MthSc 103 Test 3 Spring 2009 Version A UC , 3.1, 3.2. Student s Printed Name:

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1 Student s Printed Name: Instructor: CUID: Section # : Read each question very carefully. You are NOT permitted to use a calculator on any portion of this test. You are not allowed to use any textbook, notes, cellphone, or laptop on either portion of the test. No part of this test may be removed from the testing room. In order to receive full credit for the free response portion of the test, you must:. Show legible and logical (relevant) justification which supports your final answer. 2. Use complete and correct mathematical notation. 3. Include proper units, if necessary. 4. Give exact numerical values whenever possible. You have 90 minutes to complete the entire test. On my honor, I have neither given nor received inappropriate or unauthorized information during this test. Student s Signature: Do not write below this line. Free Response Problem Possible Earned Free Response Problem Possible a 5 5a b 2 5b 2 c 2 5c d 2 3a 5 6a 4 3b 3 6b 4 4a Earned 4b 2 8 4c 3 Free Response 54 Multiple 4d 2 Choice 46 4e 3 Test Total 00 Page of

2 Multiple Choice. There are 9 multiple choice questions. Each question is worth 2 3 points and has one correct answer. The multiple choice problems will count 46% of the total grade. Use a number 2 pencil and bubble in the letter of your response on the scantron sheet for problems 9. For your own record, also circle your choice on your test since the scantron will not be returned to you. Only the responses recorded on your scantron sheet will be graded. You are NOT permitted to use a calculator on any portion of this test.. Evaluate. 2. Given the graph of below, find the graph of. a) b) Page 2 of

3 3. Let. Find. a) b) 4. Find. 5. Given, find the equation(s) of any vertical asymptotes. a) b) 6. Find. 7. Let. Find. Suppose and are differentiable functions. Also suppose and. Use this information to answer the next 2 questions. 8. Let. Find. 9. Let. Find. Page 3 of

4 0. Let. Find. a) b). Find.. Use the graph of below to find the largest possible such that for all,. 3. Find. 4. Let a). Find the equation of the oblique asymptote. b) Page 4 of

5 5. The Intermediate Value Theorem says a) has a root because is continuous,,, and. b) has a root because is continuous everywhere. c) does not have a root because. d) has a root because is continuous,,, and. 6. Find the point where the tangent line to the curve has slope. a) b) 7. Find. Use the following information about to answer the next 2 questions. 8. Does have a horizontal asymptote? If so, what is the equation? No 9. Which of the following statements about at is true? a) is neither continuous nor differentiable at. b) is both continuous and differentiable at. c) is continuous, but not differentiable at. d) is not continuous, but is differentiable at. Page 5 of

6 Free Response. The Free Response questions will count 54% of the total grade. Read each question carefully. In order to receive full credit you must show legible and logical (relevant) justification which supports your final answer. Give answers as exact answers. You are NOT permitted to use a calculator on any portion of this test.. The graph of is given below. Awarded Correct answer (each part) -0.5 for or 2 instances of missing or inappropriate equals signs - for more than 2 instance a. (5 pts.) Find the following limits: i. lim x!" " f (x) = ii. lim f (x) = x!" iii. lim f (x) = "3 x! " iv. lim f (x) = 3 x! + v. lim x! f (x) dne b. For which values of in the interval is discontinuous? x =!, x = pt each, -0.5 for each extra answer c. For which values of in the interval is not differentiable? Discontinuities x =!, x = 0.5 pt each, -0.5 for each extra answer Vertical tangent x = 2 Sharp point x = 3 Page 6 of

7 2. (5 pts.) Consider the function. We know that. Algebraically find the greatest value for, such that for all satisfying, the inequality holds. f (x)! < 2 in this problem becomes 3x! 5! < 2 f (x) = L +! f (x) = L! " so! 2 < 3x! 5! < 2 x 0 = 2! = 2 3x " 5 = 3 2 3x! 5 = 2 2 < 3x! 5 < < 3x! 5 < < 3x < < x < 29 OR L +! = 3 2 L = L "! = 2 3x " 5 = 9 4 3x = 29 4 x = 29 3x! 5 = 4 3x = 2 4 x = 2 Then 2 2 = 24 29! = 3! 2 = 5 So! = 3 = 4 = min " 3, 5 % # $ & ' is the largest! >0 so that f (x) - < 2 whenever 0 < x ( 2 <! Awarded Correctly set up f (x) = L +! and f (x) = L! " Correctly solve f (x) = L +! and f (x) = L! ".5 Correct relationship of the 3 x-values 0.5 Correctly calculate 2 spaces Correct choice for delta -0.5 not dropping absolute values at correct step -0.5 f (x)! x 0 not f (x)! L -0.5 x! L not x! x negative delta Page 7 of

