5 Steps to a 5 AP Calculus BC 2019

Definite Integrals 255

The function fhas a relative

minimum atx=−5 andx=3, andf

has a relative maximum atx=−1 and

x=7.

(b)(See Figure 11.9-2.)

x

f′

f′′

f

incr.

–6

+– +–

concave

downward

concave

upward

concave

downward

–3 (^158)
decr. incr. decr.
concave
upward
Figure 11.9-2
The function fis concave upward on
intervals (−6,−3) and (1, 5).
(c)A change of concavity occurs at
x=−3,x=1, andx=5.

29. (Calculator) Differentiate both sides of
    9 x^2 + 4 y^2 − 18 x+ 16 y=11.
    18 x+ 8 y
    dy
    dx

− 18 + 16

dy

dx

= 0

8 y

dy

dx

+ 16

dy

dx

=− 18 x+ 18

dy

dx

(8y+16)=− 18 x+ 18

dy

dx

=

− 18 x+ 18

8 y+ 16

Horizontal tangent⇒

dy

dx

=0.

Set

dy

dx

= 0 ⇒− 18 x+ 18 =0orx= 1.

Atx=1, 9+ 4 y^2 − 18 + 16 y= 11

⇒ 4 y^2 + 16 y− 20 = 0.

Using a calculator, enter [Solve]

(4y∧ 2 + 16 y− 20 =0,y); obtainingy=− 5

ory=1.

Thus, at each of the points at (1, 1) and

(1,−5) the graph has a horizontal tangent.

Vertical tangent⇒

dy

dx

is undefined.

Set 8y+ 16 = 0 ⇒y=−2.

Aty=−2, 9x^2 + 16 − 18 x− 32 = 11

⇒ 9 x^2 − 18 x− 27 = 0.

Enter [Solve](9x^2 − 8 x− 27 =0,x) and

obtainx=3orx=−1.

Thus, at each of the points (3,−2) and

(−1,−2), the graph has a vertical tangent.

(See Figure 11.9-3.)

Figure 11.9-3

30. (Calculator)

Step 1: (See Figure 11.9-4.) LetP=x+y

wherePis the length of the pipe

andxandyare as shown. The

minimum value ofPis the

maximum length of the pipe to be

able to turn in the corner. By

similar triangles,

y

10

=

√ x

x^2 − 36

and thus,y=

√^10 x

x^2 − 36

, x> 6

P=x+y=x+

10 x

√

x^2 − 36

.