( ) . So is an even function. 72. ( ) = 1 + 3 3 − 5 , so ( − ) = 1 + 3( − ) 3 − ( − ) 5 = 1 + 3( − 3 ) − ( − 5 ) = 1 − 3 3 + 5 Since this is neither ( ) nor − ( ) , the function is neither even nor odd. 73. (i) If and are both even functions, then ( − ) = ( ) and ( − ) = ( ) . Now ( + )( − ) = ( − ) + ( − ) = ( ) + ( ) = ( + )( ) , so + is an even function. (ii) If and are both odd functions, then ( − ) = − ( ) and ( − ) = − ( ) . Now ( + )( − ) = ( − ) + ( − ) = − ( ) + [ − ( )] = − [ ( ) + ( )] = − ( + )( ) , so + is an odd function. (iii) If is an even function and is an odd function, then ( + )( − ) = ( − ) + ( − ) = ( ) + [ − ( )] = ( ) − ( ) , which is not ( + )( ) nor − ( + )( ) , s o + is neither even nor odd. (Exception: if is the zero function, then + will be odd . If is the zero function, then + will be even .) 74. (i) If and are both even functions, then ( − ) = ( ) and ( − ) = ( ) . Now ( )( − ) = ( − ) ( − ) = ( ) ( ) = ( )( ) , so is an even function. (ii) If and are both odd functions, then ( − ) = − ( ) and ( − ) = − ( ) . Now ( )( − ) = ( − ) ( − ) = [ − ( )][ − ( )] = ( ) ( ) = ( )( ) , so is an even function. (iii) If is an even function and is an odd function, then ( )( − ) = ( − ) ( − ) = ( )[ − ( )] = − [ ( ) ( )] = − ( )( ) , so is an odd function. 1.2 Mathematical Models: A Catalog of Essential Functions 1. (a) ( ) = log 2 is a logarithmic function. (b) ( ) = 4 √ is a root function with = 4 . (c) ( ) = 2 3 1 − 2 is a rational function because it is a ratio of polynomials. (d) ( ) = 1 − 1 1 + 2 54 2 is a polynomial of degree 2 (also called a quadratic function ). (e) ( ) = 5 is an exponential function. (f) ( ) = sin cos 2 is a trigonometric function. 2. (a) = is an exponential function (notice that is the exponent ). (b) = is a power function (notice that is the base ).
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SECTION 1.2 MATHEMATICAL MODELS: A CATALOG OF ESSENTIAL FUNCTIONS ¤ 19 (c) = 2 (2 − 3 ) = 2 2 − 5 is a polynomial of degree 5 . (d) = tan − cos is a trigonometric function. (e) = (1 + ) is a rational function because it is a ratio of polynomials. (f) = √ 3 − 1 (1 + 3 √ ) is an algebraic function because it involves polynomials and roots of polynomials. 3. We notice from the fi gure that and are even functions (symmetric with respect to the -axis) and that is an odd function (symmetric with respect to the origin). So (b) = 5 must be . Since is fl atter than near the origin, we must have (c) = 8 matched with and (a) = 2 matched with . 4. (a) The graph of = 3 is a line (choice ). (b) = 3 is an exponential function (choice ). (c) = 3 is an odd polynomial function or power function (choice ). (d) = 3 √ = 1 3 is a root function (choice ). 5. (a) An equation for the family of linear functions with slope 2 is = ( ) = 2 + , where is the -intercept. (b) (2) = 1 means that the point (2 1) is on the graph of . We can use the point-slope form of a line to obtain an equation for the family of linear functions through the point (2 1) . − 1 = ( − 2) , which is equivalent to = + (1 − 2 ) in slope-intercept form. (c) To belong to both families, an equation must have slope = 2 , so the equation in part (b), = + (1 − 2 ) , becomes = 2 − 3 . It is the only function that belongs to both families. 6. All members of the family of linear functions ( ) = 1 + ( + 3) have graphs that are lines passing through the point ( − 3 1) .
