Part II: The Answers
282
Answers
401–500
Find a point on the line by entering the value of a in the function. For the given value of
a, the corresponding y value is
f (
)
100
100 10
=
=
Next, find the derivative of the function to get the slope of the line:
′
′
f x
x
x
f
( )
(
)
=
=
=
=
−
1
2
1
2
100
1
2 100
1
20
1 2
Using the point-slope formula with slope m = 1
20
and point (100, 10) gives you
y y m x x
y
x
y
x
y
x
− =
−
− =
−
=
− +
=
+
1
1
10 1
20
100
1
20
5 10
1
20
5
(
)
(
)
Replacing y with L(x) gives you the linearization L x
x
( ) =
+
1
20
5.
Finally, substitute in the value x = 102 to find the estimate:
L(
)
.
102
1
20
102 5
10 1
=
( ) +
=
431.
1 035
.
To estimate the value of tan 46°, you can find a linearization using f
(x) = tan x and
a = ° =
45
4
π rad and then substitute the value 46
46
180
° = π rad into the linearization.
Find a point on the line by entering the value of a in the function. For the given value of
a, the corresponding y value is
f π
π
4
4
1
( ) ( )
=
=
tan
Next, find the derivative of the function to get the slope of the line:
′
′ ( ) ( ) ( )
f x
x
f
( ) sec
sec
=
=
=
=
2
2
2
4
4
2
2
π
π
Using the point-slope formula with slope m = 2 and point π
4
1
,
( ) gives you
y y m x x
y
x
y
x
− =
−
− =
−
=
−
+
1
1
1 2
4
2
4
1
(
)
( )
( )
π
π
Replacing y with L(x) gives you the linearization L x
x
( ) = +
−
1 2
4
π
( ) .
282
Answers
401–500
Find a point on the line by entering the value of a in the function. For the given value of
a, the corresponding y value is
f (
)
100
100 10
=
=
Next, find the derivative of the function to get the slope of the line:
′
′
f x
x
x
f
( )
(
)
=
=
=
=
−
1
2
1
2
100
1
2 100
1
20
1 2
Using the point-slope formula with slope m = 1
20
and point (100, 10) gives you
y y m x x
y
x
y
x
y
x
− =
−
− =
−
=
− +
=
+
1
1
10 1
20
100
1
20
5 10
1
20
5
(
)
(
)
Replacing y with L(x) gives you the linearization L x
x
( ) =
+
1
20
5.
Finally, substitute in the value x = 102 to find the estimate:
L(
)
.
102
1
20
102 5
10 1
=
( ) +
=
431.
1 035
.
To estimate the value of tan 46°, you can find a linearization using f
(x) = tan x and
a = ° =
45
4
π rad and then substitute the value 46
46
180
° = π rad into the linearization.
Find a point on the line by entering the value of a in the function. For the given value of
a, the corresponding y value is
f π
π
4
4
1
( ) ( )
=
=
tan
Next, find the derivative of the function to get the slope of the line:
′
′ ( ) ( ) ( )
f x
x
f
( ) sec
sec
=
=
=
=
2
2
2
4
4
2
2
π
π
Using the point-slope formula with slope m = 2 and point π
4
1
,
( ) gives you
y y m x x
y
x
y
x
− =
−
− =
−
=
−
+
1
1
1 2
4
2
4
1
(
)
( )
( )
π
π
Replacing y with L(x) gives you the linearization L x
x
( ) = +
−
1 2
4
π
( ) .