Chiang/Wainwright: Fundamental Methods of Mathematical Economics
Instructor’s Manual
CHAPTER 6
Exercise 6.2
1.
(a)
∆y
4(x + ∆x)2 + 9 − (4x2 + 9)
=
= 8x + 4∆x
∆x
∆x
(b) dy/dx = f 0 (x) = 8x
(c) f 0 (3) = 24 and f 0 (4) = 32.
2.
(a)
∆y
= 10x + 5∆x − 4
∆x
(b) dy/dx = 10x − 4
(c) f 0 (2) = 16
f 0 (3) = 26
3.
(a)
∆y
= 5; a constant function.
∆x
(b) No; dy/dx = 5.
Exercise 6.4
1. Left-side limit = right-side limit = 15. Yes, the limit is 15.
2. The function can be rewritten as q = (v 3 + 6v 2 + 12)/v = v 2 + 6v + 12 (v 6= 0). Thus
(a) lim q = 12
v→0
3.
(a) 5
(b) lim q = 28
v→2
(c) lim q = a2 + 6a + 12
v→a
(b) 5
4. If we choose a very small neighborhood of the point L + a2 , we cannot find a neighborhood of
N such that for every value of v in the N-neighborhood, q will be in the (L + a2 )-neighborhood.
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Chiang/Wainwright: Fundamental Methods of Mathematical Economics
Instructor’s Manual
Exercise 6.5
1.
(a) Adding −3x − 2 to both sides, we get −3 < 4x. Multiplying both sides of the latter by
1/4, we get the solution −3/4 < x.
(b) The solution is x < −9.
(c) The solution is x < 1/2
(d) The solution is −3/2 < x.
2. The continued inequality is 8x − 3 < 0 < 8x. Adding −8x to all sides, and then multiplying
by −1/8 (thereby reversing the sense of inequality), we get the solution 0 < x < 3/8.
(a) By (6.9), we can write −6 < x + 1 < 6. Subtracting 1 from all sides, we get −7 < x < 5
as the solution.
(b) The solution is 2/3 < x < 2.
(c) The solution is −4 ≤ x ≤ 1.
Exercise 6.6
1.
(a) lim q = 7 − 0 + 0 = 7
v→0
(b) lim q = 7 − 27 + 9 = −11
v→3
(c) lim q = 7 + 9 + 1 = 17
v→3
2.
(a) lim q = lim (v + 2) · lim (v − 4) = 1(−4) = −4
v→−1
v→−1
v→−1
(b) lim q = 2(−3) = −6
v→0
(c) lim q = 7(2) = 14
v→5
3.
(a) lim = lim (3v + 5)/ lim (v + 2) = 5/2 = 2 12
v→0
v→0
v→0
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Chiang/Wainwright: Fundamental Methods of Mathematical Economics
Instructor’s Manual
(b) lim q = (15 + 5)/(5 + 2) = 20/7 = 2 67
v→5
(c) lim q = (−3 + 5)/(−1 + 2) = 2/1 = 2
v→−1
Exercise 6.7
1. For example,
2. (a) lim q = N 2 − 5N − 2 = g(N )
v→N
(b) Yes.
(c) Yes.
3. (a) lim q = (N + 2)/(N 2 + 2) = g(N )
v→N
(b) Yes.
4.
(a) No.
(c) The function is continuous in the domain
(b) No, because f (x) is not defined at x = 4;
i.e., x = 4 is not in the domain of the function.
(c) for x 6= 4, the function reduces to y = x − 5, so lim y = −1.
x→4
5. No, because q = v + 1, as such, is defined at every value of v, whereas the given rational
function is not defined at v = 2 and v = −2. The only permissible way to rewrite is to qualify
the equation q = v + 1 by the restrictions v 6= 2 and v 6= −2.
6. Yes; each function is not only continuous but also smooth.
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