young (toy68) – HW10 – villafuerte altu – (53785)
2
If the graph of the function defined on
[

3
,
3] by
f
(
x
) =
x
2
+
ax
+
b
has an absolute minimum at (2
,
2), determine
the value of
f
(1).
1.
f
(1) = 3
correct
2.
f
(1) = 4
3.
f
(1) = 7
4.
f
(1) = 6
5.
f
(1) = 5
Explanation:
The absolute minimum of
f
on the inter
val [

3
,
3] will occur at a critical point
c
in
(

3
,
3),
i.e.
, at a solution of
f
′
(
x
) = 2
x
+
a
= 0
,
or at at an endpoint of [

3
,
3].
Thus, since
this absolute minimum is known to occur at
x
= 2 in (

3
,
3), it follows that
f
′
(2) = 0
,
f
(2) = 2
.
These equations are enough to determine the
values of
a
and
b
. Indeed,
f
′
(2) = 4 +
a
= 0
,
so
a
=

4, in which case
f
(2) = 4

8 +
b
= 2
,
so
b
= 6. Consequently,
f
(1) = 1 +
a
+
b
= 3
.
004
10.0points
Find all the critical points of
f
when
f
(
x
) =
x
x
2
+ 4
.
1.
x
=

4
,
2
2.
x
=

2
,
2
correct
3.
x
=

4
,
4
4.
x
=

2
,
4
5.
x
=

2
,
0
6.
x
= 0
,
2
Explanation:
By the Quotient Rule,
f
′
(
x
) =
(
x
2
+ 4)

2
x
2
(
x
2
+ 4)
2
=
4

x
2
(
x
2
+ 4)
2
.
Since
f
is differentiable everywhere, the only
critical points occur at the solutions of
f
′
(
x
) =
0,
i.e.
, at the solutions of
4

x
2
= 0
.
Consequently, the only critical points are
x
=

2
,
2
.
005
10.0points
Determine the absolute minimum value of
f
(
x
) =
x
radicalbig
1

x
2
+ 2
on [

1
,
1].
1.
absolute min. value =
1
2
2.
absolute min. value =
3
2
correct
3.
absolute min. value =
5
2
4.
absolute min. value = 1
5.
absolute min. value = 3