MAT 210 Test #1 Solutions, Form A

Size: px

Start display at page:

Download "MAT 210 Test #1 Solutions, Form A"

Jasper Atkins
5 years ago
Views:

1 1. Where are the following functions continuous? a. ln(x 2 1) MAT 210 Test #1 Solutions, Form A Solution: The ln function is continuous when what you are taking the log of is positive. Hence, we need x 2 1 > 0. This happens if x > 1 or x < 1. Grading: +5 points for x 2 1 > 0, +5 points for solving x 2 1 > 0 for x. Grading for common mistakes: 2 points for just x > 1; 2 points for x (which is where ln(x 2 1) is defined); +5 points total for work and an answer of x = ±1. x + 1, if x > 2; b. f(x) = 3, if x = 2; x 2 + 2, if x < 2. Solution: The function x is a polynomial and hence continuous everywhere. Hence f(x) is continuous where f(x) locally looks like x 2 +2, i.e., when x < 2. Similarly, x+1 is a continuous function, so f(x) is continuous where x > 2. To find out whether f(x) is continuous at 2, we need to see whether f(x) = f(2). To evaluate the it, we need to find two one-sided its: x 2 f(x) = (x + 1) = = 3; x 2 + x 2 + f(x) = x 2 x 2 (x2 + 2) = = 6. Since these numbers are different, x 2 f(x) does not exist, and so f(x) is not continous at 2. Thus, f(x) is continuous at all real numbers except 2. Grading: +5 points for handling all real numbers except 2, +10 points for determining whether f(x) is continuous at 2. Grading for common mistakes: +7 points (total) for a correct answer but no work. 2. Find the following its algebraically, if they exist. a. x + 3x 1 Solution: Since x is getting large in absolute value, and the function afterwards is a rational function, we need to divide the numerator and denominator by the largest power of x in the rational function, namely x 2. Then x + 3x 1 ()/x = x + (3x 1)/x 2 = x 1 x 2 x + 3 x 1 = x getting a non-zero number divided by zero, which means the it does not exist (DNE). Grading: +5 points for dividing the numerator and denominator by x 2, +5 points for determining the it DNE. Grading for common mistakes: +3 points (total) for plugging in a big number; 3 points for 1 0 = 0 or 1 0 = 1. 1 = 1 0,

2 b. x 3 3x 1 Solution: To evaluate this it, you can just substitute x = 3, since a rational function is continuous, provided the denominator is non-zero: x 3 3x 1 = (3) 1 3(3) 1 = 14 8 = 7 4. Grading: +5 points for substituting, +5 points for simplifying. Grading for common mistakes: 3 points for bad arithmetic; +5 points total for 0 0 c. x 3 + 3x 2 4x x 2 4x + 3 and nothing else. Solution: You are taking the it of a rational function as x approaches a number, so we start by trying to substitute x = 1: x 3 + 3x 2 4x x 2 4x + 3 = (1) 2 4(1) (1) 2 4(1) + 3 = 0 0. This 0 means the it might exist, or it might not. We need to factor the numerator and 0 denominator, cancel a factor of x 1, then try evaluating the it again. and x 3 + 3x 2 4x x 2 4x + 3 = x(x2 + 3x 4) (x 1)(x 3) = x(x 1)(x + 4) (x 1)(x 3) = x(x + 4) x 3, x 3 + 3x 2 4x x 2 4x + 3 = x(x + 4) = 1 5 x 3 2 = 5 2. Grading: +5 points for cancelling, +5 points for re-evaluating the it. d. x + ( 2 3e x ) Solution: We need to remember what happens to the function e x when x gets big. When this happens, e x gets closer and closer to 0. Then ( 2 3e x ) = = 2. x + Grading: +5 points for knowing that e x tends to zero as x gets big, +5 points for evaluating the it. Grading for common mistakes: +5 points (total) for DNE; 3 points for substituting a big number. e. + (3 ln x + 2 x ) Solution: This is another it where we can substitute x = 1. (Note that ln x is continuous at 1, so that we can substitute 1 for x there.) +(3 ln x + 2x ) = 3 ln = = 2. Grading: +5 points for substituting, +5 points for finding the value of the it. Grading for common mistakes: 1 point for arithmetic mistakes; 3 points for a final answer of

