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graph view rectangle function Figure graph calculator sin
GRAPHING CALCULATORS AND COMPUTERS
In this section we assume that you have access to a graphing calculator or a computer with
graphing software. We will see that the use of such a device enables us to graph more com-
plicated functions and to solve more complex problems than would otherwise be possible.
We also point out some of the pitfalls that can occur with these machines.
Graphing calculators and computers can give very accurate graphs of functions. But we
will see in Chapter 4 that only through the use of calculus can we be sure that we have
uncovered all the interesting aspects of a graph.
A graphing calculator or computer displays a rectangular portion of the graph of a func-
tion in a display window or viewing screen, which we refer to as a viewing rectangle.
The default screen often gives an incomplete or misleading picture, so it is important to
choose the viewing rectangle with care. If we choose the -values to range from a mini-
mum value of 
to a maximum value of 
and the -values to range from
a minimum of 
to a maximum of 
, then the visible portion of the graph
lies in the rectangle
shown in Figure 1. We refer to this rectangle as the
by 
viewing rectangle.
The machine draws the graph of a function much as you would. It plots points of the
form 
for a certain number of equally spaced values of between and . If an 
-value is not in the domain of , or if 
lies outside the viewing rectangle, it moves on
to the next -value. The machine connects each point to the preceding plotted point to form
a representation of the graph of .
EXAMPLE 1 Draw the graph of the function 
in each of the following view-
ing rectangles.
(a)
by 
(b)
by 
(c)
by 
(d)
by 
SOLUTION For part (a) we select the range by setting min
, max
, min
and max
. The resulting graph is shown in Figure 2(a). The display window is
blank! A moment’s thought provides the explanation: Notice that 
for all , so
for all . Thus, the range of the function 
is 
. This means
that the graph of 
lies entirely outside the viewing rectangle 
by 
.
The graphs for the viewing rectangles in parts (b), (c), and (d) are also shown in 
Figure 2. Observe that we get a more complete picture in parts (c) and (d), but in part (d)
it is not clear that the -intercept is 3.
We see from Example 1 that the choice of a viewing rectangle can make a big differ-
ence in the appearance of a graph. Often it’s necessary to change to a larger viewing rect-
angle to obtain a more complete picture, a more global view, of the graph. In the next
y
2, 2
2, 2
f
3, 
f x  x2  3
x
x 2  3  3
x
x 2  0
 2
Y
 2,
Y
 2
X
 2
X
100, 1000
50, 50
5, 30
10, 10
4, 4
4, 4
2, 2
2, 2
f x  x 2  3
f
x
f x
f
x
b
a
x
x, f x
f
c, d
a, b
a, b  c, d   x, y  a  x  b, c  y  d 
Ymax  d
Ymin  c
y
Xmax  b
Xmin  a
x
1
Thomson Brooks-Cole copyright 2007FIGURE 2 Graphs of ƒ=≈+3
(b) _4, 4 by _4, 4
(a) _2, 2 by _2, 2
2
_2
_2
2
4
_4
_4
4
(c) _10, 10 by _5, 30
30
_5
_10
10
(d) _50, 50 by _100, 1000
1000
_100
_50
50
FIGURE 1
The viewing rectangle a, b by c, d
y=d
x=a
x=b
y=c
(a, d )
(b, d )
(a, c
)
(b, c
)
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example we see that knowledge of the domain and range of a function sometimes provides
us with enough information to select a good viewing rectangle.
EXAMPLE 2 Determine an appropriate viewing rectangle for the function
and use it to graph .
SOLUTION The expression for 
is defined when
Therefore, the domain of 
is the interval 
. Also,
so the range of 
is the interval 
.
We choose the viewing rectangle so that the -interval is somewhat larger than the
domain and the -interval is larger than the range. Taking the viewing rectangle to be
by 
, we get the graph shown in Figure 3.
EXAMPLE 3 Graph the function 
.
SOLUTION Here the domain is 
, the set of all real numbers. That doesn’t help us choose a
viewing rectangle. Let’s experiment. If we start with the viewing rectangle 
by
, we get the graph in Figure 4. It appears blank, but actually the graph is so
nearly vertical that it blends in with the -axis.
If we change the viewing rectangle to 
by 
, we get the picture
shown in Figure 5(a). The graph appears to consist of vertical lines, but we know that
can’t be correct. If we look carefully while the graph is being drawn, we see that the
graph leaves the screen and reappears during the graphing process. This indicates that 
we need to see more in the vertical direction, so we change the viewing rectangle to
by 
. The resulting graph is shown in Figure 5(b). It still doesn’t
quite reveal all the main features of the function, so we try 
by 
in Figure 5(c). Now we are more confident that we have arrived at an appropriate view-
ing rectangle. In Chapter 4 we will be able to see that the graph shown in Figure 5(c)
does indeed reveal all the main features of the function.
