9/)
ECOLE HOTELIERE
LAUSANN E
Immv
0 NUMBERS AND OPERATIONS 0.1 WHOLE NUMBERS
0 NUMBERS AND OPERATIONS
The world Arithmetic comes from the Greek world arithmtik and means the science of
numbers. Arithmetic, as others sciences, started to develop in Gree
f (g (x) = f (cos x) = cos x and
g (f (x) = g x = cos x .
C01S04.017: If f (x) = sin x and g (x) = x3 , then
f (g (x) = f x3 = sin x3 = sin x3
and
3
g (f (x) = g (sin x) = (sin x) = sin3 x.
We note in passing that sin x3 and sin3 x dont mean the same thi
C01S04.010: The graph of g (x) = sin 10x resembles that of the sine function, but with much more
activity because of the factor 10. Multiply by the rapidly decreasing positive numbers 2x and you will
see the sine oscillations decreasing from the range [1,
Section 1.4
C01S04.001: Because g (x) = 2x increasesrst slowly, then rapidlyon the set of all real numbers, with
values in the range (0, + ), the given function f (x) = 2x 1 must increase in the same way, but with
values in the range (1, + ). Therefore it
gure.
1
0.8
0.6
c = 5
0.4
0.2
c=5
3
2
1
1
2
3
C01S03.037: As c increases the graph becomes wider and taller; its shape does not seem to change very
much.
C01S03.038: The length of the airfoil is approximately 1.0089 and its width is approximately 0.200
C01S03.032: The graph starts with two bends when c = 5. As c increases the bends become narrower
and narrower and disappear when c = 0. Then the graph gets steeper and steeper. See the following gure.
15
c=5
c = 5
10
5
4
2
2
4
5
10
C01S03.033: The gra
C01S03.017: The domain of f (x) = x x + 2 is the interval [2, + ), so its graph must be the one
shown in Fig. 1.3.38.
C01S03.018: The domain of f (x) = 2x x2 consists of those numbers for which 2x x2 0; that is,
x(2 x) 0. This occurs when x and 2 x have
(f + g )(x) =
f
(x) =
g
x2
1
+1+
,
4 x2
x2 + 1
4 x2 =
x2 + 1
(f g )(x) =
,
4 x2
and
4 + 3x2 x4 .
C01S03.006: The domain of f is the set of all real numbers other than 2 and the domain of g is the set
of all real numbers other than 2. Hence the domain of
Section 1.3
C01S03.001: The domain of f is R, the set of all real numbers; so is the domain of g , but g (x) = 0 when
x = 1 and when x = 3. So the domain of f + g and f g is the set R and the domain of f /g is the set of
all real numbers other than 1 and
C01S04.032: Recommended window: 5
x
5. The graph makes it evident that the equation has
exactly three solutions (approximately 3.63796, 1.86236, and 0.88947).
C01S04.033: Recommended window: 5
x
exactly one solution (approximately 1.42773).
5. The graph m
C01S04.048: We began with the viewing window 2 x 2, which showed the two smaller solutions but
not the larger solution. We rst narrowed this window to 0.9054 x 0.9052 to get the rst solution,
x 0.9053. We returned to the original window and narrowed it to
Chapter 1 Miscellaneous Problems
C01S0M.001: The domain of f (x) =
interval [4, + ).
x 4 is the set of real numbers x for which x 4
0; that is, the
C01S0M.002: The domain of f consists of those real numbers x for which 2 x = 0; that is, the set of all
rea
0,)
ECOLE HOTELIERE
LAUSANNE
HJIHJ
1 QUICK MENTAL CALCULATION 1.1 APPROXIMATIONS
1 QUICK MENTAL CALCULATION
1.1 APPROXIMATIONS
There are many times you may benet from using mental calculations and approximations
when no calculator is handy. Another approa
I,
ECOLE HOTELIERE
LAUSANNE
Since 1893
APPLIED MATHEMATICS
I
EXERCISES
Mario SOFIA
Lausanne 201 3
Haute Ecole Spcialise University of Applied Sciences
de Suisse occidentale Western Switzerland 1 QUICK MENTAL CALCULATION
1
9,)
ECDLE HOTELIBRE
LAUSANN E
QU
C01S0M.045: Suppose that a, b, and c are arbitrary real numbers. Then
a + b + c = (a + b) + c
a + b + c
a + b + c .
.
C01S0M.046: Suppose that a and b are arbitrary real numbers. Then a = (a b) + b
Therefore a b a b.
a b + b.
C
Thus the vertex of this parabola is at the point (2, 4).
C01S0M.041: The graph has a vertical asymptote at x = 5 and is shown next.