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20 ¤ CHAPTER 1 FUNCTIONS AND SEQUENCES 7. All members of the family of linear functions ( ) = − have graphs that are lines with slope − 1 . The -intercept is . 8. The vertex of the parabola on the left is (3 0) , so an equation is = ( − 3) 2 + 0 . Since the point (4 2) is on the parabola, we’ll substitute 4 for and 2 for to fi nd . 2 = (4 − 3) 2 ⇒ = 2 , so an equation is ( ) = 2( − 3) 2 . The -intercept of the parabola on the right is (0 1) , so an equation is = 2 + + 1 . Since the points ( − 2 2) and (1 − 2 5) are on the parabola, we’ll substitute − 2 for and 2 for as well as 1 for and − 2 5 for to obtain two equations with the unknowns and . ( − 2 2) : 2 = 4 − 2 + 1 ⇒ 4 − 2 = 1 (1) (1 − 2 5) : − 2 5 = + + 1 ⇒ + = − 3 5 (2) 2 · (2) + (1) gives us 6 = − 6 ⇒ = − 1 . F rom (2) , − 1 + = − 3 5 ⇒ = − 2 5 , so an equation is ( ) = − 2 − 2 5 + 1 . 9. Since ( − 1) = (0) = (2) = 0 , has zeros of − 1 , 0 , and 2 , so an equation for is ( ) = [ − ( − 1)]( − 0)( − 2) , or ( ) = ( + 1)( − 2) . Because (1) = 6 , we’ll substitute 1 for and 6 for ( ) . 6 = (1)(2)( − 1) ⇒ − 2 = 6 ⇒ = − 3 , so an equation for is ( ) = − 3 ( + 1)( − 2) . 10. (a) For = 0 02 + 8 50 , the slope is 0 02 , which means that the average surface temperature of the world is increasing at a rate of 0 02 ◦ C per year. The -intercept is 8 50 , which represents the average surface temperature in ◦ C in the year 1900. (b) = 2100 − 1900 = 200 ⇒ = 0 02(200) + 8 50 = 12 50 ◦ C 11. (a) = 200 , so = 0 0417 ( + 1) = 0 0417(200)( + 1) = 8 34 + 8 34 . The slope is 8 34 , which represents the change in mg of the dosage for a child for each change of 1 year in age. (b) For a newborn, = 0 , so = 8 34 mg. 12. (a) We are given change in pressure 10 feet change in depth = 4 34 10 = 0 434 . Using for pressure and for depth with the point ( ) = (0 15) , we have the slope-intercept form of the line, = 0 434 + 15 . (b) When = 100 , then 100 = 0 434 + 15 ⇔ 0 434 = 85 ⇔ = 85 0 434 ≈ 195 85 feet. Thus, the pressure is 100 lb in 2 at a depth of approximately 196 feet. 13. (a) (b) The slope of 9 5 means that increases 9 5 degrees for each increase of 1 ◦ C . (Equivalently, increases by 9 when increases by 5 and decreases by 9 when decreases by 5 .) The -intercept of 32 is the Fahrenheit temperature corresponding to a Celsius temperature of 0 .
