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Justifying local extrema using sign charts is now restricted as a sole response in the AP Calculus exams since 2005. Gain insight into this policy update regarding extremum justification requirements.
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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) (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.
– + + – – 4 1 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. Max at x = 3
– + + – – 4 1 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. Max at x = 3 because g' changes from positive to negative at x = 3
– + + – – 4 1 3 AB 5 (2004) (d) Find the absolute minimum value of g on the closed interval [–5,4]. Justify your answer. Absolute min is g(– 4) = –1
– + + – – 4 1 3 AB 5 (2004) (d) Find the absolute minimum value of g on the closed interval [–5,4]. Justify your answer. 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 (– 4,1)(1,2).
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