3 3. The population of a community has been found to approximate the function P (t) = where t is the number of years since a. What was the approximate population in 2000? 155, 000, e t/10, Solution: To find the population in a specific year, we need to determine what t is. Since t is the number of years since 1950, t = = 50. Then the answer is 155,000,000 P (50) = = 147,375 people e 50/10 Grading: +2 points for finding t, +3 points for substituting it. b. What do you expect to happen to the population in the long run, based on the formula given above? Solution: The long-term behavior is described by P (t). Then, since P (t) = 155,000, = 157,360, you should expect the population to grow and approach this number. Grading: +3 points for, +2 points for calculating this it. Grading for common mistakes: 2 points for the population will grow (with no indication whether it will approach a it, or grow beyond any number); 3 points for shrink ; 2 points for no work. 4. A certain computer salesman earns a base salary of $27,000 plus 4% of the total sales if sales are above $90,000 and 11% of the total sales if sales are above $200,000. Let S(x) be his salary and x the amount in dollars of computers he sells. a. Find where the function S(x) is discontinuous. Solution: Based on the description, we find out that 27,000, if x 90,000; S(x) = 27, x, if 90,000 < x 200,000; 27, x, if 200,000 < x. This is a piecewise-defined function. Since the pieces are continuous, S(x) is continuous for all x in the intervals (0, 90,000), (90,000, 200,000), and (200,000, + ). However, S(x) is not continuous at 90,000 or 200,000, since the two one-sided its have different values. Thus S(x) is discontinuous at 90,000 and 200,000. (A function s discontinuities are reported as the values of x, not y.) Grading: +5 points per discontinuity. Grading for common mistakes: +7 points (total) for 27,000 and 35,000. b. What is the significance of these discontinuities? Solution: The salesman will earn a lot of money by doing a (proportionally) small amount of work. Grading: +2 points for a statement about its; +3 points for a statement about continuity. 3

4 1. Where are the following functions continuous? a. x 2 4 MAT 210 Test #1 Solutions, Form C Solution: The function is continuous where what you are taking the square root of is positive. Hence, we need x 2 4 > 0. This happens when x > 2 or x < 2. Grading: +5 points for x 2 4 > 0, +5 points for solving x 2 4 > 0 for x. Grading for common mistakes: 2 points for just x > 2; 2 points for x (which is where x 2 4 is defined); +5 points total for work and an answer of x = ±2. x 2 + 1, if x > 1; b. f(x) = 3, if x = 1; x + 2, if x < 1. Solution: The function x + 2 is a polynomial and hence continuous everywhere. Hence f(x) is continuous where f(x) locally looks like x+2, i.e., when x < 1. Similarly, x 2 +1 is a continuous function, so f(x) is continuous where x > 1. To find out whether f(x) is continuous at 1, we need to see whether f(x) = f(1). To evaluate the it, we need to find two one-sided its: f(x) = (x 2 + 1) = = 2; + + f(x) = (x + 2) = = 3. Since these numbers are different, f(x) does not exist, and so f(x) is not continous at 1. Thus, f(x) is continuous at all real numbers except 2. Grading: +5 points for handling all real numbers except 2, +10 points for determining whether f(x) is continuous at 2. Grading for common mistakes: +7 points (total) for a correct answer but no work; +12 points (total) for a correctly-drawn graph and the answer. 2. Find the following its algebraically, if they exist. a. x + x 3 4x 2 + 4x Solution: Since x is getting large in absolute value, and the function afterwards is a rational function, we need to divide the numerator and denominator by the largest power of x in the rational function, namely x 3. Then x + x 3 4x 2 + 4x = x + ()/x 3 (x 3 4x 2 + 4x)/x 3 = x + 4 x 8 x x x + 4 x 2 = = 0. Grading: +5 points for dividing the numerator and denominator by x 3, +5 points for finding the second it. Grading for common mistakes: +3 points (total) for plugging in a big number; 3 points for 0 = + (or DNE); +7 points (total) for finding the second it to be zero, 1 concluding the first it DNE. 1

