Unformatted text preview: gives (x) of the function
b. FindThe second which valuesxofof )does (this 1)(1 x)linearizationnear xunderestimates of the
[3 points] For derivative L( f (k the function f ( + x)k(12+ so)k give = 0, where k is a positive constant.
12. 0. the local linearization is ) x = k k − local = − , x
k
√
f (x
actual you of f to use L(x) your work.) ) = (1 + x) of the number 1.1. What number should k be, and
Supposevaluewnat (x)? (Show to ﬁnd an√
approximation
!
what number should x be? Approximate (1.1.= k (k − 1).
f 0)
Solution: where k linearization gives underestimates of the actual value when f (0) >
! near x = 0, The localis a positive constant.
0. The second derivative is f (x) = k (k − 1)(1 1 x)k−2 , so
+
Since k > 0,The derivative iswhen) theksecond kfactor is positive, which is when k > 1. = 0
Solution: this is positive f (x = (1 + x) − , so the slope of the tangent line at x
is
f (0) = k (k − 1).
f (0) = k.
√
c. [2Since k >Suppose you want when the(second factor approximation of the number 1.1.1.
points] 0, this is positive to use L x) to ﬁnd an is positive, which is when k >
k = 1, the tangent line passes through the point (0, 1). Therefore, the
Since f (0) = 1
What number should k be, and what number should x be?
pointslope formula shows that the equation of the tangent line is
√
√
Solution: If k = 1 and x = 0.1, then f (0.1) = 1.1, so L(1.1) gives an approximation
√
c. [2 points] Suppose2you want to use L(x)= kx + 1. approximation of the number 1.1.
y to ﬁnd an
of
.1.
What1number should k be, and what number should x be?
!
√
Solution: If k = 1 and x = 0.1, then f (0.1) = 1.1, so L(1.1) gives an approximation
2
√
√
d. [2of 1.1. Approximate 1.1 using L(x).
points]
b. [3 points] For which values of k does this local linearization give underestimates of the !
√
actual value of f (x)? (Show your work.) 1.1 ≈ L(0.1) = 1.05.
Solution: If k and x are as above, then
√
!
1
d. [2Solution: Approximate 1.1 usinggives).underestimates of the actual value when f (0) >
points] The local linearization L(x
√
0. linear approximation to estimate the k − number
13. Use aThe second derivative is f (x) = k (given1)(1 + x)k−2 , so
e. [2Solution: What and x are asin the approximationL(0.1)part.05.
points] If k is the error above, then 1.1 ≈ from = 1 (d)?
(a) (8.06)2/3
f (0) = k (k − 1).
Solution: e. [2Since k > 0, this is positive in the the second factor is positive, which is when k > 1.
points] What is the error when approximation from part (d)? √√
1−
c. [2 points] Suppose you wantactual value) minus the approximate valuethe number 1.1 .1.
Solution: The error is the to use L(x to ﬁnd an approximation of which is
What number should k be, and what number should x be?
1.05 ≈ −0.00119.
√
√
1
√
Solution:
Solution: If k error is x actual value minus 1 1, so L(1.1) value which is 1.1 −
(b) √99.8 The = 2 andthe= 0.1, then f (0.1) = the .approximate gives an approximation
of Solution:
1−
1.05 ≈.1. 0.00119. ! 1For...
View
Full Document
 Fall '11
 fisherselin
 Derivative, Differential Calculus, Slope

Click to edit the document details