Honors 2201-4
Homework 2 Solutions
Fall 2003
Sections 1.2, 1.3, 1.4 (6)
Age is the independent variable, and the number of words is the dependent
variable. We expect that the number of words in a person’s vocabulary will
remain near zero for a short period of time, and then increase fairly rapidly
throughout childhood. Eventually, the vocabulary size will level off and stay
fairly constant throughout adulthood. The vocabulary size may start to drop
slightly when a person is very old.
This relationship is a function, as there is one vocabulary size corresponding to the person’s age. The domain of the function is the life span of the
individual. The rule is to assign to the individual’s age the number of words
in his or her vocabulary.
This is a function of time.
Sections 1.2, 1.3, 1.4 (12)
The diameter of the pizza is the independent variable, and the cost is the
dependent variable. In general, we expect that as the diameter of the pizza
increases, the cost of the pizza increases as well.
In general, this relationship is not a function, as the cost may depend
on the number and type of toppings on the pizza. Requiring that all pizzas
have the same toppings and type of crust, and be ordered from the same
restaurant would make this a function.
This is not a function of time, and unless we add the above restrictions,
this is not a function at all. With sufficient restrictions, cost is a function of
the diameter.
Section 1.3 (33)
The weight of the letter is the independent variable, and the postage is
the dependent variable. In general, this should be an increasing function
where larger weights correspond to higher postage. However, one postage
cost is often assigned to a range of weights, so the function will most likely
look like a series of steps. The relationship is a function if we assume that
all of the letters are being sent within the same country. A letter being sent
from New York to Paris will most likely cost more than a letter from New
York to Chicago. The domain is the set of positive real numbers, possibly
with some upper bound. The rule is to assign to each weight the postage
required to mail a letter of that weight by first class mail.
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Section 1.7
(17) The function is not additive. For example, f (2+1) = f (3) = 32 +7 = 16
but, f (2) + f (1) = (22 + 7) + (12 + 7) = 4 + 7 + 1 + 7 = 19. In general
f (a + b) = (a + b)2 + 7 = a2 + 2ab + b2 + 7 =
= (a2 + 7) + 2ab + b2 6= (a2 + 7) + (b2 + 7)
(18) The function is not multiplicative. For example, f (2 · 1) = f (2) =
22 + 7 = 4 + 7 = 11 but, f (2) · f (1) = (22 + 7) · (12 + 7) = (11) · (8) = 88. In
general
f (ab) = (ab)2 + 7 = a2 b2 + 7 6= a2 b2 + 7a2 + 7b2 + 49 = (a2 + 7)(b2 + 7)
(19) The function is even.
f (−a) = (−a)2 + 7 = a2 + 7 = f (a)
(20) The function is not odd. This follows directly from number 19. Also,
note that
f (−a) = (−a)2 + 7 = a2 + 7 6= −(a2 + 7) = −f (a)
(28) Logarithms turn products into sums: logB (ab) = logB (a) + logB (b).
(29) Exponential functions turn sums into products: B a+b = B a · B b .
(30) A function is called subtractive if what it does (as a process) to the
difference of any two inputs is the difference of the corresponding outputs.
In symbols f (a − b) = f (a) − f (b) for all inputs a and b in its domain. For
example, if f (x) = cx, then f (a − b) = c(a − b) = ca − cb = f (a) − f (b).
A function is called divisive if what it does (as a process) to any quotient is
to produce the quotient of its results on the individual factors. In symbols
f (a/b) = (f (a))/(f (b)) for all inputs a and b in its domain. For example, if
f (x) = xn , then f (a/b) = (a/b)n = (an )/(bn ) = (f (a))/(f (b)).
(31) Two functions that are both additive and multiplicative are f (x) = x
and g(x) = 0. To check this notice that f (a + b) = a + b = f (a) + f (b),
f (ab) = ab = f (a)f (b), g(a + b) = 0 = 0 + 0 = g(a) + g(b), and g(ab) = 0 =
0 · 0 = g(a)g(b).
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