1. Group Work Solutions
Linearization.
(1) Find the linearization L(x) of the function at a = 3
2
f (x) = √
x2 − 5
Solution:
We are trying to find the tangent line at the point a = 3. Now
−3
1
−2x
0
f (x) = 2 · −
(x2 − 5) 2 · (2x) = p
2
(x2 − 5)3
Therefore
−2(3)
6
3
f 0 (3) = p
= 3 = −
2
4
(32 − 5)3
2
2
= √
=1
f (3) = √
x2 − 5
32 − 5
and by substituting this into the equation for the tangent line at a
L(x) = f (a) + f 0 (a)(x − a)
when a = 3 we get
3
L(x) = 1 − (x − 3)
4
1
(2) Verify the given linear approximation at a = 0
√
1
4
1 + 2x ≈ 1 + x
2
Solution:
−3
1
1
f (x) =
(1 + 2x) 4 (2) = p
4
4
2 (1 + 2x)3
0
Therefore
1
1
p
=
4
3
2
2 (1 + 2(0))
p
f (0) = 4 1 + 2(0) = 1
f 0 (0) =
Therefore
1
L(x) = f (0) + f 0 (0)(x − 0) = 1 + x
2
2
(3) Find the linear approximation
√ of the
√ function f (x) =
approximate the numbers .9 and .99
Solution
f (x) ≈ 1 −
√
1 − x at a = 0 and use it to
x
2
Solution:
−1
−1
1
(1 − x) 2 (−1) = p
f (x) =
2
2 (1 − x)
0
Therefore
−1
−1
p
=
2
2 (1 − 0)
√
f (0) =
1−0 = 1
f 0 (0) =
Hence
L(x) = f (0) + f 0 (0)(x − 0) = 1 +
We now want to use this to approximate
that
f (.1) =
p
√
1 − (.1) =
.9
−1
x
2
√
√
.9 and .99. From our function, it is clear
f (.01) =
p
√
1 − (.01) =
.99
Since f (x) ≈ L(x) when x is near to 0 Then
√
.1
.9 ≈ L(.1) = 1 −
2
√
3
.9 ≈ L(.01) = 1 −
.01
2
(4) (Hard) Use a linear approximation to estimate the given number
√
3
1001
Hint: Think about what f (x) should be and then write down your line as L(x) =
f (a) + f 0 (a)(x − a). Finally evaluate at the point that you are interested in.
Solution:
√
√
If we think about the function f (x) = 3 x then its clear that f (1001) = 3 1001.
To approximate this value we can pick a point on our function that is near to 1001
and use the tangent line at that point to approximate f (1001). Picking a = 1000 we
find our tangent line as
−2
1
1
0
√
f (x) =
x3 =
3
3
3( x)2
Hence
1
1
1
=
=
3(10)2
300
3( 1000)2
√
3
f (1000) =
1000 = 10
Therefore our tangent line is
f 0 (1000) =
√
3
L(x) = f (1000) + f 0 (1000)(x − 1000) = 10 +
1
(x − 1000)
300
Hence at 1001 our approximation is
√
3
1001 ≈ L(1001) = 10 +
1
1
1
(1001 − 1000) = 10 +
(1) = 10 +
300
300
300
4