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 andsummarize 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)

g x( ) = f t( )dt−3

x

∫(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)

g x( ) = f t( )dt−3

x

∫(c) Find all values of x in the open interval (–5,4) at which g attains a relative maximum. Justify your answer.

g' x( ) =f x( )

g'

– 4 1 3

– + + –

Max at x = 3

AB 5 (2004)

g x( ) = f t( )dt−3

x

∫(c) Find all values of x in the open interval (–5,4) at which g attains a relative maximum. Justify your answer.

g' x( ) =f x( )

g'

– 4 1 3

– + + –

Max at x = 3

because g' changes from positive to negative at x = 3

Absolute min is g(– 4) = –1

AB 5 (2004)

g x( ) = f t( )dt−3

x

∫(d) Find the absolute minimum value of g on the closed interval [–5,4]. Justify your answer.

g'

– 4 1 3

– + + –

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)

g x( ) = f t( )dt−3

x

∫(d) Find the absolute minimum value of g on the closed interval [–5,4]. Justify your answer.

g'

– 4 1 3

– + + –

Mainstream Calculus I Enrollments (fall only for 2- & 4-yr colleges & universities)

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50

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150

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academic year

4-yr colleges &universities2-yr colleges

AP Calculus (AB &BC)

The Changing Face of Calculus: First-Semester Calculus as a High School Course

Featured article on the home page of the MAA:www.maa.org

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.