# 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 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

(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

(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?

(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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