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 phone, laptop, PDA, Apple 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:. 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 5 7a 5 2a 5 7b 4 2b 3 7c 4 3 9 8 4 4 6 9 7 5a 4 0 Points Earned 5b 4 Free Response 68 6 7 Multiple Choice 32 Test Total 00 Page of 5
Multiple Choice. There are 3 multiple choice questions. Each question is worth 2 3 points and has one correct answer. The multiple choice problems will count as 32% of the total grade. Use a number 2 pencil and bubble in the letter of your response on the scantron sheet for problems 3. 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.. (3 pts.) Consider the implicitly defined function 4x 2 y 2 = 2xy + y. Evaluate dy. dx (,0) a) 2 3 b) 6 c) 8 d) 8 3 e) 6 5 2. Evaluate lim tan ( 2 x). x (2 pts.) a) 0 b) π 2 c) d) e) π 2 3. Find the derivative of g(x) = csc( 3x 4 ). (2 pts.) a) g ( x)= csc( 2x 3 )cot 2x 3 ( ) b) g ( x)= 2x 3 csc( 3x 4 )cot 3x 4 ( ) c) g ( x)= ( ) sin 2x 3 d) g ( x)= csc( 3x 4 )cot( 3x 4 ) e) g ( x)=2x 3 csc( 3x 4 )cot( 3x 4 ) Page 2 of 5
4. Find all values of x where the graphed function below is NOT differentiable. (2 pts.) a) x = 3,, b) x = 3, c) x =,3 d) x = e) The function is differentiable for all x. 5. (2 pts.) Evaluate the following: lim x x 3 2x 3 x 9x 6 + 4x a) 2 b) 0 c) d) 2 e) 6. (2 pts.) Evaluate lim x 0 sin 2 6x x 2. a) 36 b) 0 c) 6 d) e) DNE Page 3 of 5
7. Below is the graph of a derivative function f ( x). Choose the graph most likely (2 pts.) to be the original function f ( x). a) b) c) d) e) There is intentionally no choice (e) due to spacing. Page 4 of 5
8. If f (x) = 2x 3 + cos x, find f ( π). (3 pts.) a) 0 b) 2π c) 2π + d) 2π e) 2π + 2 2 The rest of the problems on this page rely on the table below. 9. (3 pts.) If k(x) = ( f! g) ( x)= f ( g( x) ), find k ( 5). a) 9 b) 3 c) 2 d) 0 e) 2 0. (3 pts.) If b(x) = f ( 2x), find b ( 2). x 2 a) 2 b) 3 2 c) 3 d) 24 e) 3 2 Page 5 of 5
. The average rate of change of f (x) = x 3 x on the interval [ 2,2] is 3. (3 pts.) Find all x value(s) within this interval for which the instantaneous rate of change matches the average rate of change. a) x = 3 2 + 37 b) x = 6 c) x = ± 2 3 d) x = 2 3 e) There are no such x-values. 2. Using the graphs of F(x)and G( x) below, (3 pts.) if H ( x)= G( x) F( x), find H ( ). Clarity: F( )= 3 2, G( )= 2 3, G( 2)= 3 a) 5 9 b) 9 c) 0 27 d) 5 6 e) 9 Page 6 of 5
3. State the derivative of f (x) = arctan( 2x) arcsin(x). (2 pts.) a) f ( x)= + 4x b) f ( x)= 2 2 x 2 + 4x + 2 x 2 c) f ( x)= 2 + 2x d) f ( x)= 2 2 x 2 + 2x + 2 x 2 e) f ( x)= 2 + 4x 2 x 2 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 7 of 5
Free Response. The Free Response questions will count as 68% 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.. (5 pts.) Sketch a graph of a function f ( x) that satisfies all of the following conditions. You do not need to give a formula for the function. Indicate all asymptotes with dotted lines. Minus ½ pt. f ( x) is continuous for all x except x = 3,0,3 each ½ pt. f ( 0) is undefined pt. lim x pt. lim f (x) = 2 x ½ pt. lim f (x) = x 3 ½ pt. ½ pt. ½ pt. ½ pt. lim f (x) = x 3 + lim f (x) = 3 x 0 lim f (x) = x 3 lim f (x) = x 3 + - each instance of not a function (fails vertical line test) answer is variable OK to simply be flat at edges instead of HA -0.5 excessive asymptotes Page 8 of 5
2. Consider the function f (x) = x. ( ) f ( 2) f x a. (5 pts.) Use lim to find the slope of the tangent line to the graph of f (x) = x 2 x 2 x at the point a = 2. Detailed limit work must be shown. Use of any shortcut derivatives rules will receive no credit. lim x 2 ( ) f ( 2) f x x 2 = lim x 2 x 2 x 2 x 2 = lim 2x x 2 x 2 = lim 2 x x 2 2x x 2 = lim x 2 2x = 4 Set up 0.5 Common Denominator and Simplify numerator 2 Simplify.5 Limit evaluation (award only if correct) - including 0/0 in work - missing equals - missing limits -0.5 limit notation carried too far up to - for notational errors (such as omitting the parentheses or poor limit notation) -5 using any derivative rule besides the definition b. (3 pts.) State the equation of the line normal to f (x) = x at x = 2. from part a m tan = 4 so m nor = +4 so equation of normal line to f (x) = x at x = 2 is y 2 = 4( x 2) Points Awarded Function evaluation 0.5 Point placed properly in tangent line SLOPE placed properly in.5 tangent line - missing ( ) Page 9 of 5
