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Math 141
Fall 1998
No Work-No Credit
1)
Exam 4
Name
SS#
Use Newton's method to approximate the nearest root of
f ( x) =
1
− tan x
x2
using
x1 = 3.8 as the initial guess. List each approximation until the
approximation is repeated to 6 decimal places.
X1=3.8
x2=3.369190
x5=3.236755
2)
If
24
is
x3=3.239422
x4=3.236756
x6=x5
A softball diamond has the shape of a square with sides 60 feet long.
a player is running from second base to third base at a speed of
ft/sec, at what rate is her distance from home plate changing when she
20 ft from third?
Want
dL
dt
at the instant y = 20. Also given
L2 = x 2 + y 2
Then
then
dy
( 20)( −24)
dL
= − 7.6
= dt =
2
2
dt
L
( 20) + (60)
y
dy
= − 24 .
dt
differentiate
to
get:
3)
For
f ( x ) = sin x − cos x
a) critical numbers
determine:
b) relative maximum
2 , when x =
Set f’(x) = 0
f’(x) = cos x - sin x = 0
x=
3π
4
relative minimum

 7π
,− 2


 4
3π 7π
,
4 4
c) intervals where f(x) is
increasing:
 3π   7π

,2π 
 0,  ∪ 
 4  4

d) intervals where f(x) is
concave up:
 π   5π

 0,  ∪  ,2π 
 4  4

decreasing:
concave down:
 3π 7π 
 , 
 4 4
 π 5π 
 , 
4 4 
4) Determine the absolute maximum and absolute minimum values of
f ( x ) = x 2 − 8 ln x
over the interval
[1,e] .
8 2 x2 − 8
f '( x ) = 2 x − =
=0
x
x
Critical Number(s):
So x = +/- 2, but only x = 2 is in the interval of interest.
So we need: f(1) = 1, f(e) = -0.61, and f(2) = -1.545
So the absolute max is 1 when x = 1
and the absolute min is -1.545 when x = 2.
5) Evaluate:
Rewrite
lim (1 + 3x )
1 2x
x→ 0
(1 + 3x ) = e
ln( 1+ 3 x )
Then
(1 + 3x )
1
2x
=e
 ln(1+ 3 x ) 

2 x
so the original limit may be written as:
lim (1 + 3x )
1 2x
x→ 0
=
e
lim  ln( 1+ 3 x ) 2 x 
x→ 0
.
Now consider just the exponent.
ln(1 + 3x )
3
3
LR lim
= ,
x→ 0
=
x → 0 2 (1 + 3 x )
2x
2
lim
So
lim (1 + 3x )
1 2x
x→ 0
=
e
( 32 )