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

MA 3972-MA-Book April 11, 2018 16:32

372 STEP 5. Build Your Test-Taking Confidence

(C) Note that

lim

x→ 1

2 f(x)

x+ 1

=

2 f( 1 )

1 + 1

=

2 f( 1 )

2

=f( 1 ).

Also,

∫ 2

1

f′(x)dx=f( 2 )− f( 1 ), with

f( 2 )=3. Since the area of regionCis 6,

and region∫ Cis below thex-axis, you have

2

1

f′(x)dx=−6. Therefore,

f( 2 )− f( 1 )=−6, or 3−f( 1 )=−6,

which yieldsf( 1 )=9. Thus,

lim

x→ 1

2 f(x)

x+ 1

=f( 1 )=9.

(D) By substitutingx=−2, you have

lim

x→− 2

f(x)+ 1

x+ 2

=

f(− 2 )+ 1

0

. To find the
value of∫ f(− 2 ), note that
2

− 2

f′(x)dx=f( 2 )−f(− 2 )= 3 −f(− 2 ).

Also, using regions∫ BandC, you have

2

− 2

f′(x)dx=( 10 )+(− 6 )=4. Therefore,

3 −f(− 2 )=4, or f(− 2 )=−1. Since

f(− 2 )=−1, lim

x→− 2

f(x)+ 1

x+ 2

leads to

− 1 + 1

− 2 + 2

=

0

0

, an indeterminate form.

ApplyingL'Hôpital'sRule, you have

lim

x→− 2

f′(x)

1

= f′(− 2 ). Note that

f′(− 2 )=0, and thus lim

x→− 2

f(x)+ 1

x+ 2

=0.

6. (A) The equation of the curve is
   14 y−y^2 =x^2 +24. Using implicit
   differentiation, you have

14

dy

dx

− 2 y

dy

dx

= 2 x. Solve for

dy

dx

and

obtain

dy

dx

=

2 x

14 − 2 y

,or

dy

dx

=

x

7 −y

.

(B) The slope of the tangent at the point

(−3, 11)is

dy

dx

∣∣

∣∣

(−3, 11)

=

− 3

7 − 11

=

3

4

. Using

the formula of a line,y−y 1 =m

(

x−x 1

)

,

you havey− 11 =

3

4

(x+ 3 ),or

y=

3

4

x+

53

4

.

(C) The slope of all tangents to the curve is

derived in part (A) as

dy

dx

=

x

7 −y

.

Remember that a vertical tangent has no

slope. Therefore,

dy

dx

is undefined for all

vertical tangents, and

dy

dx

is undefined

when 7−y=0ory=7. Substitutingy= 7

in the equation 14y−y^2 =x^2 +24, you

have 14(7)−(7)^2 =x^2 +24. Solve forx

and obtainx=±5. Therefore, the points

of tangency are (5, 7) and (−5, 7), and the

equations of the vertical tangents are

x=−5 andx=5.

(D) From part (A),

dy

dx

=

x

7 −y

. Using the
quotient rule, you have:
d^2 y
dx^2

=

(1)(7−y)−(x)

(

−

dy

dx

)

(7−y)^2

At the point (0, 2) on the curve,

dy

dx

=

0

7 −(2)

=0. Therefore,

d^2 y

dx^2

∣∣

∣∣

(0, 2)

=

(7−2)− 0

( 7 − 2 )^2

=

1

5

.