AP calculus AB student samples and commentary from the 2019 AP exam administration: free response question 6

AP Calculus AB Student Samples and Commentary from the 2019 AP Exam Administration Free Response Question 6 2019 AP ® Calculus AB Sample Student Responses and Scoring Commentary © 2019 The College Boa[.]

2019 AP Calculus AB ® Sample Student Responses and Scoring Commentary Inside: Free Response Question R Scoring Guideline R Student Samples R Scoring Commentary © 2019 The College Board College Board, Advanced Placement, AP, AP Central, and the acorn logo are registered trademarks of the College Board Visit the College Board on the web: collegeboard.org AP Central is the official online home for the AP Program: apcentral.collegeboard.org AP® CALCULUS AB 2019 SCORING GUIDELINES Question (a) h   : answer (b) a x   x h x   3x3h x   : form of product rule  :  : a x   : a  a    22 h    23 h   36   24   160 (c) Because h is differentiable, h is continuous, so lim h x   h   x2 Also, lim h x   lim x2  x 4 x 4 , so lim   x   f  x  3  f  x  2 x 2     Because lim x   0, we must also have lim   f  x  3  x2 x2 Thus lim f  x    x2  4  : xlim    f  x  3  :  : f  2  : L’Hospital’s Rule   : f   x2 Because f is differentiable, f is continuous, so f    lim f  x   x 2 Also, because f is twice differentiable, f  is continuous, so lim f  x   f   exists x2 Using L’Hospital’s Rule, x2  2x  lim   lim x    f  x  3 x  3  f  x   f  x  3 1  f   Thus f     (d) Because g and h are differentiable, g and h are continuous, so lim g  x   g    and lim h x   h   x2 : continuous with justification x2 Because g  x   k  x   h x  for  x  3, it follows from the squeeze theorem that lim k  x   x2 Also,  g    k    h   4, so k    Thus k is continuous at x  © 2019 The College Board Visit the College Board on the web: collegeboard.org © 2019 The College Board Visit the College Board on the web: collegeboard.org © 2019 The College Board Visit the College Board on the web: collegeboard.org © 2019 The College Board Visit the College Board on the web: collegeboard.org © 2019 The College Board Visit the College Board on the web: collegeboard.org © 2019 The College Board Visit the College Board on the web: collegeboard.org © 2019 The College Board Visit the College Board on the web: collegeboard.org AP® CALCULUS AB 2019 SCORING COMMENTARY Question Note: Student samples are quoted verbatim and may contain spelling and grammatical errors Overview This problem introduces three twice-differentiable functions f , g , and h It is given that g    h   4, and the line y   ( x  2) is tangent at x  to both the graph of g and the graph of h In part (a) students were asked to find h  A response should demonstrate the interpretation of the derivative as the slope of a tangent line and answer with the slope of the line y   ( x  2) In part (b) the function a given by a x   3x3h x  is defined, and students were asked for an expression for a x  and the value of a  A response should demonstrate facility with the product rule for differentiation x2  for x  and that lim h x  can be x2   f  x  3 evaluated using L’Hospital’s Rule Students were then asked to find f   and f   A response should observe In part (c) it is given that the function h satisfies h x     that the differentiability of h implies that h is continuous so that lim h x   h   Because lim x   0,  x2  x2 and lim h x  can be evaluated, as is given, it must be that lim   f  x  3  0, as well Using properties of x2 x2 x2  , x    f  x  3 limits, students could conclude that lim f  x   Finally, an application of L’Hospital’s Rule to lim x2 combined with the chain rule to differentiate  f  x  3 , yields an equation that can be solved for f   In part (d) students were given that g  x   h x  for  x  and that k is a function satisfying g  x   k  x   h  x  for  x  Students were asked to decide, with justification, whether k is continuous at x  A response should observe that the differentiability of g and h implies that these functions are continuous, so the limits as x approaches of each of g and h match the value g    h   From the inequality g    k    h  it follows that k    4, and the squeeze theorem applies to show that k is continuous at x  For part (a) see LO CHA-2.C/EK CHA-2.C.1 For part (b) see LO FUN-3.B/EK FUN-3.B.1 For part (c) see LO LIM-2.A/EK LIM-2.A.2, LO LIM-4.A/EK LIM-4.A.2 For part (d) see LO LIM-1.E/EK LIM.1.E.2 This problem incorporates the following Mathematical Practices: Practice 1: Implementing Mathematical Processes, Practice 3: Justification, and Practice 4: Communication and Notation Sample: 6A Score: The response earned points: point in part (a), points in part (b), points in part (c), and point in part (d) In part (a) the response earned the point for the value of h   in line In part (b) the response earned both the first and second points