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1 Student s Printed Name: Instructor: CUID: Section # : You are not permitted to use a calculator on any portion of this test. You are not allowed to use any textbook, notes, cell phone, laptop, PDA, smart watch, or any technology on either portion of this test. All devices must be turned off while you are in the testing room. During this test, any communication with any person (other than the instructor or a designated proctor) in any form, including written, signed, verbal, or digital, is understood to be a violation of academic integrity. No part of this test may be removed from the testing room. Read each question very carefully. In order to receive full credit for the free response portion of the test, you must: 1. 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 at any time before or during this test. Student s Signature: Do not write below this line. Free Response Problem Possible Points Points Earned Free Response Problem Possible Points 1 4 8b 5 Points Earned 2 5 8c a b Free Response 70 8a 4 Multiple Choice 30 Test Total 100 Page 1 of 12

2 Multiple Choice. There are 13 multiple choice questions. Each question is worth 2 3 points and has one correct answer. The multiple choice problems will count as 30% of the total grade. Use a number 2 pencil and bubble in the letter of your response on the scantron sheet for problems 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. 1. Suppose the position of an object is given by the equation (2 pts.) s(t) = 4t ( + 2t + 1. Find the speed of the object at time t = 2. a) 11 b) 2 c) -14 d) Evaluate lim (2 pts.) a) ((0) b) 3 7 c) 9 7 d) 3 :; 3. An equation of the line tangent to the graph of f(x) at the point (1,9) is y = 6x + 3. (2 pts.) Find f(1) and f C (1). a) f(1) = 9 and f C (1) = 6 b) f(1) = 6 and f C (1) = 9 c) f(1) = 3 and f C (1) = 9 d) f(1) = 9 and f C (1) = 3 4. On what interval(s) is the function f(x) = 0D EF0E9 (2 pts.) 0E7 continuous? a) (-, -7) U (-7, ) b) (-, 7) U (7, ) c) (-7, ) d) (-, ) Page 2 of 12

5. Below is the graph of f(x) that satisfies lim f(x) = 2. Use the graph to find the 0 3 (2 pts.) largest δ such that x 3 < δ means f(x) 2 < 0.45 a) 1.89 b) 0.93 c) 0.45 d) 0.96 6.

3 5. Below is the graph of f(x) that satisfies lim f(x) = 2. Use the graph to find the 0 3 (2 pts.) largest δ such that x 3 < δ means f(x) 2 < 0.45 a) 1.89 b) 0.93 c) 0.45 d) Given f C (0) = π and g C (0) = 2. Calculate h C (0) given h defined as follows: (3 pts.) h(x) = : (f(x) + g(x) sin x) (O a) h C (0) = OE( (O b) h C (0) = OE: (O c) h C (0) = 0 d) h C (0) does not exist 7. Let f(x) = sin x + cos x. At which of the following x-values in [0, π] will f(x) (3 pts.) have a horizontal tangent? a) x = 0 b) x = O ( c) x = O ; d) x = 3O ; Page 3 of 12

4 8. Based on the graph of f(x) below, choose the best possible graph of f (x). (2 pts.) a) b) c) d) 9. The function f(x) = (0WF)D W: has a hole at x = 4. Find a such that setting f(4) = a (2 pts.) 0W; would remove the hole. a) a = 6 b) a = 10 c) a = 2 d) no such a exists Page 4 of 12

5 10. Find the derivative of g(x) = x Z + e 0 + cos π. (3 pts.) a) g C (x) = ex ZW: + xe 0W: + cos π b) g (x) = ex ZW: + e 0 c) g C (x) = ex ZW: + xe 0W: sin π d) g (x) = ex ZW: + xe 0W: 11. On which of the following intervals does f(x) = x ( 4x + 1 have a root? (2 pts.) (Hint: Use the Intermediate Value Theorem.) a) [2, 3] b) [1,2] c) [-1, 0] d) [0, 1] 12. (2 pts.) Evaluate lim ] 0^W(0D W_0 0 ; 0W;. a) 24 = 2 6 b) 0 c) 6 d) 8 = Find the derivative of s(t) = 4 t + 9 b t:2. (3 pts.) a) s C (t) = 4.5t 2.F 9 b D 10tc b) s C (t) = ( t b b D c) s C (t) = ( b 9 b 10tc d) s C (t) = ( b 9 b D 10tc The Free Response section follows. PLEASE TURN OVER YOUR SCANTRON while you work on the Free Response questions. You are welcome to return to the Multiple Choice section at any time. Page 5 of 12

