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

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

Graphs of Functions and Derivatives 169

8. Set f′′(x)=0. Thus,x^2 (x+3)(x−5)=
   0 ⇒x=0,x=−3orx=5.
   (See Figure 8.8-1.)
   Thus,fhas a change of concavity at
   x=−3 and atx=5.

Change of

concavity

Change of

concavity

Concave

downward

Concave

downward

Concave

upward

Concave

upward

+

• 30 5

––+

f

f′′

x

Figure 8.8-1

9. See Figure 8.8-2.
   Thus,fis increasing on [2, 3] and
   concave downward on (0, 1).

Concave

downward

Concave

upward

[]

[ ]

incr. incr.

incr.

decr.

decr. decr.

• 3013
  
  
  • 3023
  
  
  
  

Concave

upward

+ – +

• – +

f′′

f′

f′

f

f

x

x

Figure 8.8-2

10. The correct answer is (A).
    f(−1)=0;f′(0)<0 since fis decreasing
    and f′′(−1)<0 since fis concave
    downward. Thus,f(−1) has the largest
    value.


11. Step 1: Domain: all real numbers.
    Step 2: Symmetry: Even function
    (f(x)=f(−x)); symmetrical with
    respect to they-axis.
    Step 3: f′(x)= 4 x^3 − 2 xand
    f′′(x)= 12 x^2 −2.
    Step 4: Find the critical numbers:
    f′(x) is defined for all real
    numbers. Setf′(x)=
    4 x^3 − 2 x= 0 ⇒ 2 x(2x^2 −1)= 0
    ⇒x=0orx=±

√

1 /2.

Possible points of inflection:

f′′(x) is defined for all real

numbers. Setf′′(x)= 12 x^2 − 2 = 0

⇒2(6x^2 −1)= 0

⇒x=±

√

1 /6.

Step 5: Determine the intervals:

––1/2 1/6 0 1/6 1/2

The intervals are:

(

−∞,−

√

1 / 2

)

,

(

−

√

1 /2,−

√

1 / 6

)

,

(

−

√

1 /6, 0

)

,

(

0,

√

1 / 6

)

,

(√

1 /6,

√

1 / 2

)

(√ , and

1 /2,∞

)

.

Sincef′(x) is symmetrical with

respect to they-axis, you only

need to examine half of the

intervals.

Step 6: Set up a table (See Table 8.8-1).

The function has an absolute

minimum value of (−1/4) and no

absolute maximum value.