# 屏幕快照 2018-05-29 下午9.53.57.png - AP Calculus BC...

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Unformatted text preview: AP“ Calculus BC Free-Response Scoring Guidelines Question 4 Graph of f Let f be the function given by f (x) = (1n x)(sin x). The ﬁgure above shows the graph of f for 0 < x S 21:. The function g is deﬁned by 30:) = fﬂt) :1! for 0 < x S 21:. (a) (b) (c) (d) (a) (b) ((1) Find 3(1) and 311). On what intervals, if any, is 3 increasing? Justify your answer. For 0 < x S 23:, ﬁnd the value of x at which 3 has an absolute minimum. Justify your answer. For 0 < x < 23:, is there a value of x at which the graph of g is tangent to the x—axis? Explain why or why not. 3(1)=jllf(r)dt= o and 3’0) = f0) = 0 Since g’(x) = f(x), 3 is increasing on the interval 15x53: because f(x):>0 for l<x<r:. For 0 < x 4 21:, g’(x) = f(x) = 0 when x =1, 1:. g’ = f changes from negative to positive only at x = l. The absolute minimum must occur at x = l or at the right endpoint. Since 3(1) = O and 3(25) =j12”f(:)d: =j1”f(:)dr+j:"f(r)d: 4. 0 by comparison of the two areas, the absolute minimum occurs at x = 21:. Yes, the graph of g is tangent to thex-axis at x = 1 since 3(1) = 0 and g’(l) : 0. t—A—xl—A—nr—A—x .—.— ~— :30) :31!) : interval : reason : identiﬁes l and 2:: as candidates .- or - indicates that the graph of 3 decreases, increases, then decreases :justiﬁes g(21:) < 3(1) : answer : answer of “yes” with x = l : explanation ...
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• Summer '16
• Robert Tuskey
• Calculus, absolute minimum, Free-Response Scoring Guidelines

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