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

MA 3972-MA-Book April 11, 2018 17:21

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

Table 8.8-1

INTERVALS X= 0 (0,

√

1 /6) X=

√

1 /6(

√

1 /6,

√

1 /2) X=

√

1 /2(

√

1 /2,∞)

f(x)0 − 5 / 36 − 1 / 4

f′(x)0−− − 0 +

f′′(x) −− 0 +++

conclusion rel. max. decr. decr. decr. rel. min. incr.

concave pt. of concave concave

downward inflection upward upward

Step 7: Sketch the graph.

(See Figure 8.8-3.)

x

y

f

rel. max

(0, 0)

Abs. min Abs. min

pt of

infl.

( )––^12 ,^14 ( )––^12 ,^14

( )–– 61 , 365 ( )– 61 ,– 365

pt of

infl.

Figure 8.8-3

12. Step 1: Domain: all real numbersx=4.

Step 2: Symmetry: none.

Step 3: Findf′and f′′.

f′(x)=

( 1 )(x− 4 )−( 1 )(x+ 4 )

(x− 4 )^2

=

− 8

(x− 4 )^2

, f′′(x)=

16

(x− 4 )^3

Step 4: Find the critical numbers:

f′(x)=0 and

f′(x) is undefined atx=4.

Step 5: Determine the intervals.

4

Intervals are (−∞, 4) and (4,∞).

Step 6: Set up a table as below:

INTERVALS (−∞, 4) (4,∞)

f′ −−

f′′ −+

conclusion decr. concave incr. concave

downward upward

Step 7: Horizontal asymptote:

lim

x→±∞

x+ 4

x− 4

=1. Thus,y=1isa

horizontal asymptote.

Vertical asymptote:

lim

x→ 4 +

x+ 4

x− 4

=∞and

xlim→ 4 −

x+ 4

x− 4

=−∞; Thus,x=4isa

vertical asymptote.

Step 8: Determine the intercepts:

x-intercept: Set f′(x)= 0

⇒x+ 4 =0;x=−4.

y-intercept: Setx= 0

⇒f(x)=−1.