ASSIGNMENT 5  SOLUTIONS
Problem 1.
Done in class.
Problem 2.
Take
f
(
x
) =
x
2
, whose graph is a parabola. Drop it by 1 (so now the function is
g
(
x
) =
x
2

1). Slide it to the right by 1 (the function becomes
h
(
x
) = (
x

1)
2

1 =
x
2

2
x
.)
This graph has all the required properties.
Problem 3.
By definition, concave upwards means increasing first derivative. If
f
and
g
are
increasing, then so is their sum
f
+
g
= (
f
+
g
) . So the first statement is true. But the second
is false.
For example, both
f
(
x
) =
x
2
and
g
(
x
) =
x
4
+
x
2
are concave upwards on [0
,
1] yet
f
(
x
)

g
(
x
) =

x
4
is concave downwards on [0
,
1].
Problem 4.
To start off, the domain of
f
(
x
) =
x
√
9

x
is (
∞
,
9].
The first derivative is
f
(
x
) =
x
√
9

x
+
x
(
√
9

x
) =
√
9

x
+
x
1
2
1
√
9

x
(9

x
) =
√
9

x

x
2
√
9

x
=
2(9

x
)

x
2
√
9

x
=
18

3
x
2
√
9

x
=
3
2
6

x
√
9

x
The critical numbers of
f
(
x
) are
x
= 6 (
f
(6) = 0) and
x
= 9 (
f
(
x
) DNE). Both points are in
the domain of
f
(
x
).
i) Let’s use the 1st derivative test. At
x
= 6,
f
(
x
) changes sign from + to

, so
f
(6) = 6
√
3
is a relative maximum. At
x
= 9 there is no change in the sign of
f
(
x
) because we cannot go
past 9. So, strictly speaking, the 1st derivative test does not apply at this point. However, we
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 Spring '10
 MARYAMNAMAZI
 Calculus, Logic, Derivative, Mathematical analysis, Convex function

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