EXAMPLE 4 Graph the function 
in an appropriate viewing rectangle.
SOLUTION Figure 6(a) shows the graph of produced by a graphing calculator using the
viewing rectangle 
by 
. At first glance the graph appears to be rea-
sonable. But if we change the viewing rectangle to the ones shown in the following parts
of Figure 6, the graphs look very different. Something strange is happening.
1.5, 1.5
12, 12
f
f x  sin 50x
V
FIGURE 5 y=˛-150x
(a)
(c)
(b)
1000
_1000
_20
20
500
_500
_20
20
20
_20
_20
20
1000, 1000
20, 20
500, 500
20, 20
20, 20
20, 20
y
5, 5
5, 5

y  x 3  150x
1, 4
3, 3
y
x
[0, 2s2]
f
0  s8  2x 2  s8  2s2  2.83
2, 2
f
 &?
  x   2
 &?  2  x  2
 8  2x 2  0 &? 2x 2  8 &? x 2  4
f x
f
f x  s8  2x 2
2 ■ GRAPHING CALCULATORS AND COMPUTERS
Thomson Brooks-Cole copyright 2007FIGURE 3
4
_1
_3
3
5
_5
_5
5
FIGURE 4
Play the Video
V
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In order to explain the big differences in appearance of these graphs and to find an
appropriate viewing rectangle, we need to find the period of the function 
We know that the function 
has period 
and the graph of 
is com-
pressed horizontally by a factor of 50, so the period of 
is
This suggests that we should deal only with small values of 
in order to show just 
a few oscillations of the graph. If we choose the viewing rectangle
by
, we get the graph shown in Figure 7.
Now we see what went wrong in Figure 6. The oscillations of
are so rapid
that when the calculator plots points and joins them, it misses most of the maximum and
minimum points and therefore gives a very misleading impression of the graph.
We have seen that the use of an inappropriate viewing rectangle can give a misleading
impression of the graph of a function. In Examples 1 and 3 we solved the problem by
changing to a larger viewing rectangle. In Example 4 we had to make the viewing rect-
angle smaller. In the next example we look at a function for which there is no single view-
ing rectangle that reveals the true shape of the graph.
EXAMPLE 5 Graph the function 
.
SOLUTION Figure 8 shows the graph of produced by a graphing calculator with viewing
rectangle 
by 
. It looks much like the graph of 
, but per-
haps with some bumps attached. 
If we zoom in to the viewing rectangle 
by 
, we can see much
more clearly the shape of these bumps in Figure 9. The reason for this behavior is that the
second term,
, is very small in comparison with the first term,
. Thus we
really need two graphs to see the true nature of this function.
sin x
1
100 cos 100x
0.1, 0.1
0.1, 0.1
FIGURE 9
0.1
_0.1
_0.1
0.1
FIGURE 8
1.5
_1.5
_6.5
6.5
y  sin x
1.5, 1.5
6.5, 6.5
f
f x  sin x  1
100 cos 100x
V
y  sin 50x
1.5, 1.5
0.25, 0.25
x
2
50


25
 0.126
y  sin 50x
y  sin 50x
2
y  sin x
y  sin 50x.
(a)
(b)
(c)
(d)
FIGURE 6
Graphs of ƒ=sin 50x
in four viewing rectangles
1.5
_1.5
_10
10
1.5
_1.5
_12
12
1.5
_1.5
_9
9
1.5
_1.5
_6
6
GRAPHING CALCULATORS AND COMPUTERS ■ 3
Thomson Brooks-Cole copyright 2007FIGURE 7
ƒ=sin 50x
1.5
_1.5
_.25
.25
Play the Video
V
■ ■ The appearance of the graphs in Figure 6
depends on the machine used. The graphs you
get with your own graphing device might not
look like these figures, but they will also be
quite inaccurate.
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Thomson Brooks-Cole copyright 20074 ■ GRAPHING CALCULATORS AND COMPUTERS
EXAMPLE 6 Draw the graph of the function 
.
SOLUTION Figure 10(a) shows the graph produced by a graphing calculator with viewing
rectangle 
by 
. In connecting successive points on the graph, the calculator
produced a steep line segment from the top to the bottom of the screen. That line seg-
ment is not truly part of the graph. Notice that the domain of the function 
is 
. We can eliminate the extraneous near-vertical line by experimenting with 
a change of scale. When we change to the smaller viewing rectangle 
by
on this particular calculator, we obtain the much better graph in Figure 10(b).