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SECTION 1.2 MATHEMATICAL MODELS: A CATALOG OF ESSENTIAL FUNCTIONS ¤ 21 14. (a) Assuming is a linear function of we can sketch the graph of ( ) by plotting the points (150 0 35) and (50 0 14) and drawing the straight line that passes through both these points. 50 150 V (150, 0.35) (50, 0.14) A 100 0 0.1 0.4 0.3 0.2 (b) The slope is = (150) − (50) 150 − 50 = 0 35 − 0 14 100 = 0 0021 min − 1 This represents the rate of change of absorption rate with respect to volume. The slope of 0 0021 means that increases by 0 0021 mL min for each 1 mL increase in (c) The -intercept of 0 035 mL min is the absorption rate corresponding to a cerebrospinal fl uid volume of 0 mL. 15. (a) Using in place of and in place of , we fi nd the slope to be 2 − 1 2 − 1 = 80 − 70 173 − 113 = 10 60 = 1 6 . So a linear equation is − 80 = 1 6 ( − 173) ⇔ − 80 = 1 6 − 173 6 ⇔ = 1 6 + 307 6 307 6 = 51 16 . (b) The slope of 1 6 means that the temperature in Fahrenheit degrees increases one-sixth as rapidly as the number of cricket chirps per minute. Said differently, each increase of 6 cricket chirps per minute corresponds to an increase of 1 ◦ F . (c) When = 150 , the temperature is given approximately by = 1 6 (150) + 307 6 = 76 16 ◦ F ≈ 76 ◦ F . 16. (a) Using in place of and in place of , we fi nd the slope to be 2 − 1 2 − 1 = 460 − 380 800 − 480 = 80 320 = 1 4 . So a linear equation is − 460 = 1 4 ( − 800) ⇔ − 460 = 1 4 − 200 ⇔ = 1 4 + 260 . (b) Letting = 1500 we get = 1 4 (1500) + 260 = 635 . The cost of driving 1500 miles is $635. (c) The slope of the line represents the cost per mile, $0 25 . (d) The -intercept represents the fi xed cost, $260. (e) A linear function gives a suitable model in this situation because you have fi xed monthly costs such as insurance and car payments, as well as costs that increase as you drive, such as gasoline, oil, and tires, and the cost of these for each additional mile driven is a constant. 17. (a) The data appear to be periodic and a sine or cosine function would make the best model. A model of the form ( ) = cos( ) + seems appropriate. (b) The data appear to be decreasing in a linear fashion. A model of the form ( ) = + seems appropriate. 18. (a) The data appear to be increasing exponentially. A model of the form ( ) = · or ( ) = · + seems appropriate. (b) The data appear to be decreasing similarly to the values of the reciprocal function. A model of the form ( ) = seems appropriate.
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22 ¤ CHAPTER 1 FUNCTIONS AND SEQUENCES 19. (a) A linear model does seem appropriate. (b) Using the points (4000 14 1) and (60 , 000 8 2) , we obtain − 14 1 = 8 2 − 14 1 60 , 000 − 4000 ( − 4000) or, equivalently, ≈ − 0 000105357 + 14 521429 . (c) Using a computing device, we obtain the least squares regression line = − 0 0000997855 + 13 950764 . The following commands and screens illustrate how to fi nd the least squares regression line on a TI-84 Plus. Enter the data into list one (L1) and list two (L2). Press to enter the editor. Find the regession line and store it in Y 1 . Press . Note from the last fi gure that the regression line has been stored in Y 1 and that Plot1 has been turned on (Plot1 is highlighted). You can turn on Plot1 from the Y= menu by placing the cursor on Plot1 and pressing or by pressing . Now press to produce a graph of the data and the regression line. Note that choice 9 of the ZOOM menu automatically selects a window that displays all of the data. (d) When = 25 , 000 , ≈ 11 456 ; or about 11 5 per 100 population. (e) When = 80 , 000 , ≈ 5 968 ; or about a 6% chance.
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SECTION 1.2 MATHEMATICAL MODELS: A CATALOG OF ESSENTIAL FUNCTIONS ¤ 23 (f) When = 200 , 000 , is negative, so the model does not apply. 20. (a) (b) Using a computing device, we obtain the least squares regression line = 4 856 − 220 96 . (c) When = 100 ◦ F , = 264 7 ≈ 265 chirps min. 21. (a) 35 55 140 185 L H (b) Using a calculator to perform a linear regression gives = 1 8807 + 82 6497 where is the height in centimeters and is the femur length in centimeters. This line, having slope 1.88 and -intercept 82.65, is plotted below. 35 55 140 185 L H (c) The height of a person with = 53 is (53) = (1 8807)(53) + 82 6497 ≈ 182 3 cm . 22. (a) Using a calculator to perform a linear regression gives = 0 0188 + 0 3048 (b) The plot shows that the data is approximately linear. A higher degree polynomial fi t, such as a cubic, may better model the data. 0 3000 60 (c) The -intercept represents the percentage of mice that developed tumors without any asbestos exposure. 23. If is the original distance from the source, then the illumination is ( ) = − 2 = 2 . Moving halfway to the lamp gives us an illumination of 1 2 = 1 2 − 2 = (2 ) 2 = 4( 2 ) , so the light is 4 times as bright. 24. (a) Set = 90 in and solve for : 90 = 30 6 0 3952 ⇐⇒ 90 30 6 = 0 3952 ⇐⇒ = 90 30 6 1 0 3952 ≈ 15 33 lb (b) Set = 300 lb and calculate: = 30 6 (300) 0 3952 ≈ 291 5 in (c) According to the model, a 300 lb ostrich needs a wingspan of 292 in to fl y. Therefore, an ostrich with a 72 in wingspan cannot generate enough lift for fl ight.
