Barrons AP Calculus

Since Q(1) = 0.8Q 0 , 0.8Q 0 = Q 0 ek·1, 0.8 = ek, and Q(t) = Q 0 (0.8)t. We seek t when Q(t) ...

AB/BC 1. Part A This is the graph of f ′(x). (a) f is increasing when f ′(x) > 0. The graph show ...

AB/BC 2. Since the graph of f changes concavity at p, q, and r, there are three points of inflection. (a) Sin ...

AB 3. Part B (a) The slope field for the twelve points indicated is: The slopes are given in the table below. B ...

AB 4. Since f(x) passes through (1, 1), it must be true that . Thus C 2 = 5, and the positive sq ...

AB 5. (b) (i) Use washers; then (ii) See the figure above. (a) f(x) = e^2 x(x^2 − 2), f ′(x) = e^2 x(2x) ...

AB/BC 6. (c) As x approaches ±∞, f(x) = e^2 x(x^2 − 2) also approaches ±∞. There is no global maximum. ( ...

1. 2. 3. 4. 5. 6. 7. 8. Part A (D) as x → 2. (A) Divide both numerator and denominator by . (D) Since eln u ...

Since and . Substituting yields . (B) For f(x) = , this limit represents f ′( ...

(C) The winning times are positive, decreasing, and concave upward. (D) , where represents ...

(B) Speed is the magnitude of velocity; its graph is shown above. (B) The average rate o ...

(D). (A). (B) From the Riemann Sum, we see , then . Notice that the term involving k in t ...

31. 32. Part B (D) We solve the differential equation by separation: If s = 1 when t = 0, we have ...

Solving on a calculator gives k (or x) equal to 8. (C) If N is the number of bacteria at time ...

(D) G′(x) = f(3x − 1) · 3. (B) Since f changes from positive to negative at t = ...

At x = 3, the answer is 2[ 2(−2) + 5^2 ] = 42. (A) The object is at rest when v(t) ...

45. (B) Section II Free-Response ...

AB/BC 1. (b) (c) (d) AB 2/BC 4. (b) Part A (a) . . To work with g(x) = f −^1 (x), interchange x and y ...

Therefore . (NOTE: Although the shells method is not a required AP topic, another correct integral for ...

AB 3. (b) (c) AB 4. Part B (a) Refer back to the figure for Part B. The average value of a function on an ...