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.
- (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
- (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