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24 ¤ CHAPTER 1 FUNCTIONS AND SEQUENCES 25. (a) Using a computing device, we obtain a power function = , where ≈ 3 1046 and ≈ 0 308 . (b) If = 291 , then = ≈ 17 8 , so you would expect to fi nd 18 species of reptiles and amphibians on Dominica. 26. (a) = 1 000 431 227 1 499 528 750 (b) The power model in part (a) is approximately = 1 5 . Squaring both sides gives us 2 = 3 , so the model matches Kepler’s Third Law, 2 = 3 . 27. (a) Using a calculator to perform a 3rd-degree polynomial regression gives = 0 0155 3 − 0 3725 2 + 3 9461 + 1 2108 where is age and is length. This polynomial is plotted along with a scatterplot of the data. 0 15 30 A L (b) A 5-year old rock bass has a length of (5) = (0 0155)(5) 3 − (0 3725)(5) 2 + (3 9461)(5) + 1 2108 ≈ 13 6 in (c) Using computer algebra software to solve for in the equation 20 = 0 0155 3 − 0 3725 2 + 3 9461 + 1 2108 gives ≈ 10 88 years. Alternatively, the graph from part (a) can be used to estimate the age when = 20 by drawing a horizontal line at = 20 to the curve and observing the age at this point. 1.3 New Functions from Old Functions 1. (a) If the graph of is shifted 3 units upward, its equation becomes = ( ) + 3 . (b) If the graph of is shifted 3 units downward, its equation becomes = ( ) − 3 . (c) If the graph of is shifted 3 units to the right, its equation becomes = ( − 3) . (d) If the graph of is shifted 3 units to the left, its equation becomes = ( + 3) . (e) If the graph of is re fl ected about the -axis, its equation becomes = − ( ) . (f) If the graph of is re fl ected about the -axis, its equation becomes = ( − ) . (g) If the graph of is stretched vertically by a factor of 3 , its equation becomes = 3 ( ) . (h) If the graph of is shrunk vertically by a factor of 3 , its equation becomes = 1 3 ( ) . 2. (a) To obtain the graph of = ( ) + 8 from the graph of = ( ) , shift the graph 8 units upward. (b) To obtain the graph of = ( + 8) from the graph of = ( ) , shift the graph 8 units to the left. (c) To obtain the graph of = 8 ( ) from the graph of = ( ) , stretch the graph vertically by a factor of 8 . (d) To obtain the graph of = (8 ) from the graph of = ( ) , shrink the graph horizontally by a factor of 8 . (e) To obtain the graph of = − ( ) − 1 from the graph of = ( ) , fi rst re fl ect the graph about the -axis, and then shift it 1 unit downward. (f) To obtain the graph of = 8 ( 1 8 ) from the graph of = ( ) , stretch the graph horizontally and vertically by a factor of 8 . 3. (a) (graph 3) The graph of is shifted 4 units to the right and has equation = ( − 4) . (b) (graph 1) The graph of is shifted 3 units upward and has equation = ( ) + 3 . (c) (graph 4) The graph of is shrunk vertically by a factor of 3 and has equation = 1 3 ( ) .