20
10
8
6
4
2
10
20
C01S0M.042: The graph has vertical asymptotes at x = 2 and is shown next.
6
4
2
4
2
2
4
2
4

y 2 = 8 + 2x x2 ;
x2 2x + 1 + y 2 = 9;
(x 1)2 + (y 0)2 = 32 .
The last is the equation of a circle with center (1, 0) and radius 3. But y
half of that circle, and it is shown in Fig. 1.MP.10.
0, so the graph of f is the upper
C01S0M.031: Given: f (x) = 2x
y (5) =
1
(x 0);
2
alternatively,
2y + 10 = x.
C01S0M.022: The equation 3x 2y = 4 of the other line may be written in the form y = 3 x 2, revealing
2
that it and L have slope 3 . Hence an equation of L is
2
y (3) =
3
(x 2);
2
that is,
y=
3
x 6.
2
C01S0M.0
Answer:
V
A(V ) = 4
1 /3
V
r=
and so
A = 4
V
2/3
.
2 /3
,
0 < V < + .
It is permissible, and sometimes desirable, to use instead the domain 0
V < + .
C01S0M.017: The following gure shows an equilateral triangle with sides of length 2x and an altitude
of l
C01S0M.012: If 70
F
32 + 9 C
5
90, then 70
90. Hence
70 32
9
5C
90 32;
38
9
5C
58;
190
9C
290;
190
9
C
290
9.
Answer: The Celsius temperature ranged from a low of about 21.1 C to a high of about 32.2 C.
C01S0M.013: Because 25 < R < 50, 25I < IR < 50I , so
the vertical axis are degrees Fahrenheit. Remember that these are average daily temperatures; it is not
uncommon for a winter low in Athens to be below 28 F and for a summer high to be as much as 92 F.
80
60
40
20
50
100
150
200
12
250
300
350
average of these is 1.65 (to two places) and should be a good estimate of the true value of c. Alternatively,
you can use a computer algebra program to nd c; in Mathematica, for example, the command Fit will t
given data points to a sum of constant multip
[ x 8] = 1, $1.60 if [ x 8] = 2, and so on. But this isnt quite rightwe are using the Floor function of
Section 1.1, whereas we should really be using the Ceiling function. By the result of Problem 51 of that
section, we see that instead of cost
C (x) = 8
C01S01.048: Recall that A(x) = x(50 x), 0
at some special numbers in its domain:
x0
A0
5
225
10
400
15
525
x
20
600
50. Here is a table of a few values of the function A
25
625
30
600
35
525
40
400
45
225
50
0
It appears that when x = 25 (so the rectangle
Because x > 0 and y > 0 (the box has positive volume), but because y can be arbitrarily close to zero (as
well as x), we see also that 0 < x < + . We use the equation x2 y = 324 to eliminate y from Eq. (1) and
thereby nd that
C (x) = 3x2 +
1296
,
x
0 < x
r=
1
2
S
;
V=
Answer: V (S ) =
1
6
S
43
4
1
r =
3
3
8
0
3 /2
3 /2
S < + .
S
=
1
6
S
.
3 /2
,
C01S01.039: To avoid decimals, we note that a change of 5 C is the same as a change of 9 F, so when
the temperature is 10 C it is 32 + 18 = 50 F; when the temper
we require that x = 1 so that the fraction is dened. In addition we require that the fraction be nonnegative
so that its square root will be dened. These conditions imply that both numerator and denominator be
positive or that both be negative; moreover,
C01S01.026: Given g (t) = 3 t + 4 = (t + 4)1/3 , we note that t + 4 is dened for every real number t and
the cube root of t + 4 is dened for every possible resulting value of t + 4. Therefore the domain of g is the
set R of all real numbers.
C01S01.027:
f (x) = 0
if
0
x < 1,
3
f (x) = 1
if
1
3
x < 2,
3
f (2) = 2
if
2
3
x < 1;
moreover,
f (x) = 3
if
1
x < 2,
3
f (x) = 2
if
2
3
x < 1,
3
f (x) = 1
if
1
3
x < 0.
f (x) = 3m
if
m
x < m + 1,
3
f (x) = 3m + 1
if
m+
1
3
x < m + 2,
3
f (x) = 3m + 2
if
m+
2
3
x < m
f (a + h) f (a) = [1 2(a + h)] [1 2a] = 1 2a 2h 1 + 2a = 2h.
C01S01.013: If f (x) = x2 , then
f (a + h) f (a) = (a + h)2 a2
= a2 + 2ah + h2 a2 = 2ah + h2 = h (2a + h).
C01S01.014: If f (x) = x2 + 2x, then
f (a + h) f (a) = [(a + h)2 + 2(a + h)] [a2 + 2a]