18. B (1997 AB15) The left and right limits as are both equal to 2. The limit as does not exist since the limits are not equivalent; therefore, A, C and D cannot be true either. 19. A (2003 AB13/BC13) The graph of is continuous at ; however, since the graph has a sharp turn at , the function is not differentiable at . 20. D (2003 AB79) The one-sided limits as are equivalent for the graphs of in I and II but not for III. 21. C (2008 AB77) and exist; however, since they are not equivalent, the does not exist. 22. A (2003 BC81 appropriate for AB) Free Response 23. 2009B AB3abc/BC3abc (a) 1: sets up difference quotient at 1: answer with justification Since the one-sided limits do not agree is not differentiable at . (b) when . There are two values of for which 1: expression for average rate of a this is true. change 1: answer with reason (c) Yes, . The function is differentiable on the interval and continous on . Also, . 1: answers and identifies “yes” By the Mean Value Theorem, there is a 1: justification value , such that . c, 24. 2003 AB6ac (a) is continuous at because 1: answer and equates the f “yes” Therefore, values of the left- and limits 1: explanation involving limits (c) Since is continuous at 1: g 1: and 1: values for and k m Since these two limits exist and is g differentiable at the two limits are equal. Thus
18. B (1997 AB15) The left and right limits as are both equal to 2. The limit as does not exist since the limits are not equivalent; therefore, A, C and D cannot be true either. 19. A (2003 AB13/BC13) The graph of is continuous at ; however, since the graph has a sharp turn at , the function is not differentiable at . 20. D (2003 AB79) The one-sided limits as are equivalent for the graphs of in I and II but not for III. 21. C (2008 AB77) and exist; however, since they are not equivalent, the does not exist. 22. A (2003 BC81 appropriate for AB) Free Response 23. 2009B AB3abc/BC3abc (a) Since the one-sided limits do not agree is not differentiable at . 1: sets up difference quotient at 1: answer with justification (b) when . There are two values of a for which this is true. 1: expression for average rate of change 1: answer with reason (c) Yes, . The function is differentiable on the interval and continous on . Also, . By the Mean Value Theorem, there is a value c, , such that . 1: answers “yes” and identifies 1: justification 24. 2003 AB6ac (a) f is continuous at because Therefore, 1: answer “yes” and equates the values of the left- and limits 1: explanation involving limits (c) Since g is continuous at and Since these two limits exist and g is differentiable at the two limits are equal. Thus 1: 1: 1: values for k and m