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SECTION 1.3 NEW FUNCTIONS FROM OLD FUNCTIONS ¤ 25 (d) (graph 5) The graph of is shifted 4 units to the left and re fl ected about the -axis. Its equation is = − ( + 4) . (e) (graph 2) The graph of is shifted 6 units to the left and stretched vertically by a factor of 2 . Its equation is = 2 ( + 6) . 4. (a) To graph = ( ) − 2 , we shift the graph of , 2 units downward.The point (1 2) on the graph of corresponds to the point (1 2 − 2) = (1 0) . (b) To graph = ( − 2) , we shift the graph of , 2 units to the right.The point (1 2) on the graph of corresponds to the point (1 + 2 2) = (3 2) . (c) To graph = − 2 ( ) , we re fl ect the graph about the -axis and stretch the graph vertically by a factor of 2 . The point (1 2) on the graph of corresponds to the point (1 − 2 · 2) = (1 − 4) . (d) To graph = ( 1 3 ) + 1 , we stretch the graph horizontally by a factor of 3 and shift it 1 unit upward. The point (1 2) on the graph of corresponds to the point (1 · 3 2 + 1) = (3 3) . 5. (a) To graph = (2 ) we shrink the graph of horizontally by a factor of 2 . The point (4 − 1) on the graph of corresponds to the point 1 2 · 4 − 1 = (2 − 1) . (b) To graph = 1 2 we stretch the graph of horizontally by a factor of 2 . The point (4 − 1) on the graph of corresponds to the point (2 · 4 − 1) = (8 − 1) . (c) To graph = ( − ) we re fl ect the graph of about the -axis. The point (4 − 1) on the graph of corresponds to the point ( − 1 · 4 − 1) = ( − 4 − 1) . (d) To graph = − ( − ) we re fl ect the graph of about the -axis, then about the -axis. The point (4 − 1) on the graph of corresponds to the point ( − 1 · 4 − 1 · − 1) = ( − 4 1) .
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26 ¤ CHAPTER 1 FUNCTIONS AND SEQUENCES 6. (a) The graph of = 2 sin can be obtained from the graph of = sin by stretching it vertically by a factor of 2 . (b) The graph of = 1 + √ can be obtained from the graph of = √ by shifting it upward 1 unit. 7. = 1 + 2 : Start with the graph of the reciprocal function = 1 and shift 2 units to the left. 8. = ( − 1) 3 : Start with the graph of = 3 and shift 1 unit to the right. 9. = − 3 √ : Start with the graph of = 3 √ and re fl ect about the -axis. 10. = 2 + 6 + 4 = ( 2 + 6 + 9) − 5 = ( + 3) 2 − 5 : Start with the graph of = 2 , shift 3 units to the left, and then shift 5 units downward.
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SECTION 1.3 NEW FUNCTIONS FROM OLD FUNCTIONS ¤ 27 11. = √ − 2 − 1 : Start with the graph of = √ , shift 2 units to the right, and then shift 1 unit downward. 12. = 4 sin 3 : Start with the graph of = sin , compress horizontally by a factor of 3 , and then stretch vertically by a factor of 4 . 13. = sin( 2) : Start with the graph of = sin and stretch horizontally by a factor of 2 . 14. = 2 − 2 : Start with the graph of = 1 , stretch vertically by a factor of 2 , and then shift 2 units downward. 15. = − 3 : Start with the graph of = 3 and re fl ect about the -axis. Note: Re fl ecting about the -axis gives the same result since substituting − for gives us = ( − ) 3 = − 3 .
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28 ¤ CHAPTER 1 FUNCTIONS AND SEQUENCES 16. = 1 − 2 √ + 3 : Start with the graph of = √ , shift 3 units to the left, stretch vertically by a factor of 2 , re fl ect about the -axis, and then shift 1 unit upward. 17. = 1 2 (1 − cos ) : Start with the graph of = cos , re fl ect about the -axis, shift 1 unit upward, and then shrink vertically by a factor of 2. 18. = | | − 2 : Start with the graph of = | | and shift 2 units downward. 19. = 1 − 2 − 2 = − ( 2 + 2 ) + 1 = − ( 2 + 2 + 1) + 2 = − ( + 1) 2 + 2 : Start with the graph of = 2 , re fl ect about the -axis, shift 1 unit to the left, and then shift 2 units upward.
