Math 191
FINAL EXAM SOLUTIONS
Fall 2000
1.
(a) Using l’Hopital’s rule, the limit is 1.
(b) The only critical point is
x
=
e

1
. (
f
0
(
x
) = (1 + ln
x
)
e
x
ln
x
)
(c) The absolute minimum occurs when
x
=
e

1
where the value is
e

1
/e
. There is
no absolute maximum since lim
x
→∞
=
∞
.
(d) The linearization of
f
at
x
= 1 is the function
l
(
x
) = 1 + (
x

1) =
x
.
(e) The approximation from Newton’s method is
x
1
= 1
.
01 = 1


0
.
01
1
.
2. Separate variables and use partial fractions:
Z
dt
y
2

4
=
Z
sin
xdx
to obtain
1
4
ln(
y

2
y
+ 2
) =

cos
x
+
C
Substitute the initial condition to compute
C
= 1 +
1
4
ln
1
5
.
3.
(a)
1
4
(
1
2
+ 2
4
5
+ 2 + 2
4
5
+
1
2
) =
31
20
(b) This problem is more difficult than intended.
The approximation is an under
estimate of 2 tan

1
(1)
≈
1
.
57.
The function
1
1+
t
2
is concave downwards on the
interval (

1
/
√
3
,
1
√
3) and concave upwards for

t

>
1
√
3). The parts that are
above the trapezoids are larger than those that are below. If one evaluates the
difference between
f
(
±
1
/
4) and
f
(
±
3
/
4) and the points on the secants, then at
±
1
/
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 Fall '07
 BERMAN
 Critical Point, Derivative, Continuous function, Convex function

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