Barrons AP Calculus - David Bock
(E) The graph of f ′(x) jumps at each corner of the graph of f (x), namely, at x equal to −3, −1, 1,2, and 5. ...
(D) On the interval (−6,−3), f (x) = ...
(B) Verify that all choices but (B) are true. The graph of f ′(x) has five (not four) jump discontinuities. ...
(C) The best approximation to f ′(0.10) is ...
103. (D) The average rate of change is represented by the slope of secant segment There appear to be 3 points at which the tange ...
Answers Explained (D) Substituting y = 2 yields x = 1. We find y ′ implicitly. 3 y^2 y ′ − (2 xyy ′ + y^2 ) = 0; (3 y^2 − 2xy)y ...
(A) 2 yy ′ − (xy ′ + y) − 3 = 0. Replace x by 0 and y by −1; solve for y ′. ...
(E) Find the slope of the curve at The equation is ...
(B) Since y ′ = e−x (1 − x) and e−x > 0 for all x, y ′ = 0 when x = 1. ...
(D) The slope y ′ = 5x^4 + 3x^2 − 2. Let g = y ′. Since g ′(x) = 20x^3 + 6x = 2x(10x^2 + 3), g ′(x) = 0 only if x = 0. Since g ...
(C) Since 2x − 2yy ′ = 0, At (4, 2), y ′ = 2. The equation of the tangent at (4, 2) is y − 2 = 2(x − 4). ...
(D) Since the tangent is vertical for x = 2y. Substitute in the given equation and solve for y. ...
(D) Since therefore, dV = 4πr^2 dr. The approximate increase in volume is dV ≈ 4π(3^2 ) (0.1) in^3. ...
(C) Differentiating implicitly yields 4x − 3y^2 y ′ = 0. So The linear approximation for the true value of y when x changes fro ...
(B) We want to approximate the change in area of the square when a side of length e increases by 0.01e. The answer is A ′(e)(0. ...
(D) Since V = e^3 , V ′ = 3e^2. Therefore at e = 10, the slope of the tangent line is 300. The change in volume is approximatel ...
(E) f ′(x) = 4x^3 − 8x = 4x(x^2 − 2). f ′ = 0 if x = 0 or f ′′(x) = 12x^2 − 8; f ′′ is positive if x = negative if x = 0. ...
(C) Since f ′′(x) = 4(3x^2 − 2), it equals 0 if Since f ′′ changes sign from positive to negative at and from negative to posit ...
(A) The domain of y is {x | x 2}. Note that y is negative for each x in the domain except 2, where y = 0. ...
(B) f ′(x) changes sign (from negative to positive) as x passes through zero only. ...
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