1 QUICK MENTAL CALCULATION 1.1 APPROXIMATIONS
1 QUICK MENTAL CALCULATION
There are many times you may benet from using mental calculations and approximations
when no calculator is handy. Another approa
Lausanne 201 3
Haute Ecole Spcialise University of Applied Sciences
de Suisse occidentale Western Switzerland 1 QUICK MENTAL CALCULATION
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|.
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.
C01S0M.042: The graph has vertical asymptotes at x = 2 and is shown next.
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) =
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
that it and L have slope 3 . Hence an equation of L is
y (3) =
A(V ) = 4
A = 4
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
C01S0M.012: If 70
32 + 9 C
90, then 70
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
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
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
C01S04.032: Recommended window: 5
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
exactly one solution (approximately 1.42773).
5. The graph m
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
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,
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
c = 5
C01S03.037: As c increases the graph becomes wider and taller; its shape does not seem to change very
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.
c = 5
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) =
x2 + 1
4 x2 =
x2 + 1
(f g )(x) =
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
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
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.
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
x(t) = 60 60t
if 0 t 0.5,
if 0.5 < t 1,
if 1 < t 3.
The graph of x(t) is shown next.
C01S02.077: Initially we work in units of pages and cents (to avoid decimals and fractions). The graph
of C , as a func
t = 1, as the automobile cannot suddenly jump from one position to a completely dierent position in an
instant. Hence 45 = 75 + C , so that C = 30. Therefore
if 0 t
if 1 < t
To see the graph of x(t), plot in Mathematica
x[t ] := If
C01S02.066: Recall that the area of the rectangle is given by y = A(x) = x(50 x). To maximize A(x) we
nd the vertex of the parabola: y = 50x x2 = (x2 50x) = (x2 50x + 625 625) = (x 25)2 + 625.
Because the vertex of the parabola is at (25, 625) and x = 25
f [ x ] := Floor [ 2x ];
Plot [ f [ x ], cfw_x, 3.5, 3.5 , AspectRatio > Automatic, PlotRange > cfw_ cfw_ 3.5, 3.5 , cfw_4.5, 4.5 ];
Mathematica will draw vertical lines connecting points that it shouldnt, making the graph look like treads
and risers of
C01S02.048: Given: f (x) = | x 3 |. If x 0 then f (x) = x 3, so the graph of f consists of the straight
line through (3, 0) with slope 1 for x 3. If x < 0 then f (x) = x + 3, so the graph of f consists of that
part of the straight line with slope 3 and y
C01S02.039: Note that f (x) > 0 for all x other than x = 1, where f is not dened. If | x | is large, then
f (x) is near zero, so the graph of f almost coincides with the x-axis for such x. If x is very close to 1, then
f (x) is the reciprocal of a very sm
C01S02.035: To graph f (x) = x2 9, note that there is no graph for 3 < < 3, that f (3) = 0, and
that f (x) > 0 for x < 3 and for x > 3. If x is large positive, then x2 9 x2 = x, so the graph of f
has x-intercept (3, 0) and rises as x increases, nearly