for the derivative a x   h x   x3  x h x  in line The response earned the third © 2019 The College Board Visit the College Board on the web: collegeboard.org AP® CALCULUS AB 2019 SCORING COMMENTARY Question (continued) point for the numerical expression a     23   22  in line Numerical simplification is not required In part (c) the response earned the first point for the verbal connection “ lim h  x  MUST equal ” in x2 lines and on the right The response would have earned the second point for the equation lim   f  x  3  x2 in line on the left and presenting f    in line on the left The response correctly restates f    in line on the left and earned the second point with the final restatement of f    in the circled statement in the 2x last line The response earned the third point for L’Hospital’s Rule with the expression lim in x  3  f  x    f  x  1 line in the middle The response would have earned the fourth point with “ f   MUST equal ” in line on 1 the right The response earned the fourth point with the restatement of f    in the circled statement in the last line In part (d) the response earned the point for concluding that k  x  is continuous at x  in line with the following justification: stating g and h are continuous in line 1, evaluating lim g  x   in line and x2 lim h  x   in line 5, concluding lim k  x   in line 6, and concluding k    because g     h  x2 x2 in lines and Although not required, the response correctly states use of the squeeze theorem Sample: 6B Score: The response earned points: point in part (a), points in part (b), points in part (c), and no point in part (d) in line on the right In part (b) the response In part (a) the response earned the point for the value of h   earned the first and second points for the derivative a x   h x  x  3x3 h x  in line The response     earned the third point for the numerical expression a     2     3    in line Numerical x2  4 x    f  x  3 in line on the left Although not required for the first point, application of L’Hospital’s Rule resulting in the 2x equation lim  in line on the right would also have earned the first point The response did x  3  f  x  2 f  x  simplification is not required In part (c) the response earned the first point for the equation lim not earn the second point because no value is given for f   The response earned the third point for the 2x in line on the right The response did not earn the fourth point because no expression lim x  3  f  x  2 f   x  value is given for f   In part (d) the response correctly concludes that k is continuous at x  The response did not earn the point because the justification is not sufficient The response does not state that g and h are continuous, does not evaluate lim g  x   and lim h x   4, and does not conclude lim k  x    k   x2 x2 © 2019 The College Board Visit the College Board on the web: collegeboard.org x2 AP® CALCULUS AB 2019 SCORING COMMENTARY Question (continued) Sample: 6C Score: The response earned points: point in part (a), points in part (b), no points in part (c), and no point in part (d) In part (a) the response earned the point for the value of h   In part (b) the response earned the first and 3 second points for the derivative a x   3x  h x    x  h x   in line The response did not earn the third       point because of an incorrect simplification in the numerical expression of a  with rewriting  3 in line as in line In part (c) the response did not earn the first point because there is no connection between lim h x  and The response did not earn the second point because no value is given for f   Although the x2 response attempts to apply L’Hospital’s Rule in line 1, the response did not earn the third point because of a lack of limit notation: The use of lim is not presented with the quotient The response did not earn the fourth point x2 because no value is given for f   In part (d) the response correctly concludes that k is continuous at The response did not earn the point because the justification is not sufficient The response attempts to apply both the “IVT” (Intermediate Value Theorem) and the “MVT” (Mean Value Theorem) rather than the squeeze theorem © 2019 The College Board Visit the College Board on the web: collegeboard.org  ... line The response earned the third © 2019 The College Board Visit the College Board on the web: collegeboard.org AP? ? CALCULUS AB 2019 SCORING COMMENTARY Question (continued) point for the numerical... 2019 The College Board Visit the College Board on the web: collegeboard.org AP? ? CALCULUS AB 2019 SCORING COMMENTARY Question Note: Student samples are quoted verbatim and may contain spelling and. ..   x2 x2 © 2019 The College Board Visit the College Board on the web: collegeboard.org x2 AP? ? CALCULUS AB 2019 SCORING COMMENTARY Question (continued) Sample: 6C Score: The response earned