MATH1110 Lecture tasks  Worked solutions
Differentiation (IV)
derivative at various points, we find the following:
x

2

1
0
1
2
3
.
817
4
f
00
(
x
)
134
27
DNE

6
DNE
22
27
0

2
675
Concavity
∪
∩
∪
—
∩
Nature
conc. up
conc. down
conc. up
POI
conc. down
We find
f
is concave down for

1
< x <
0, and therefore the statement is TRUE.
Task 5.
Using the Chain Rule, we have
f
0
(
x
) =
e
1

2
x
2
·
(

4
x
) =

4
xe
1

2
x
2
.
To compute the second derivative, we will need to use the Product Rule to obtain
f
00
(
x
) =
e
1

2
x
2
·
(

4) + (

4
x
)
·

4
xe
1

2
x
2
= (16
x
2

4)
e
1

2
x
2
.
The stationary points of
f
occur when
f
0
(
x
) = 0. That is, when

4
xe
1

2
x
2
= 0
=
⇒
x
= 0
.
We find
f
00
(0) =

4
e <
0, and so by the second derivative test,
f
has a local maximum
stationary point at (0
, f
(0)) = (0
, e
).
Task 6.
f
is the black curve (B),
f
0
is the blue curve (A), and
f
00
is the red curve (C). To
see this, note that the stationary points of
f
occur at the same
x
values as the
x
intercepts
of
f
0
, and the stationary points of
f
0
occur at the same
x
values as the
x
intercepts of
f
00
.
It's also possible to make additional observations about the various curves being
increasing/decreasing on certain intervals, and similarly regarding their concavity, but
these are not necessary to uniquely identify which curve corresponds to which function.
Task 7.
Using the Product Rule (and also the Chain Rule), we obtain
h
0
(
t
) =
√
t
+ 3
·
1 +
t
·
1
2
√
t
+ 3
=
3
t
+ 6
2
√
t
+ 3
.
The stationary points of
h
will occur when
h
0
(
t
) = 0. That is, when
3
t
+ 6
2
√
t
+ 3
= 0
=
⇒
3
t
+ 6 = 0
=
⇒
t
=

2
.
Using the first derivative test, we obtain
x

2
.
1

2

1
.
9
h
0
(
x
)
3(

2
.
1) + 6
2
√

2
.
1 + 3
<
0
0
3(

1
.
9) + 6
2
√

1
.
9 + 3
>
0
Slope
\
—
/
Nature
decreasing
min.
increasing
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