6 Free Response. The Free Response questions will count as 70% 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. All work to be graded must be on this test paper. Follow these special directions for all limits in the Free Response: Show all work. You will not be given any credit for using L Hopital s Rule to find the limits. Infinite limits should be answered with = or =, whichever is appropriate. If the limit does not exist (and cannot be answered as or ), state DNE. 1. (4 pts.) State at which x-values the following graph has discontinuities. Also state whether the discontinuity at each x-value is a jump discontinuity, a removable discontinuity, or an infinite discontinuity. x-value type of discontinuity x = 2 removable x = 1 jump x = 1 infinite 1 pt each x-value and reason 1 pt no extras 2. (5 pts.) Find the second derivative of f(x) = tan x. Use an appropriate notation for each derivative. Simplify your answers including changing any negative exponents, combining like terms, cancellations, reducing fractions, and factoring. HINT: Since we have not covered the chain rule yet, you will need to use product rule in part of this question. f (x) = sec ( x = sec x sec x f " (x) = sec x sec x tan x + sec x tan x sec x = 2sec ( x tan x First derivative correct 2 Product rule 2 Simplification 1-2 incorrectly labeling derivatives or not labeling derivatives Page 6 of 12

7 3. (5 pts.) Suppose that the height of a falling object t seconds after being dropped from a height of 64 feet is given by s(t) = 16t ( + 64 feet. i(b)wi(() Use lim to find the instantaneous velocity of the falling object after 2 seconds. Detailed b ( bw( limit work must be shown. Use of any shortcut derivatives rules will receive no credit. Be sure to include units with your final answer. s(t) s(2) 16t ( (t ( 4) 16(t 2)(t + 2) lim = lim = lim = lim b ( t 2 b ( t 2 b ( t 2 b ( t 2 = lim b ( 16(t + 2) = 16 4 = 64 ft/s Set up for this function 1 Factor 2 Cancel 1 Limit evaluation (only award if correct) 1-1 including 0/0 in work -1 missing equals (-0.5 for only one missing equals) -1 missing limits (-0.5 for only one missing) Excused 1 missing = sign ONLY if it was at a line break -0.5 limit notation carried too far up to -1 for notational errors -0.5 lack of limit rewrite after cancellation Page 7 of 12

8 4. a. (5 pts.) Use the limit definition of the derivative to show the derivative of f(x) = x + 2 is f C (x) = :. Detailed limit work must be shown. Use of any shortcut derivatives rules ( 0E( will receive no credit. f C x + h + 2 x + 2 x + h x + 2 (x) = lim j 2 h x + h x + 2 x + h + 2 (x + 2) = lim j 2 h( x + h x + 2) = lim h j 2 h( x + h x + 2) 1 = lim j 2 x + h x + 2 = 1 2 x + 2 b. (4 pts.) Find the equation of the normal line to the function f(x) = x + 2 at x = 7. f(7) = = 3 1 f C (x) = 2 x m bmn 0o7 = f C (6) = = 1 6 m npq 0o7 = 6 So equation of normal line at x = 6 is y 3 = 6(x 7) Points Awarded Conjugate 1 Correctly simplifies numerator 2 Correctly eliminates 0/0 issue 1 Correct limit (award only if correct) 1-5 only set up, no work -3 definition reversed -2 pt for no limit notation -1 pt for poor limit notation -1/2 missing equals -1/2 pt for limit notation carried too far -2 dropping the denominator -1 each skipping a step -1 final answer wrong sign work that jumped from a correct point to the correct answer without showing work in between - lose points for whatever steps were not shown y value 1 Slope of tangent 1 Slope of normal 1 Putting line together properly 1-3 answer is not linear ie used derivative function instead of number for slope of tangent -1 simplified tangent line incorrectly -3 took derivative again and it is incorrect Page 8 of 12

9 5. (5 pts.) For the graph below, state all the x-values in ( 10,7) for which the function is not differentiable. For each point, state the reason why the function is not differentiable. x-value x = 5 reason not continuous x = 4 x = 2 x = 3 x = 5 1 pt each x-value and reason -1 pt each extra sharp point cusp not continuous not continuous 6. (5 pts.) Given the graph of f(x) below, sketch the graph of f (x) on the empty axes. Derivative graph of linear piece 1 Derivative graph of parabola piece 2 Derivative graph of constant piece 1 All open circles at junctions 1 Middle linear part of derivative need not have proper slope as long as positive. Page 9 of 12

