APCalculus Development Committee  Janet L. Beery, University of Redlands; Redlands, California  David M. Bressoud (Committee Chair), Macalester College;

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Presentation transcript:

APCalculus Development Committee  Janet L. Beery, University of Redlands; Redlands, California  David M. Bressoud (Committee Chair), Macalester College; St. Paul, Minnesota  David Lomen, University of Arizona; Tucson, Arizona  Guy Mauldin, Science Hill High School, Johnson City, Tennessee  Carol Miller, Glenbrook North High School; Northbrook, Illinois  Monique Morton, Woodrow Wilson Senior High School; Washington, D.C.

Sign charts can provide a useful tool to investigate and summarize the behavior of a function. We commend their use as an investigative tool. However, the Development Committee has recommended and the Chief Reader concurs that sign charts, by themselves, should not be accepted as a sufficient response when a problem asks for a justification for the existence of either a local or an absolute extremum at a particular point in the domain. This is a policy that will take effect with the 2005 AP Calculus exams and Reading. New Policy on Use of Sign Charts to Justify Local Extrema AP Calculus AB Home Page, Exam Information: “On the role of sign charts …”

AB 5 (2004) (c) Find all values of x in the open interval (–5,4) at which g attains a relative maximum. Justify your answer. (d) Find the absolute minimum value of g on the closed interval [–5,4]. Justify your answer.

AB 5 (2004) (c) Find all values of x in the open interval (–5,4) at which g attains a relative maximum. Justify your answer. – 413 –++– Max at x = 3

AB 5 (2004) (c) Find all values of x in the open interval (–5,4) at which g attains a relative maximum. Justify your answer. – 413 –++– Max at x = 3 because g' changes from positive to negative at x = 3

Absolute min is g(– 4) = –1 AB 5 (2004) (d) Find the absolute minimum value of g on the closed interval [–5,4]. Justify your answer. – 413 –++–

Absolute min is g(– 4) = –1 because g' changes from negative to positive at x = – 4, g' is negative on (–5,–4) (so g(–5) > g(– 4) ), and g(4) = g(2) > g(– 4) because g' ≥ 0 on (– 4,2). AB 5 (2004) (d) Find the absolute minimum value of g on the closed interval [–5,4]. Justify your answer. – 413 –++–

The Changing Face of Calculus: First-Semester Calculus as a High School Course Featured article on the home page of the MAA: First-semester calculus has become a high school topic for most of our strongest students. This has several implications: 1.We should ensure that students who take calculus in high school are prepared for the further study of mathematics. 2.We should address the particular needs of those students who arrive in college with credit for calculus. 3.We should recognize that the students who take first-semester calculus in college may need more support and be less likely to continue with further mathematics than those of a generation ago.