Barrons AP Calculus - David Bock
(B) Let t = x − 1; then t = −1 when x = 0, t = 5 when x = 6, and dt = dx. ...
(B) The required area, A, is given by the integral ...
(B) The average value is The definite integral represents the sum of the areas of a trapezoid and a rectangle: (8 + 3)(6) = 4(7 ...
(A) Solve the differential equation by separation of variables: yields y = ce^2 x. The initial condition yields 1 = ce2 · 2; s ...
(C) Changes in values of f ′′ show that f ′′′ is constant; hence f ′′ is linear, f ′ is quadratic, and f must be cubic. ...
(B) By implicit differentiation, so the equation of the tangent line at (3,0) is y = −9(x−3). ...
24. (A) ...
(D) The graph shown has the x-axis as a horizontal asymptote. ...
(B) Since to render f (x) continuous at x = 1 f (1) must be defined to be 1. ...
(B) f ′(x) = 15x^4 − 30x^2 ; f ′′(x) = 60x^3 − 60x = 60x(x + 1)(x − 1); this equals 0 when x = −1, 0, or 1. Here are the signs ...
(C) Note that so f has a critical value at x = −4. As x passes through −4, the sign of f ′ changes from − to +, so f has a loca ...
Part B (B) We are given that (1) f ′(a) > 0; (2)f ′′(a) < 0; and (3) G ′(a) < 0. Since G ′(x) = 2 f (x) · f ′(x), ther ...
(E) Since it equals 0 for When x = 3, c = 9; this yields a minimum since f ′′(3) > 0. ...
(E) Use your calculator to graph velocity against time. Speed is the absolute value of velocity. The greatest deviation from v ...
(D) Because f ′ changes from increasing to decreasing, f ′′ changes from positive to negative and thus the graph of f changes c ...
(D) We evaluate this definite integral by finding the area of a trapezoid (negative) and a triangle: so the tangent line passes ...
(D) The distance is approximately 14(6) + 8(2) + 3(4). ...
(D) R(x)dx = 166.396. ...
(A) Selecting an answer for this question from your calculator graph is unwise. In some windows the graph may appear continuous ...
37. (D) ...
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