EXAMPLE 7 Graph the function 
.
SOLUTION Some graphing devices display the graph shown in Figure 11, whereas others
produce a graph like that in Figure 12. We know from Section 1.2 (Figure 8) that the
graph in Figure 12 is correct, so what happened in Figure 11? The explanation is that
some machines compute the cube root of using a logarithm, which is not defined if 
is
negative, so only the right half of the graph is produced.
You should experiment with your own machine to see which of these two graphs is
produced. If you get the graph in Figure 11, you can obtain the correct picture by graph-
ing the function 
Notice that this function is equal to 
(except when 
).
To understand how the expression for a function relates to its graph, it’s helpful to graph
a family of functions, that is, a collection of functions whose equations are related. In the
next example we graph members of a family of cubic polynomials.
EXAMPLE 8 Graph the function 
for various values of the number . How
does the graph change when is changed?
SOLUTION Figure 13 shows the graphs of 
for 
,
,
,
, and 
. We
see that, for positive values of , the graph increases from left to right with no maximum
or minimum points (peaks or valleys). When 
, the curve is flat at the origin. When
is negative, the curve has a maximum point and a minimum point. As decreases, the
maximum point becomes higher and the minimum point lower.
c
c
c  0
c
2
1
0
1
c  2
y  x 3  cx
c
c
y  x 3  cx
V
x  0
s
3 x
f x 
x
 x    x 
1	3
FIGURE 11
2
_2
_3
3
FIGURE 12
2
_2
_3
3
x
x
y  s
3 x
(a)
(b)
9
_9
_9
9
4.7
_4.7
_4.7
4.7
FIGURE 10
4.7, 4.7
4.7, 4.7
x  x  1
y  1	1  x
9, 9
9, 9
y 
1
1  x
Play the Video
V
■ ■ Another way to avoid the extraneous line
is to change the graphing mode on the calcu-
lator so that the dots are not connected.
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GRAPHING CALCULATORS AND COMPUTERS ■ 5
EXAMPLE 9 Find the solution of the equation 
correct to two decimal places.
SOLUTION The solutions of the equation 
are the -coordinates of the points of
intersection of the curves 
and 
. From Figure 14(a) we see that there is
only one solution and it lies between 0 and 1. Zooming in to the viewing rectangle 
by 
, we see from Figure 14(b) that the root lies between 0.7 and 0.8. So we zoom in
further to the viewing rectangle 
by 
in Figure 14(c). By moving the
cursor to the intersection point of the two curves, or by inspection and the fact that the 
-scale is 0.01, we see that the solution of the equation is about 0.74. (Many calculators
have a built-in intersection feature.)
0.7, 0.8 by 0.7, 0.8
x-scale=0.01
(c)
0, 1 by 0, 1
x-scale=0.1
(b)
_5, 5 by _1.5, 1.5
x-scale=1
(a)
0.8
0.7
0.8
y=x
1
0
1
y=x
1.5
_1.5
_5
5
y=x
y=cos x
FIGURE 14
Locating the roots
of cos x=x
y=cos x
y=cos x
x
0.7, 0.8
0.7, 0.8
0, 1
0, 1
y  x
y  cos x
x
cos x  x
cos x  x
FIGURE 13
Several members of the family of
functions y=˛+cx, all graphed
in the viewing rectangle _2, 2
by _2.5, 2.5
(a) y=˛+2x
(b) y=˛+x
(c) y=˛
(d) y=˛-x
(e) y=˛-2x
Thomson Brooks-Cole copyright 2007EXERCISES
1. Use a graphing calculator or computer to determine which of
the given viewing rectangles produces the most appropriate
graph of the function 
.
(a)
by 
(b)
by 
(c)
by 
2. Use a graphing calculator or computer to determine which of
the given viewing rectangles produces the most appropriate
graph of the function 
.
(a)
by 
(b)
by 
(c)
by 
(d)
by 
3–14 Determine an appropriate viewing rectangle for the given
function and use it to draw the graph.
3.
4.
5.
6.
7.
8. f x 
x
x 2  100
f x  x 2 
100
x
f x  s0.1x  20
f x  s
4 81  x 4
f x  x 3  30x 2  200x
f x  5  20x  x 2
50, 50
5, 5
50, 50
50, 50
10, 10
10, 10
3, 3
3, 3
f x  x 4  16x 2  20
0, 10
0, 10
0, 2
0, 10
5, 5
5, 5
f x  sx 3  5x 2
9.