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SECTION 1.3 NEW FUNCTIONS FROM OLD FUNCTIONS ¤ 29 20. = 1 4 tan( − 4 ) : Start with the graph of = tan , shift 4 units to the right, and then compress vertically by a factor of 4 . 21. This is just like the solution to Example 4 except the amplitude of the curve (the 30 ◦ N curve in Figure 9 on June 21) is 14 − 12 = 2 . So the function is ( ) = 12 + 2 sin 2 365 ( − 80) . March 31 is the 90 th day of the year, so the model gives (90) ≈ 12 34 h. The daylight time (5:51 AM to 6:18 PM ) is 12 hours and 27 minutes, or 12 45 h. The model value differs from the actual value by 12 45 − 12 34 12 45 ≈ 0 009 , less than 1% . 22. Using a sine function to model the brightness of Delta Cephei as a function of time, we take its period to be 5 4 days, its amplitude to be 0 35 (on the scale of magnitude), and its average magnitude to be 4 0 . If we take = 0 at a time of average brightness, then the magnitude (brightness) as a function of time in days can be modeled by the formula ( ) = 4 0 + 0 35 sin 2 5 4 . 23. Let ( ) be the water depth in meters at hours after midnight. Apply the following transformations to the cosine function: • Vertical stretch by factor 5 since the amplitude needs to be 12 − 2 2 = 5 m • Horizontal stretch by factor 12 2 = 6 since the period needs to be 12 h • Vertical shift 7 units upward since the function ranges between 2 and 12 which has a midpoint of 12+2 2 = 7 m • Horizontal shift 6 75 units to right to position the maximum at = 6 75 h (6:45AM) Combining these transformations gives the water depth function ( ) = 5 cos 6 ( − 6 75) + 7 24. Let ( ) be the total volume of air in mL after seconds. Because the respiratory cycle is periodic, a sine function can be used as a model by applying the following transformations: • Vertical stretch by factor 250 since the amplitude needs to be 500 2 = 250 mL • Horizontal stretch by factor 4 2 = 2 since the period needs to be 4 s • Vertical shift 2250 units upward since the function ranges between 2000 and 2500 which has a midpoint of 2000+2500 2 = 2250 mL Combining these transformations gives the volume function ( ) = 250 sin 2 + 2250 25. Let ( ) be the gene frequency after years. The gene frequency dynamics can be modeled using a sine function with the following transformations: • Vertical stretch by factor 30 since the amplitude needs to be 80 − 20 2 = 30% • Horizontal stretch by factor 3 2 since the period needs to be 3 years • Vertical shift 50 units upward since the function ranges between 80 and 20 which has a midpoint of 80+20 2 = 50 Combining these transformations gives the gene frequency function ( ) = 30 sin 2 3 + 50 26. Let ( ) be the density of neutrophils in cells/ L after days. The density is periodic and can be modeled using a cosine function with the following transformations: • Vertical stretch by factor 1000 since the amplitude needs to be 2000 − 0 2 = 1000 • Horizontal stretch by factor 21 2 since the period needs to be 21 days (or 3 weeks) • Vertical shift 1000 units upward since the function ranges between 0 and 2000 which has a midpoint of 2000+0 2 = 1000 Combining these transformations gives the density function ( ) = 1000 cos 2 + 1000