3. Consider the function f (x) = 3x2 + 3x 6 x 2 3x + 2 = 3( x + 2) ( x ) ( x ) ( x 2) Work must be shown on the limits to earn full credit no shortcuts. a. (2 pts.) Evaluate lim f (x). x + 3( x + 2) ( x ) lim x + ( x ) ( x 2) = lim 3 ( x + 2 ) x + x 2 = 3 3 = 9 Uses factored form (or gets to the factored form) Correct limit (award only if correct) -0.5- poor notation b. (2 pts.) Evaluate lim x 2 + f (x). 3( x + 2) ( x ) lim x 2 + ( x ) ( x 2) = lim 3( x + 2) x 2 + x 2 = 3 4 small pos = + c. (3 pts.) Evaluate lim f (x). x ± 3x 2 + 3x 6 lim x ± x 2 3x + 2 x 2 x 2 = lim x ± 3+ 3 x 6 x 2 3 x + 2 x 2 = 3+ 0 0 0 + 0 = 3 Shows evidence of understanding that numerator approaches constant and denominator is small Recognizes limit goes to (correct sign) infinity d. ( pt.) Does f( x ) have any horizontal asymptotes? If so, state the equation(s). yes y = 3 right or wrong (NOT based on a-c) do not have to say yes just show equation, except -/2 if not equation just number or variable wrong, -0.5 extraneous solutions e. ( pt.) Does f( x ) have any vertical asymptotes? If so, state the equation(s). yes x = 2 right or wrong (NOT based on a-c) do not have to say yes just show equation, except -/2 if not equation just number or variable wrong, -0.5 extraneous solutions Points Awarded -0.5- poor notation (-0.5 if 0 is ever written in denominator; - if no limit is written in problem) - specific x-values plugged in inline with the work - incorrect sign on infinity Correct WORK for a limit going to.5 infinity Limit as go to infinity applied 0.5 Correct answer (award only if correct) - including inf/inf in work - missing equals (-0.5 for only one missing equals) - missing limits (-0.5 for only one missing) -0.5 limit notation carried too far Page 0 of 5
4. (6 pts.) Let f (x) = 2x + 3. Use the limit definition of the derivative to show the derivative of f (x) = 2x + 3 is f '(x) = Detailed limit work must be shown. Use of any 2x + 3 shortcut derivatives rules will receive no credit. 2x + 2h + 3 2x + 3 2x + 2h + 3 + 2x + 3 f ( x)= lim h 0 h 2x + 2h + 3 + 2x + 3 2x + 2h + 3 ( 2x + 3) 2h = lim h 0 h 0 h 2x + 2h + 3 + 2x + 3 h 2x + 2h + 3 + 2x + 3 ( ) = lim ( ) = lim h 0 2 2x + 2h + 3 + 2x + 3 = 2 2 2x + 3 = 2x + 3 Set up 0.5 Conjugate 0.5 Simplify numerator 2.5 Cancel h.5 limit evaluation (only if correct) - including 0/0 in work - missing equals -2 missing limits -0.5 limit notation carried too far up to - for notational errors (such as omitting the parentheses in the denominator or poor limit notation) -2 Bad cancellation -5 using derivative in place of f(x) Page of 5
5. Suppose a stone is thrown vertically upward from the edge of a 00-foot cliff on Planet X. If the initial velocity of the stone is 24 ft/s, then the stone s height above ground level (the base of the cliff) is given by the function s(t) = 6t 2 + 24t +00 feet where t is in seconds. a. (4 pts.) State the velocity function v( t) and the acceleration function a( t) of the stone at any time t. Be sure to include units with your answers. v( t)= 2t + 24 ft/s a( t)= 2 ft/s 2 Correct velocity.5 Velocity units 0.5 Correct acceleration.5 Acceleration units 0.5 b. (4 pts.) What is the speed of the stone when it is 28 feet above the ground. Be sure to include units with your final answer. s(t) = 6t 2 + 24t +00 = 28 6t 2 + 24t + 72 = 0 6( t 2 4t 2)= 0 6( t 6) ( t + 2)= 0 t = 6 t = 2 v( 6)= 2( 6)+ 24 = 72 + 24 = 48 ft/s so the speed is 48 ft/s. Position = 0 0.5 Solve by factoring.5 Eliminate the negative solution 0.5 Velocity evaluated Speed 0.5-0.5 lack of units Page 2 of 5
