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 (– 4,1)(1,2).
–
The Changing Face of Calculus:
First-Semester Calculus as a
High School Course
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