Studying Workbooks

AP Calculus by Shirley O. Hockett, David Bock

By Shirley O. Hockett, David Bock

Either Calculus AB and Calculus BC are coated during this complete AP try out practise guide. potential try out takers will locate 4 perform assessments in Calculus AB and 4 extra in Calculus BC, with all questions spoke back and options defined. The guide additionally presents an in depth 10-chapter assessment masking subject matters for either assessments. The authors additionally provide an summary of the AP Calculus checks, inclusive of suggestion to scholars on making top use in their graphing calculators.

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Extra resources for AP Calculus

Sample text

The object reverses direction when the velocity changes sign, that is, when the graph crosses the x-axis. There are two such reversals—at x = a and at x = b. 31. (D) The sign diagram shows that f changes from increasing to decreasing f inc 0 f inc 2 + dec 4 + 5 – and thus f has a maximum at x = 4. Because f increases to the right of x = 0 and decreases to the left of x = 5, there are minima at the endpoints. 32. (D) Since f Ј decreases, increases, then decreases, f Љ changes from negative to positive, then back to negative.

507. 507. (c) As the object moves to the left (with v(t) negative), the speed of the object increases when its acceleration v¢(t) is also negative, that is, when v(t) is decreasing. This is true when, for example, t = 2. 37 Diagnostic Test Calculus AB 7_3679_APCalc_01DiagnosticAB Diagnostic Test Calculus AB 7_4324_APCalc_01DiagnosticAB 38 10/4/09 1:22 PM Page 38 AP Calculus (d) The displacement of an object from time t1 to time t2 is equal to t2 ∫t v (t ) dt = 1 ∫1 ( t + e ) sin ( t 3 t 2 ) dt.

Find c in [0,1] such Δx 1 that, f Ј(c) = e – 2, or f Ј(c) – (e – 2) = 0. Since f Ј(x) = ex – 2x, c can be calculated by solving 0 = ex Ϫ 2x Ϫ (e Ϫ 2). 351. 34. (B) y y y = ex y = ex y=2 y=2 (x,y) 1 1 x 0 (x,y) 0 Use disks; then ⌬V = ␲R2H = ␲(ln y)2 ⌬y. Note that the limits of the definite integral are 1 and 2. 592 2 2 1 Alternatively, use shells*; then ΔV = 2πRHT = 2πx(2 – ex) Δx. Here, the upper limit of integration is the value of x for which ex = 2, namely, ln 2. 592 0 *No question requiring the use of the shells method will appear on the AP exam.

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