6. (7 pts.) Find the derivative of the following function. Use appropriate notation to denote each derivative. Simplify by combining like terms, reducing fractions, and removing negative exponents from answers. y = 3( 7 x )+ 4ex 3 x 4 ( 4x) + 2 4 x + cos ( 4x)= 3( 7 x )+ 4 3 ex x 4 6x + 2 4 x + cos ( 4x) = 3( 7 x )+ 4 3 ex 6x + 6 4 x + cos ( 4x)= 3( 7 x )+ 4 3 ex 6 x 6 + 4 x + cos ( 4x) y = 3 ln 7 7 x + 4 3 ex 6 ( 6)x 7 ln 4 4 x = 3 ln 7 7 x + 4 3 ex + 3 8x 7 ln 4 4 x 4 6x 2 4 ( 4x) 2 First derivative of first term First derivative of second term First derivative of third term.5 Derivative of third term simplified 0.5 First derivative of fourth term (not required to put 4^x in denominator) Proper sign on fourth term derivative 0.5 First derivative of fifth term Proper sign on fifth term derivative 0.5-2 incorrectly labeling derivatives or not labeling derivatives Page 3 of 5
7. Find the first derivative of the following functions. Use appropriate notation to denote the derivative. DO NOT SIMPLIFY. a. (5 pts.) y = tan 6 πx + 7e 6 x π 6 + x e ( ) y = sec 2 ( 6 πx + 7e 6 x π 6 + x e ) π ln6 6 πx + 42e 6 x 0 + e x e Derivative of outside, keep inside Derivative of inside first term Derivative of inside second term Derivative of inside third term Derivative of inside fourth term -2 incorrectly labeling derivative or not labeling derivative at least -3 for derivative of outside at derivative of inside - missing ( ) -5 treating as a product ( ) 3 b. (4 pts.) f ( x)= x 4 4 x 5 + 9 ( ) 3 ( ) 3 4 f ( x)= x 4 4 x 5 + 9 = x 4 x 5 + 9 ( ) f ( x)= x 4 3 4 x5 + 9 ( ) 3 4 4 5x 4 + 4x 3 x 5 + 9 Keep the first 0.5 Derivative of second : (2 pts) Derivative of outside, keep inside Derivative of inside Derivative of first term Keep the second 0.5-2 incorrectly labeling derivative or not labeling derivative at least -2 for derivative of outside at derivative of inside - missing ( ) -3 f *g -3 only term of product rule c. (4 pts.) g(y) = sec 5 ( cot y) g(y) = sec 5 ( cot y)= sec( cot y) 4 g ( y)= 5 sec( cot y) sec( coty)tan( cot y) csc 2 y 5 ( ) Derivative of outside, keep the inside Derivative of first inside.5 Derivative of second inside.5-2 incorrectly labeling derivative or not labeling derivative at least -2 for derivative of outside at derivative of inside -/2 missing ( ) that make it look like subtraction (*-cscx OK) -4 treating as a product Page 4 of 5
8. (4 pts.) Let f (x) = 2 3 x3 9 2 x2 8x + 7. At what x - value(s) does the graph of f have a horizontal tangent line? f ( x)= 2x 2 9x 8 horizontal tangent when f ( x)= 0 2x 2 9x 8 = 0 ( 2x + 3) ( x 6)= 0 x = 3 2 x = 6 9. (7 pts.) Find dy dx for 5x3 y π sin y = 3π 2 5x 3 dy dx +5x2 y π cos y dy dx = 0 5x 3 dy dy π cos y dx dx = 5x2 y dy dx 5x3 π cos y = 5x2 y dy dx = 5x2 y 5x 3 π cos y Correct derivative.5 Derivative = 0 0.5 Solves for x 2-0.5 to - algebra mistakes or not simplifying answer -4 setting function = 0 only received last point of the 2 points for solving if correct -0.5 each extra solution - each missing solutions Derivative of first term 2 Derivative second term.5 Derivative of third term isolate dy/dx 2.5 (follow work as long as implicit differentiation used) OK to use y -0.5 notational error ( time deduction for any and all mistakes including equating derivative to a number (slope)) 0. ( 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: is bubbled with firm marks so that the form can be machine read; is not damaged and has no stray marks (the form can be machine read); has 3 bubbled in answers; has MATH 040 and my Section number written at the top; has my Instructor s name written at the top; has Test No. 3 written at the top; has Test Version A both written at the top and bubbled in below my CUID; and shows my correct CUID both written and bubbled in (bubble in a 0 in place of the C). Page 5 of 5