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30 ¤ CHAPTER 1 FUNCTIONS AND SEQUENCES 27. ( ) = 3 + 2 2 ; ( ) = 3 2 − 1 . = R for both and . (a) ( + )( ) = ( 3 + 2 2 ) + (3 2 − 1) = 3 + 5 2 − 1 , = R . (b) ( − )( ) = ( 3 + 2 2 ) − (3 2 − 1) = 3 − 2 + 1 , = R . (c) ( )( ) = ( 3 + 2 2 )(3 2 − 1) = 3 5 + 6 4 − 3 − 2 2 , = R . (d) ( ) = 3 + 2 2 3 2 − 1 , = | 6 = ± 1 √ 3 since 3 2 − 1 6 = 0 . 28. ( ) = √ 3 − , = ( −∞ 3] ; ( ) = √ 2 − 1 , = ( −∞ − 1] ∪ [1 ∞ ) . (a) ( + )( ) = √ 3 − + √ 2 − 1 , = ( −∞ − 1] ∪ [1 3] , which is the intersection of the domains of and . (b) ( − )( ) = √ 3 − − √ 2 − 1 , = ( −∞ − 1] ∪ [1 3] . (c) ( )( ) = √ 3 − · √ 2 − 1 , = ( −∞ − 1] ∪ [1 3] . (d) ( ) = √ 3 − √ 2 − 1 , = ( −∞ − 1) ∪ (1 3] . We must exclude = ± 1 since these values would make unde fi ned. 29. ( ) = 2 − 1 , = R ; ( ) = 2 + 1 , = R . (a) ( ◦ )( ) = ( ( )) = (2 + 1) = (2 + 1) 2 − 1 = (4 2 + 4 + 1) − 1 = 4 2 + 4 , = R . (b) ( ◦ )( ) = ( ( )) = ( 2 − 1) = 2( 2 − 1) + 1 = (2 2 − 2) + 1 = 2 2 − 1 , = R . (c) ( ◦ )( ) = ( ( )) = ( 2 − 1) = ( 2 − 1) 2 − 1 = ( 4 − 2 2 + 1) − 1 = 4 − 2 2 , = R . (d) ( ◦ )( ) = ( ( )) = (2 + 1) = 2(2 + 1) + 1 = (4 + 2) + 1 = 4 + 3 , = R . 30. ( ) = − 2 ; ( ) = 2 + 3 + 4 . = R for both and , and hence for their composites. (a) ( ◦ )( ) = ( ( )) = ( 2 + 3 + 4) = ( 2 + 3 + 4) − 2 = 2 + 3 + 2 . (b) ( ◦ )( ) = ( ( )) = ( − 2) = ( − 2) 2 + 3( − 2) + 4 = 2 − 4 + 4 + 3 − 6 + 4 = 2 − + 2 . (c) ( ◦ )( ) = ( ( )) = ( − 2) = ( − 2) − 2 = − 4 . (d) ( ◦ )( ) = ( ( )) = ( 2 + 3 + 4) = ( 2 + 3 + 4) 2 + 3( 2 + 3 + 4) + 4 = ( 4 + 9 2 + 16 + 6 3 + 8 2 + 24 ) + 3 2 + 9 + 12 + 4 = 4 + 6 3 + 20 2 + 33 + 32 31. ( ) = 1 − 3 ; ( ) = cos . = R for both and , and hence for their composites. (a) ( ◦ )( ) = ( ( )) = (cos ) = 1 − 3 cos . (b) ( ◦ )( ) = ( ( )) = (1 − 3 ) = cos(1 − 3 ) . (c) ( ◦ )( ) = ( ( )) = (1 − 3 ) = 1 − 3(1 − 3 ) = 1 − 3 + 9 = 9 − 2 . (d) ( ◦ )( ) = ( ( )) = (cos ) = cos(cos ) [Note that this is not cos · cos .] 32. ( ) = √ , = [0 ∞ ) ; ( ) = 3 √ 1 − , = R . (a) ( ◦ )( ) = ( ( )) = 3 √ 1 − = 3 √ 1 − = 6 √ 1 − . The domain of ◦ is { | 3 √ 1 − ≥ 0 } = { | 1 − ≥ 0 } = { | ≤ 1 } = ( −∞ 1] . (b) ( ◦ )( ) = ( ( )) = ( √ ) = 3 1 − √ . The domain of ◦ is { | is in the domain of and ( ) is in the domain of } . This is the domain of , that is, [0 ∞ ) . (c) ( ◦ )( ) = ( ( )) = ( √ ) = √ = 4 √ . The domain of ◦ is { | ≥ 0 and √ ≥ 0 } = [0 ∞ ) .
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