10 7. (5 pts.) Use the δ ε definition of a limit to prove lim(2x 10) = Step 1 Find δ This will be considered scratch work and will not be graded. (You will not gain or lose credit based on what you write or don t write here.) 2x 10 4 < ε 2x 6 < ε 2 x 3 < ε x 3 < 1 2 ε Step 2 Proof Let f(x) = 2x 10 Let ε > 0 be given and assume 0 < x 3 < δ where δ = (and then complete the proof below) then 2x 10 4 = 2x 6 = 2 x 3 < 2δ = 2 : ( = ε : ( ε (Fill in the blank with your choice of δ) 8. Find the first derivative of the following functions. Use an appropriate notation for each derivative. Do not expand terms such as (a + b) n. Otherwise, simplify your answers including changing any negative exponents, combining like terms, cancellations, reducing fractions, and factoring. a. (4 pts.) k(x) = 4x 7 2x + e 3 cot x k C (x) = 28x csc ( x=28x csc ( x Proper delta 2 Proof 3-1/2 for each unique invalid statement like = or > instead of < switching epsilon and delta dropping constants algebra mistakes dropping absolute value assuming conclusion -1.0 for breakdown in showing "< epsilon" but heading in right direction -2.0 for failure to show "< epsilon" First derivative of first term 1 First derivative of second term 1 First derivative of third term 1 First derivative of fourth term 1-1 incorrectly labeling derivatives or not labeling derivatives Page 10 of 12

11 b. (5 pts.) h(x) = :W0D :E0 D h C (x) = (1 + x( ) ( 2x) (1 x ( ) (2x) (1 + x ( ) ( = 2x 2x3 2x + 2x 3 (1 + x ( ) ( = 4x (1 + x ( ) ( c. (3 pts.) g(x) = x 3 (4 x W3 ) = 4x 3 1 g C (x) = 12x ( 9. (5 pts.) Given that g(1) = 4 and g C (1) = 5, find w w0 xy(0) ^ z at x = 1. 0 y = g(x) ^ x ^ y C = x ^ y C (1) = 1 = g C (x) g(x) 1 3 xw( 3 ^ { x ( g C (1) g(1) 1 3 1W( 3 ^ { 1 ( = = 11 3 Keep bottom 0.5 Correct derivative of top 1.5 Hold top 0.5 Correct derivative of bottom 1.5 Denominator squared 1-1 incorrectly labeling derivative or not labeling derivative -4 f /g if f and g correct -5 f /g if f and/or g incorrect -3 top of quotient rule backwards -4.0 for "factoring" the quotient incorrectly and applying quotient rule -3 denominator squared completely missing -0.5 each for missing parentheses -0.5 for "+" in the numerator, all else correct -0.5 for (1+x^2)^2 = 1+x^4-0.5 to -1.0 for incorrect algebra in simplifying depending on severity -2.5 for giving two different h'(x) and one was correct Function transformed 1 derivative of first term 1 derivative of second term 1-1 incorrectly labeling derivatives or not labeling derivatives Keep bottom 0.5 Correct derivative of top 0.5 Hold top 0.5 Correct derivative of bottom 0.5 Denominator squared 0.5 Evaluation started properly 1 Evaluation simplified properly incorrectly labeling derivative or not labeling derivative -1 function = evaluation -3 f /g if f and g correct -2 top of quotient rule backwards -2 denominator squared completely missing -1 erroneous simplification Page 11 of 12

12 10. (4 pts.) Use the given graph to find w {2g(x) 3f(x) at x = 0. w0 f(x) g(x) Slope of f at 0 correct 1 Slope of g at 0 correct 1 Derivative 1 Proper evaluation and simplification 1-1 function = evaluation based on slope f C (0) = 2 and g C (0) = 1 2 y = 2g(x) 3f(x) y C = 2g C (x) 3f C (x) y C (0) = 2g C (0) 3f C (0) = 2 } 1 ~ 3 2 = 1 6 = (5 pts.) Find an equation of the tangent line of s(t) = te b when t = 1. s(1) = 1 e : = e s C (t) = t e b + 1 e b s C (1) = 1 e : + 1 e : = 2e So the equation of the tangent line at t = 1 is y e = 2e(x 1) 12. (5 pts.) A stone is dropped off a cliff, and the height, in feet, of the stone above the ground is given by s(t) = 16t ( + 32t + 48, where t is in seconds. Find the velocity of the stone at the instant that it hits the ground. Be sure to include units on your final answer. Hits the ground when s(t) = 0 16t ( + 32t + 48 = 0 16(t ( 2t 3) = 0 16(t 3)(t + 1) = 0 t = 3 t = 1 Not a valid solution v(t) = 32t + 32 v(3) = = = 64 ft/s y value 1 derivative 1.5 slope of tangent (evaluation) 1 Putting line together properly answer is not linear ie used derivative function instead of number for slope of tangent -1 simplified tangent line incorrectly -1/2 function = evaluation Position = 0 1 Factoring 1 Proper solutions 1 Velocity 1 Velocity at valid point 1-1 function = evaluations -1 final answer was speed (positive) -0.5 incorrect unit (but accept m/s) 13. (1 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. My scantron:.... Page 12 of 12

Student s Printed Name: _KEY Grading Guidelines CUID:

Student s Printed Name: _KEY Grading Guidelines CUID: Instructor: Section # : You are not permitted to use a calculator on any portion of this test. You are not allowed to use any textbook, notes, cell