5 Steps to a 5 AP Calculus AB 2019 - William Ma

MA 3972-MA-Book April 11, 2018 15:14

226 STEP 4. Review the Knowledge You Need to Score High

dy

dx

∣∣

∣∣

y= 2

=

1

3(2)^2 −6(2)+ 3

=

1

3

23. Substitutingx=100 into the expression
    x− 100
    √
    x− 10
    would lead to

0

0

. Apply

L’Hôpital’sRule, and you have limx→ 100

1

1

2 x

−^12

or

1

1

2 (100)

−^12 =^20 .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.

24. (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

endpointsf(−1)<0 and f(8)= 1 /2,

f has an absolute minimum atx=−1.

(c) Atx=5,fhas a change of concavity

and f′exists atx=5.

(d) f′′<0on(−1, 5), fis concave

downward on (−1, 5).

f′′>0 on (5, 8), f is concave

upward on (5, 8).

(e) A possible graph of fis given in

Figure 10.8-1.

• 123456780 1

(2, 3)

(8, 1⁄ 2)

(^3) f
y
x
Figure 10.8-1

25. (a) v(t)=3 ft/sec att=6. The tangent
    line to the graph ofv(t)att=6 has a
    slope of approximatelym=1. (The
    tangent line passes through the points
    (8, 5) and (6, 3); thusm=1.)
    Therefore, the acceleration is 1 ft/sec^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.