8 3. Let. a. (5 pts.) Use the limit definition of the derivative to find. 3(x + h) 2 # 4(x + h) + 7 # (3x 2 # 4x + 7) f!(x) = lim h"0 h 3(x 2 + 2xh + h 2 ) # 4x # 4h + 7 # 3x 2 + 4x # 7 = lim h"0 h 3x 2 + 6xh + 3h 2 # 4x # 4h + 7 # 3x 2 + 4x # 7 = lim h"0 h 6xh + 3h 2 # 4h = lim h"0 h = lim h"0 ( ) h h 6x + 3h # 4 = lim( 6x + 3h # 4) = 6x # 4 h"0 Correctly states limit definition of derivative with correct substitution Awarded Correctly multiplies out the numerator Correctly simplifies numerator Correctly eliminates h/h Correctly evaluates limit - pt for no limit notation -/2 to - pt for poor limit notation -/2 pt for limit notation carried too far - dropping the denominator -4 pt for taking the derivative correctly without using the limit definition award pt for writing the difference quotient without substituting in for the specific function and not going any further work that jumped from a correct point to the correct answer without showing work in between lost points for whatever steps were not shown b. Find an equation for the normal line to the curve at the point m tan = f!(2) = 6("2) " 4 = " " 4 = "6 so m nor = " m tan = " "6 = 6 so equation of normal line is y " 27 = ( 6 x " "2 ) ( ) y " 27 = 6 x + 2 Awarded Correctly finds the slope of the tangent Uses tangent and normal relationship 0.5 Correct slope of normal line 0.5 Correctly use the slope in the equation 0.5 Correctly uses the point if change to slope-intercept form incorrectly - for solving y = mx + b incorrectly for b and never writing a correct equation Page 8 of

9 4. Consider the function. f (x) = x2 + x! 6 x 2! x! 2 = x + 3 x! 2 ( )( x! 2) ( )( x + ) = x + 3 x + x " 2 a. State the coordinates ( - and - values) of any intercepts. If there are any, state if the point is an or intercept. Awarded x = 0 gives y =!6 Correct (each intercept) = 3 so ( 0,3) is the y-intercept!2-0.5 for x=2 given as second x-intercept y = 0 gives x each for not stated as coordinate point = 0 so (!3,0) is the x-intercept -0.5 each for x and y coordinates reversed in ordered pair x+ b. Use limit(s) to show that has a vertical asymptote at. x + 3 lim x!" " x + = 2 small " = "# OR lim x!" + x + 3 x + = 2 small + = +# c. Use limit(s) to show that has a removable discontinuity at. Define so that is continuous at (x + 3)(x " 2) lim x!2 (x " 2)(x + ) = lim x + 3 x!2 x + = 5 3 f (2) = 5 3 would make f (x) continuous at x = 2. Awarded Correctly sets up limit 0.5 Shows evidence of understanding the 0.5 denominator gets small (implicit or explicit) Recognizes limit goes to infinity - incorrect sign on infinity Awarded Correctly sets up limit 0.5 Correctly finds limit 2 Correct summary conclusion but no supporting limit - including 0/0 in work d. Use limit(s) to find any horizontal asymptotes or to show that there are none. x 2 + x # 6 lim x!±" x 2 # x # 2 = lim x 2 + x # 6 x!±" x 2 # x # 2 $ + = lim x # 6 x # 0 x!±" # x # 2 = # 0 # 0 = x 2 x 2 x 2 Awarded Correctly sets up limit Correctly finds limit Summary statement Need not show work on limit -0.5 horizontal asymptote not shown as an equation so the horizontal asymptote is y = Page 9 of

10 e. Use the information gathered to graph. Plot all asymptotes, intercepts and discontinuities. Awarded Intercepts drawn and used each Asymptotes drawn and used each Hole drawn and used Left half placed properly with at least one point (not a guess) Based on previous work -/2 ends of graph tend away from the asymptotes -/2 added extra stuff that wasn t found earlier and shouldn t be there 5. Let. a. ( pt.) Find. f (2) = 5(2)! (2) 2 = 0! 4 = 6 b. Find. right or wrong # lim f (x) = lim 4x + x!2 " x!2 " $ % 3 c & ' ( = c This guideline applies to both parts b and c Awarded Choose correct piece Evaluate correctly -0.5 for missing, poorly used, or inappropriate limit notation -0.5 for missing or poorly used equals -0.5 if negative error during evaluation -2 chose wrong piece c. Find. ( ) = 0 " 4 = 6 lim f (x) = lim 5x " x 2 x!2 + x!2 + d. For what value of the constant is continuous everywhere? c = 6 3 c =!2 c =!6 Awarded Correctly sets up equality Correctly finds c Page 0 of

11 6. (4 pts. each) Find the first derivative of the following functions. Do NOT simplify. a. = ( ) = f! x 2 3x x 2! 4x + 5 ( x 2 " 4x + 5)3# 2 2 x" " 3 x 2x " 4 ( x 2 " 4x + 5) 2 ( ) Awarded Keep denominator 0.5 Correct derivative of numerator Minus 0.5 Keep numerator 0.5 Correct derivative of denominator Denominator squared pts for incorrectly labeling derivative or not labeling derivative -4 g /h even if g and h are correct b. y! = ( 6x " 4) ( 3e x + 3x ) 2 + 6( 3e x + x ) 3 Awarded Correct derivative of first function Hold the second function Correct derivative of second function Hold the first function Product rule correct -4 for g * h, even if g and h are correct -2 pts for incorrectly labeling derivative or not labeling derivative 7. (4 pts.) Find the coordinates ( - and - values) for any points where has a horizontal tangent. horizontal tangent when y! = 0 y! = 6x " 4 = 0 6x = 4 x = 2 3 # y 2 & $ % 3' ( = 3 # 2 & $ % 3 ' ( 2 # " 4 2 & $ % 3' ( + 8 = 3 # 4 & $ % 9 ' ( " = 4 3 " = 20 3 # 2 so the point where the tangent is horizontal is 3, 20 & $ % 3 ' ( 8. ( pt.) Check to make sure your Scantron form meets the following criteria. If any of the items are NOT satisfied when your Scantron is handed in and/or when your Scantron is processed one point will be subtracted from your test total.... rest removed for space on key Awarded Correct derivative Set derivative =0 0.5 Solve for x Correct evaluates f(x) for their x Summary 0.5 Page of

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