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AP Calculus BC Name Suppose n_x_ is a twice-differentiable

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AP Calculus BC Name Suppose n_x_ is a twice-differentiable Powered By Docstoc
					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).

				
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