5 Steps to a 5 AP Calculus BC 2019

More Applications of Derivatives 205

Another approach to solve the problem is

as follows: Multiply both numerator and

denominator by the conjugate of the

denominator (

√

x+10):

xlim→ 100

(x−100)

(√

x− 10

)·

(√

x+ 10

)

(√

x+ 10

)=

xlim→ 100

(x−100)

(√

x+ 10

)

(x−100)

xlim→ 100 (

√

x+10)= 10 + 10 = 20.

An alternative solution is to factor the

numerator:

xlim→ 10

(

√

x−10)

(√

x+ 10

)

(√

x− 10

) = 20.

29. (a) f′>0on(−1, 2),f is increasing on
    (−1, 2) f′<0 on (2, 8), fis
    decreasing on (2, 8).
    (b)Atx=2,f′=0 and f′′<0, thus at
    x=2,f has a relative maximum.
    Since it is the only relative extremum
    on the interval, it is an absolute
    maximum. Since fis a continuous
    function on a closed interval and at its
    endpoints f(−1)<0 andf(8)= 1 /2,
    fhas an absolute minimum at
    x=−1.
    (c)Atx=5,fhas a change of concavity
    and f′exists atx=5.
    (d) f′′<0on(−1, 5),f is concave
    downward on (−1, 5).
    f′′>0 on (5, 8), fis concave
    upward on (5, 8).
    (e)A possible graph off is given in
    Figure 9.9-1.

–1 0 1 2 3 4 5 6 7 8

(2,3)

(8,^1 ⁄ 2 )

(^3) f
y
x
Figure 9.9-1

30. (a) v(t)=3 ft/s att=6. The tangent line
    to the graph ofv(t)att=6 has a slope
    of approximatelym=1. (The tangent
    passes through the points (8, 5) and
    (6, 3); thusm=1.) Therefore the
    acceleration is 1 ft/s^2.
    (b) The acceleration is a minimum
    att=0, since the slope of the tangent to
    the curve ofv(t) is the smallest att=0.


31. To convertr=4 cosθto a Cartesian
    representation, recall thatr=

√

x^2 +y^2

and tanθ=

y

x

.Then,

√

x^2 +y^2 =4 cos

(

tan−^1

y

x

)

. Since
cos

(

tan−^1

y

x

)

=

x

√

x^2 +y^2

, the equation

becomes

√

x^2 +y^2 =

4 x

√

x^2 +y^2

. Multiply

through by

√

x^2 +y^2 to produce

x^2 +y^2 = 4 x. Completing the square

produces (x−2)^2 +y^2 =4.