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)    3

x

 

dt

(c) Find all values of

x g

in the open interval (–5,4) at which attains a relative maximum. Justify your answer.

(d) Find the absolute minimum value of

g

interval [–5,4]. Justify your answer.

on the closed

AB 5 (2004)    3

x

 

dt

(c) Find all values of

x g

in the open interval (–5,4) at which attains a relative maximum. Justify your answer.

g

'  Max at

x

= 3

g

' – + + – – 4 1 3

AB 5 (2004)    3

x

 

dt

(c) Find all values of

x g

in the open interval (–5,4) at which attains a relative maximum. Justify your answer.

g

' 

g

' – – 4 + 1 + 3 – Max at

x

= 3 because

g

' changes from positive to negative at

x

= 3

AB 5 (2004)    3

x

 

dt

(d) Find the absolute minimum value of

g

interval [–5,4]. Justify your answer.

on the closed

g

' – + + Absolute min is

g

(– 4) = –1 – 4 1 3 –

AB 5 (2004)    3

x

 

dt

(d) Find the absolute minimum value of

g

interval [–5,4]. Justify your answer.

on the closed

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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