Barrons AP Calculus - David Bock
(A) If ...
(D) represents the area of the same region as translated one unit to the left. ...
(D) According to the Mean Value Theorem, there exists a number c in the interval [1,1.5] such that Use your calculator to solve ...
(E) Here are the relevant sign lines: We see that f ′ and f ′′ are both positive only if x > 1. ...
(E) Note from the sign lines in Question 63 that f changes from decreasing to increasing at x = 1, so f has a local minimum the ...
(C) The derivatives of ln (x + 1) are The nth derivative at ...
(C) The absolute-value function f (x) = |x| is continuous at x = 0, but f ′(0) does not exist. ...
(C) Let F ′(x) = f (x); then F ′(x + k) = f (x + k); Or let u = x + k. Then dx = du, when x = 0, u = k; when x = 3, u = 3 + k. ...
(E) See the figure. The equation of the generating circle is (x − 3)^2 + y^2 = 1, which yields ...
(D) Note that f (g(u)) = tan−1 (e^2 u); then the derivative is ...
(D) Let Then cos (xy) [xy′ + y] = y ′. Solve for y ′. ...
71. (E) ...
(C) About the x-axis; see the figure. Washer. ...
(C) By the Mean Value Theorem, there is a number c in [1, 2] such that ...
(D) The enclosed region, S, is bounded by y = sec x, the y-axis, and y = 4. It is to be rotated about the y-axis. Use disks; t ...
(C) If Q is the amount at time t. then Q = 40e−kt. Since Q = 20 when t = 2, k = −0.3466. Now find Q when t = 3, from Q = 40e−(0 ...
(A) The velocity v(t) is an antiderivative of a(t), where So arctan t C. Since v (1) = 0, C = − π. ...
(D) Graph y = tan x and y = 2 − x in [−1, 3] × [−1, 3]. Note that The limits are y = 0 and y = b, where b is the ordinate of t ...
(E) Center the ellipse at the origin and let (x, y) be the coordinates of the vertex of the inscribed rectangle in the first qu ...
79. (B) ...
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