10.
11.
12.
13.
14.
■
■
■
■
■
■
■
■
■
■
■
■
■
15. Graph the ellipse 
by graphing the functions
whose graphs are the upper and lower halves of the ellipse.
16. Graph the hyperbola 
by graphing the functions
whose graphs are the upper and lower branches of the
hyperbola.
17–19
Find all solutions of the equation correct to two decimal
places.
17.
18.
19.
■
■
■
■
■
■
■
■
■
■
■
■
■
20. We saw in Example 9 that the equation 
has exactly
one solution.
(a) Use a graph to show that the equation 
has
three solutions and find their values correct to two decimal
places.
cos x  0.3x
cos x  x
x 2  sin x
x 3  4x  1
x 3  9x 2  4  0
y 2  9x 2  1
4x 2  2y 2  1
y  x 2  0.02 sin 50x
y  10 sin x  sin 100x
f x  sec20x
f x  sin sx
f x  cos0.001x
f x  sin21000x
Click here for answers.
A
Click here for solutions.
S
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Thomson Brooks-Cole copyright 20076 ■ GRAPHING CALCULATORS AND COMPUTERS
(b) Find an approximate value of 
such that the equation
has exactly two solutions.
21. Use graphs to determine which of the functions 
and 
is eventually larger (that is, larger when 
is
very large).
22. Use graphs to determine which of the functions
and 
is eventually larger.
23. For what values of 
is it true that 
?
24. Graph the polynomials 
and
on the same screen, first using the viewing rect-
angle 
by [
] and then changing to 
by
. What do you observe from these
graphs?
25. In this exercise we consider the family of root functions
, where 
is a positive integer.
(a) Graph the functions 
,
, and 
on 
the same screen using the viewing rectangle 
by
.
(b) Graph the functions 
,
, and 
on the
same screen using the viewing rectangle 
by 
.
(See Example 7.)
(c) Graph the functions 
,
,
, and 
on the same screen using the viewing rectangle 
by
.
(d) What conclusions can you make from these graphs?
26. In this exercise we consider the family of functions
, where 
is a positive integer.
(a) Graph the functions 
and 
on the same
screen using the viewing rectangle 
by 
.
(b) Graph the functions 
and 
on the same
screen using the same viewing rectangle as in part (a).
(c) Graph all of the functions in parts (a) and (b) on the same
screen using the viewing rectangle 
by 
.
(d) What conclusions can you make from these graphs?
27. Graph the function 
for several values 
of . How does the graph change when changes?
28. Graph the function 
for various values of .
Describe how changing the value of affects the graph.
29. Graph the function 
,
, for 
,
and 6. How does the graph change as 
increases?
n
n  1, 2, 3, 4, 5
x  0
y  x n2x
c
c
s1  cx 2
f x 
c
c
f x  x 4  cx 2  x
1, 3
1, 3
y  1	x 4
y  1	x 2
3, 3
3, 3
y  1	x 3
y  1	x
n
f x  1	x n
1, 2
1, 3
y  s
5 x
y  s
4 x
y  s
3 x
y  sx
2, 2
3, 3
y  s
5 x
y  s
3 x
y  x
1, 3
1, 4
y  s
6 x
y  s
4 x
y  sx
n
f x  s
n x
10,000, 10,000
10, 10
2, 2
2, 2
Qx  3x 5
Px  3x 5  5x 3  2x
 sin x  x   0.1
x
tx  x 3
f x  x 4  100x 3
x
tx  x 3	10
f x  10x 2
cos x  mx
m
30. The curves with equations
are called bullet-nose curves. Graph some of these curves to
see why. What happens as 
increases?
31. What happens to the graph of the equation 
as 
varies?
32. This exercise explores the effect of the inner function on a
composite function 
.
(a) Graph the function 
using the viewing rect-
angle 
by 
. How does this graph differ
from the graph of the sine function?
(b) Graph the function 
using the viewing rectangle
by 
. How does this graph differ from the
graph of the sine function?
33. The figure shows the graphs of 
and 
as
displayed by a TI-83 graphing calculator.
The first graph is inaccurate. Explain why the two graphs
appear identical. [Hint: The TI-83’s graphing window is 95
pixels wide. What specific points does the calculator plot?]
34. The first graph in the figure is that of 
as displayed
by a TI-83 graphing calculator. It is inaccurate and so, to help
explain its appearance, we replot the curve in dot mode in the
second graph.
What two sine curves does the calculator appear to be plotting?
Show that each point on the graph of 
that the 
TI-83 chooses to plot is in fact on one of these two curves.
