The Answers for All Econ1202 Lectures

Topics: Derivative, Trigraph, Antiderivative Pages: 46 (8508 words) Published: June 18, 2013
UNSW Econ 1202/2291

Quantitative Analysis For Business and Economics Examples Covered in Lectures 2011

WARNING! 1. These examples were given as part of a Lecture. To look at them outside of their original context would be reckless. Be sure to understand these examples in the context of the lecture material. 2. Since they are Lecture examples, they are subject to the constraints of lectures: they do not aim to illustrate all the techniques that you are expected to master in the course. 3. Lecture examples intend to demonstrate a particular application of a method. You will not be testing your ability to recognise which kind of question you have before you by simply reading these through (they are often labelled!). Life does not usually come in such neat packages. It is recommended that you construct your own tests from textbook and tutorial problems to enhance your problem recognition and solution skills. Happy studying!

Lecture 1
1.1 Q: A linear function Consider what is meant by the simple linear function f (x) = 1 + 0.5x. A: “ef-of-x is equal to 1 plus 0.5 times x.” Each x value put into function f will give exactly one f (x) value. If we are dealing with continuous inputs (x can take any value between 0 and 5 say), then we normally draw a line to represent the functional relationship between f (x) and x. 6 5 4

1.2 Q: A:
Dependent, Independent Identify the dependent and independent terms, and the value and argument of the function H(a, b, c) = a2 + 2b + 3. • H depends on a and b (the independent variables); • H is the value, whilst a and b are the arguments.

1.3 Q: A:
Find the domain of the function, y(x) = Factorising gives, y(x) = 2 , (x − 1)(x + 4) 2 . x2 +3x−4

f (x)

3 2 1 0

*
0

*
1

*
2

*

*

*

which implies, x = 1 or -4. Hence, the domain of y is the Real 3 4 5 6

x

numbers except 1, -4.

1

1.4 Q: A:
Combining functions Suppose f (x) = 2x2 −3x−2 and g(x) = x − 2, and let h(x) = f (x), then show that (h − g)(x) = x + 3. g We begin by finding h(x),

2.2 Q: Reverse simple interest With the same interest rate as in the previous example, how many years would it take to gain our target ($3,200)? A: We solve for t, S = = = = P + I = P + P rt (P ) + (P r)x = c + mx S−P Pr 3200 − 1500 (1500)(0.05) 22 years, 8 months ! f (x) ∴ t

2x2 − 3x − 2 (2x + 1)(x − 2) f = = 2x + 1 , h(x) = (x) = g x−2 x−2 now, (h − g)(x) = = h(x) − g(x) 2x + 1 − (x − 2) = x + 3 .

1.5 Q:
Composite functions Let p(x) = x2 − 2, and h(x) = (for x ≥ 0). Find (p ◦ h)(2). A: We have, (p ◦ h) = p(h(x)) solving by substitution, p(h) ∴ ∴ p((h(x)) p((h(2)) = = = h2 − 2 √ ( 5x + 1)2 − 2 (5)(2) − 1 = 9 √ 5x + 1

3000

S($)

2000 1000 0 0

t(years)

1

2

3

2.3 Q: Compound Interest Find the final value of a $1000 investment, invested for 5 years at the nominal rate of 8% compounded quarterly. A: First, find the periodic rate of interest: since the period is quarterly, this means, r= 0.08 = 0.02 4

1.6 Q: A:
Suppose f (x) = First, let
x +1 , 5
2

find f −1 (x). x2 + 1 , 5

y = f (x) = now, solve for x in terms of y, ∴ 5y − 1 ∴ x = =

Next, find the total number of periods, n = 5 × 4 = 20; solve, S = = = P (1 + r)n 1000(1 + 0.02)20 $1, 485.95

x2 5y − 1 5x − 1

f −1 (x) =



2.4 Q:
Option 2: reverse compound interest Returning to our problem, how long would it take to obtain $3,200 with an initial investment of $1,500, if interest is compounded every two months at a nominal rate of 5%? A: We have, S = P (1 + r)n rearranging (using our log laws!), S P ln(S/P ) ∴ n = = = (1 + r)n n ln(1 + r) ln(S/P ) ln(1 + r)

Lecture 2
2.1
= $1500, r = 5%p.a and t = 2yrs. How much interest (calculated simply) would we gain? What would be the final value of our investment? A: In the first case we have, I ∴ I = = = P rt (1500)(0.05)(2) $150

Q: Simple Interest Suppose we invest (as in our case), P



So, correcting the interest rate, we obtain, n = = = ln( 3200 ) 1500 ln(1 + 0.05 )...
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