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.
New Policy on Use of Sign Charts to
Justify Local Extrema
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.
AP Calculus AB Home Page, Exam Information:
“On the role of sign charts …”
AB 5 (2004)
x
gx   3 f t dt
(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)
x
gx   3 f t dt
(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
Max at x = 3
+
+
1
–
3
AB 5 (2004)
x
gx   3 f t dt
(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
Max at x = 3
+
+
1
–
3
because g' changes from
positive to negative at x = 3
AB 5 (2004)
x
gx   3 f t dt
(d) Find the absolute minimum value of g on the closed
interval [–5,4]. Justify your answer.
g'
Absolute min is g(– 4) = –1
–
–4
+
+
1
–
3
AB 5 (2004)
x
gx   3 f t dt
(d) Find the absolute minimum value of g on the closed
interval [–5,4]. Justify your answer.
g'
Absolute min is g(– 4) = –1
–
+
+
–4
1
3
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).
–
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.