Document Sample

AP Calculus BC Name _______________________ Suppose n(x) is a twice-differentiable function for all x such that n′′(x) < 0. The table below shows the function values for n at specific values of x. x 0 2 3 6 9 10 n(x) 78 91 96 99 85 78 Let x = c be some value such that 0 < c < 10. Answer the following. (a) Use the conditions and conclusions of the Mean Value Theorem to prove there is at least one value 0 < c < 10 where n′(c) = 0. (9 points) Since n(x) is continuous on [0, 10], differentiable on (0, 10), and n(10) = n(0) = 78, then there n(10) − n(0) exists at least one c on the interval such that n′( c ) = = 0. 10 − 0 (b) Can it be determined with certainty whether n(c) is a relative minimum or relative maximum value? Explain your reasoning. (6 points) Yes; since n′(c) = 0 and n′′(c) < 0, there is a relative minimum at x = c. Consider the function f (x) and its derivatives below defined on the closed interval [−8, 8]. 4x + 2 4x − 4 f ( x) = 3 x1/3 ( x + 2 ) f ′( x) = f ′′( x) = x 2/3 3 x 5/3 Showing your work that leads to your conclusions, answer the following. (a) Identify all relative extrema for f on the closed interval. (10 points) f ′(x) = 0 when x = −1/2 and f ′(x) is undefined when x = 0, so those are the critical points of f (x). Test the intervals for increasing and decreasing behavior: f f′ − + + −8 −1/2 0 8 Since f ′(x) changes from negative to positive at x = −1/2, there is a relative minimum at (−1/2, −3.572). There is no relative maximum. Alternatively, f ′(−1/2) = 0 and f ′′(−1/2) > 0, so there is a relative minimum when x = −1/2. AP Calculus BC Name _______________________ Consider the function f (x) and its derivatives below defined on the closed interval [−8, 8]. 4x + 2 4x − 4 f ( x) = 3 x1/3 ( x + 2 ) f ′( x) = f ′′( x) = x 2/3 3 x 5/3 Showing your work that leads to your conclusions, answer the following. (b) Find the absolute maximum and minimum values of f on the closed interval. (10 points) Since f (x) is continuous on a closed interval, test values at critical points and endpoints (although the work in part (a) could be used to simplify the work here). f (−8) = 36 f (−1/2) = −3.572 f (0) = 0 f (8) = 60 The absolute maximum is f (8) = 60 and the absolute minimum is f (−1/2) = −3.572. (c) State the coordinates for all points of inflection in f. (10 points) f ′′(x) = 0 when x = 1 and f ′′(x) is undefined when x = 0, so those are the critical points of f ′(x). Test the intervals for concavity: f ∪ ∩ ∪ f ′′ + − + −8 0 1 8 Since concavity changes at x = 0 and x = 1, there are points of inflection at (0, 0) and (1, 9). (d) Describe the behavior in the graph of f at x = 0. Explain your reasoning. (6 points) f (x) is continuous at x = 0 but f ′(x) is undefined there, so there must be a corner, cusp, or vertical tangent at x = 0. Since lim f ′( x ) = ∞ , f ′(x) unbounded in the positive direction at x = 0 and there must be a x →0 vertical tangent at x = 0. AP Calculus BC Name _______________________ Suppose g(x) is a continuous function defined for all x. g′ Its derivative, g′(x), is shown at the right and has a vertical asymptote at x = 2, a horizontal asymptote at y = 5, and a horizontal tangent at x = 4. Answer the following. (a) For what interval(s) of the domain is g decreasing? Justify your answer. (8 points) (1, 2) ∪ (3, 5) since g′(x) < 0. (b) Where does the graph of g have upward concavity? Justify your answer. (8 points) (4, ∞) since g′(x) is increasing (which means g′′(x) > 0). (c) Describe the behavior in the graph of g at x = 2. Explain your reasoning. (8 points) Since g(x) is continuous and g′(x) is unbounded in opposite directions at x = 2, there must be a cusp in the graph of g at x = 2. x Suppose the function h ( x ) = . Answer the following. x +2 2 (a) Use limits to find and verify all horizontal asymptotes in h. (10 points) x x x Since lim = lim = lim = lim ( −1) = −1 , there is an asymptote at y = −1. x →−∞ − x x →−∞ x2 + 2 x →−∞ x2 x →−∞ x x x Since lim = lim = lim = lim(1) = 1 , there is an asymptote at y = 1. x →∞ x2 + 2 x →∞ x2 x →∞ x x →∞ AP Calculus BC Name _______________________ x Suppose the function h ( x ) = . Answer the following. x2 + 2 (b) Show that h is always increasing. (10 points) The algebra involved in the differentiation is easier if h(x) = x (x 2 + 2)−1/2. ⎡ 1 ⎤ h′( x ) = 1 ⋅ ( x 2 + 2 ) + x ⋅ ⎢− ( x2 + 2 ) ⋅ 2 x ⎥ −1/2 −3/2 ⎣ 2 ⎦ = ( x2 + 2) − x2 ( x2 + 2) −1/2 −3/2 = ( x2 + 2) ⎡( x 2 + 2 ) − x 2 ⎤ −3/2 ⎣ ⎦ 2 = (x + 2) 2 3/2 Since h′(x) > 0 for all x, then h is always increasing. (c) State the range for h(x). (5 points) Because h is continuous and always increasing, the curve must be bounded by the horizontal asymptotes. Therefore, the range must be (−1, 1).

DOCUMENT INFO

Shared By:

Categories:

Tags:
AP Calculus BC, Calculus BC, AP Calculus, Differential Equations, definite integrals, Advanced Placement, AP Calculus AB, graphing calculator, definite integral, parametric equations

Stats:

views: | 47 |

posted: | 3/25/2010 |

language: | English |

pages: | 4 |

OTHER DOCS BY akgame

How are you planning on using Docstoc?
BUSINESS
PERSONAL

By registering with docstoc.com you agree to our
privacy policy and
terms of service, and to receive content and offer notifications.

Docstoc is the premier online destination to start and grow small businesses. It hosts the best quality and widest selection of professional documents (over 20 million) and resources including expert videos, articles and productivity tools to make every small business better.

Search or Browse for any specific document or resource you need for your business. Or explore our curated resources for Starting a Business, Growing a Business or for Professional Development.

Feel free to Contact Us with any questions you might have.