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

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

(e) function is concave upward on

intervals (0, 0.491),

(

π

2

,2. 651

)

,

(

3 .632,

3 π

2

)

, and (5.792, 2π), and

concave downward on the intervals(

0 .491,

π

2

)

, (2.651, 3.632), and

(

3 π

2

,5. 792

)

.

8.9 Solutions to Cumulative Review Problems

21. (x^2 +y^2 )^2 = 10 xy

2

(

x^2 +y^2

)(

2 x+ 2 y

dy

dx

)

= 10 y+( 10 x)

dy

dx

4 x

(

x^2 +y^2

)

+ 4 y

(

x^2 +y^2

)dy

dx

= 10 y+( 10 x)

dy

dx

4 y

(

x^2 +y^2

)dy

dx

−( 10 x)

dy

dx

= 10 y− 4 x

(

x^2 +y^2

)

dy

dx

(

4 y

(

x^2 +y^2

)

− 10 x

)

= 10 y− 4 x

(

x^2 +y^2

)

dy

dx

=

10 y− 4 x

(

x^2 +y^2

)

4 y(x^2 +y^2 )− 10 x

=

5 y− 2 x

(

x^2 +y^2

)

2 y(x^2 +y^2 )− 5 x

22. Substituting√ x=0 in the expression
    x+ 9 − 3
    x
    leads to

0

0

, an indeterminant

form. ApplyL'Hôpital’sRule and you have

lim

x→ 0

1

2 (x+9)

−^12 (1)

1

,or

1

2

(0+9)−

(^12)

1

6

.

Alternatively,

limx→ 0

√

x+ 9 − 3

x

=

limx→ 0

(√

x+ 9 − 3

)

x

·

(√

x+ 9 + 3

)

(√

x+ 9 + 3

)

=limx→ 0

(x+9)− 9

x

(√

x+ 9 + 3

)

=limx→ 0

x

x

(√

x+ 9 + 3

)

=limx→ 0

1

√

x+ 9 + 3

=

1

√

0 + 9 + 3

=

1

3 + 3

=

1

6

23. y =cos(2x)+ 3 x^2 − 1
    dy
    dx
    =−sin(2x)+ 6 x=
    −2 sin(2x)+ 6 x
    d^2 y
    dx^2
    =−2(cos(2x))(2)+ 6 =
    −4 cos(2x)+ 6


24. (Calculator) The functionf is continuous
    everywhere for all values ofkexcept
    possibly atx=1. Checking with the three
    conditions of continuity atx=1:

(1) f(1)=(1)^2 − 1 = 0

(2) lim

x→ 1 +

(2x+k)= 2 +k, lim

x→ 1 −

(

x^2 − 1

)

=0;

thus, 2+k= 0 ⇒k=−2. Since

lim

x→ 1 +

f(x)=lim

x→ 1 −

f(x)=0, therefore,

lim

x→ 1

f(x)=0.

(3) f(1)=lim

x→ 1

f(x)=0. Thus,k=−2.

25. (a) Since f′>0on(−1, 0) and (0, 2),
    the functionfis increasing on the
    intervals [−1, 0] and [0, 2]. Since
    f′<0 on (2, 4), f is decreasing on
    [2, 4].