5 Steps to a 5 AP Calculus BC 2019

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

9. (See Figure 7.9-2.)
   Thus,fis increasing on [2, 3] and
   concave downward on (0, 1).

Figure 7.9-2

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: 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 intervals:

––√√1/2 1/6 0 √√1/6 1/2

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 (Table 7.9-1).

The function has an absolute

minimum value of (−1/4) and no

absolute maximum value.

Table 7.9-1

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

concave

downward

decr

pt. of

inflection

decr concave

upward

rel min incr

concave

upward