Graphs of Functions and Derivatives 147
7.10 Solutions to Cumulative Review Problems
- (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
- Substituting√ x=0 in the expression
x+ 9 − 3
x
leads to
0
0
, an indeterminate
form. ApplyL’Hoˆpital’sRule and you
have limx→ 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
- 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 - (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) xlim→ 1 +(2x+k)= 2 +k, limx→ 1 −
(
x^2 − 1
)
=0;
thus, 2+k= 0 ⇒k=−2. Since
xlim→ 1 +f(x)=xlim→ 1 −f(x)=0, therefore,
limx→ 1 f(x)=0.
(3) f(1)=limx→ 1 f(x)=0. Thus,k=−2.
- (a) Sincef′>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].
(b) The absolute maximum occurs at
x=2, since it is a relative maximum
and it is the only relative extremum on
(−1, 4). The absolute minimum occurs
atx=−1, sincef(−1)< f(4) and the
function has no relative minimum on
[−1, 4].
(c) A change of concavity occurs atx=0.
However,f′(0) is undefined, which
impliesfmay or may not have a
tangent atx=0. Thus,f may or may
not have a point of inflection atx=0.
(d) Concave upward on (−1, 0) and
concave downward on (0, 4).