5 b. x 2 x 3 4x 2 + 4x Solution: You are taking the it of a rational function as x approaches a number, so we start by trying to substitute x = 2: x 3 4x 2 + 4x = 4(2)2 8(2) (2) 2 + 4(2) = 3 0, getting a non-zero number divided by zero, which means the it does not exist (DNE). Grading: +5 points for cancelling, +5 points for re-evaluating the it. Grading for common mistakes: 3 points for bad arithmetic (which led to a numeric answer); +5 points (total) for getting 0 0 and doing nothing else. c. x 1 + x 2 + 3x 4 x 3 + x 2 2x Solution: A rational function is continuous except where the denominator is zero, so we try plugging in x = 1: x 2 + 3x 4 x 1 + x 3 + x 2 2x = ( 1)2 + 3( 1) 4 ( 1) 3 + ( 1) 2 2( 1) = 6 2 = 3. This means the it is 3. Grading: +5 points for substituting, +5 points for simplifying. Grading for common mistakes: +5 points total for 0 and nothing else. (NOTE: The problem was intended to have x 0 approach 1.) d. x + (2 + 3ex ) Solution: We need to remember what happens to the function e x when x gets big. When this happens, e x gets bigger and bigger, so x + ex does not exist. Then (2 + x + 3ex ) = x + ex does not exist, either. Grading: +5 points for knowing that e x gets big (or the it DNE) as x gets big, +5 points for stating the it DNE. Grading for common mistakes: +3 points for 2 + 3e 100 ; +5 points for = 2, or dividing 3e x by e x (which changes the function you re finding the it of), or = 5. e. (3x ln x 4) Solution: This is another it where we can substitute x = 1. (Note that ln x is continuous at 1, so that we can substitute 1 for x there.) +(3x ln x 4) = 3 1 ln 1 4 = = 4. Grading: +5 points for substituting, +5 points for finding the value of the it. Grading for common mistakes: 1 point for arithmetic mistakes. 2

6 3. The population of a community has been found to approximate the function P (t) = where t is the number of years since a. What was the approximate population in 1990? 145, 000, e t/11, Solution: To find the population in a specific year, we need to determine what t is. Since t is the number of years since 1950, t = = 40. Then the answer is 145,000,000 P (50) = = 104,516 people e 40/11 Grading: +2 points for finding t, +3 points for substituting it. b. What do you expect to happen to the population in the long run, based on the formula given above? Solution: The long-term behavior is described by P (t). Then, since P (t) = 145,000, = 128,889, you should expect the population to grow and approach this number. Grading: +3 points for, +2 points for calculating this it. Grading for common mistakes: 2 points for the population will grow (with no indication whether it will approach a it, or grow beyond any number); 3 points for shrink ; 2 points for no work. 4. A certain computer salesman earns a base salary of $28,000 plus 6% of the total sales if sales are above $80,000 and 12% of the total sales if sales are above $250,000. Let S(x) be his salary and x the amount in dollars of computers he sells. a. Find where the function S(x) is discontinuous. Solution: Based on the description, we find out that 28,000, if x 80,000; S(x) = 28, x, if 80,000 < x 250,000; 28, x, if 250,000 < x. This is a piecewise-defined function. Since the pieces are continuous, S(x) is continuous for all x in the intervals (0, 80,000), (80,000, 250,000), and (250,000, + ). However, S(x) is not continuous at 80,000 or 250,000, since the two one-sided its have different values. Thus S(x) is discontinuous at 80,000 and 250,000. (A function s discontinuities are reported as the values of x, not y.) Grading: +5 points per discontinuity. Grading for common mistakes: +7 points (total) for 27,000 and 35,000. b. What is the significance of these discontinuities? Solution: The salesman will earn a lot of money by doing a (proportionally) small amount of work. Grading: +2 points for a statement about its; +3 points for a statement about continuity. 3

Sec 2.2: Infinite Limits / Vertical Asymptotes Sec 2.6: Limits At Infinity / Horizontal Asymptotes

Sec 2.2: Infinite Limits / Vertical Asymptotes Sec 2.6: Limits At Infinity / Horizontal Asymptotes Sec 2.2: Infinite Limits / Vertical Asymptotes Sec 2.6: Limits At Infinity / Horizontal Asymptotes Infinite