(The TI-83’s graphing window is 95 pixels wide.)
y  sin 45x
0
2π
0
2π
y  sin 45x
y=sin 96x
0
2π
y=sin 2x
0
2π
y  sin 2x
y  sin 96x
1.5, 1.5
5, 5
y  sinx 2 
1.5, 1.5
0, 400
y  sin(sx )
y  f tx
t
c
y 2  cx 3  x 2
c
y 
 x 
sc  x 2
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GRAPHING CALCULATORS AND COMPUTERS ■ 7
ANSWERS
1. (c)
3.
5.
7.
9.
11.
13.
2
_2
_
π
25
π
25
11
_11
2π
_2π
1.5
_1.5
100
0
1.5
0
0.01
_0.01
250
50
20
20
4
1
4
4
150
_50
30
_10
15.
17. 9.05
19. 0, 0.88
21.
23.
25. (a)
(b)
(c)
(d) Graphs of even roots are similar to 
, graphs of odd roots are
similar to 
. As n increases, the graph of 
becomes steeper
near 0 and flatter for 
.
27.
If 
, the graph has three humps: two minimum points and
a maximum point. These humps get flatter as c increases until at
two of the humps disappear and there is only one mini-
mum point. This single hump then moves to the right and
approaches the origin as c increases.
29. The hump gets larger and moves to the right.
31. If 
, the loop is to the right of the origin; if 
, the loop
is to the left. The closer c is to 0, the larger the loop.
c 	 0
c  0
c  1.5
c  1.5
2
_4
_2.5
2.5
_1.5
-1 -2 -3
1
x 	 1
y  s
n x
s
3 x
sx
3
_1
_1
4
œ̂„x
$œ„x
œ„x
2
_2
_3
3
x
%œ„x
Œ„x
3
_1
_1
4
œ̂„x
$œ„x
œ„x
0.85  x  0.85
t
1
_1
_1
1
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8 ■ GRAPHING CALCULATORS AND COMPUTERS
Thomson Brooks-Cole copyright 2007SOLUTIONS
1. f(x) =
√
x3 − 5x2
(a) [−5, 5] by [−5, 5]
(There is no graph shown.)
(b) [0, 10] by [0, 2]
(c) [0, 10] by [0, 10]
The most appropriate graph is produced in viewing rectangle (c).
3. Since the graph of f(x) = 5+ 20x− x2 is a parabola opening downward, an appropriate viewing rectangle should include the
maximum point.
5. f(x) = 4
√
81− x4 is defined when
81− x4 ≥ 0 ⇔ x4 ≤ 81 ⇔
|x| ≤ 3, so the domain of f is [−3, 3]. Also 0 ≤ 4√81− x4 ≤ 4√81 = 3, so the range is
[0, 3].
7. The graph of f(x) = x2 + (100/x) has a vertical asymptote of x = 0. As you zoom out, the graph of f looks more and more
like that of y = x2.
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GRAPHING CALCULATORS AND COMPUTERS ■ 9
Thomson Brooks-Cole copyright 20079. The period of g(x) = sin(1000x) is 2π
1000 ≈ 0.0063 and its range is
[−1, 1]. Since f(x) = sin2(1000x) is the square of g, its range is
[0, 1] and a viewing rectangle of [−0.01, 0.01] by [0, 1.5] seems
appropriate.
11. The domain of y =
√
x is x ≥ 0, so the domain of f(x) = sin√x is [0,∞)
and the range is [−1, 1]. With a little trial-and-error experimentation, we
find that an Xmax of 100 illustrates the general shape of f , so an appropriate
viewing rectangle is [0, 100] by [−1.5, 1.5].
13. The first term, 10 sinx, has period 2π and range [−10, 10]. It will be the
dominant term in any “large” graph of y = 10 sinx+ sin 100x, as shown in
the first figure. The second term, sin 100x, has period 2π
100
= π
50
and range
[−1, 1]. It causes the bumps in the first figure and will be the dominant term
in any “small” graph, as shown in the view near the origin in the second
figure.
15. We must solve the given equation for y to obtain equations for the upper and
lower halves of the ellipse.
4x2 + 2y2 = 1 ⇔ 2y2 = 1− 4x2 ⇔ y2 = 1− 4x
2
2
⇔
y = ±
u
1− 4x2
2
17. From the graph of f(x) = x3 − 9x2 − 4, we see that there is one solution
of the equation f(x) = 0 and it is slightly larger than 9. By zooming in or
using a root or zero feature, we obtain x ≈ 9.05.
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10 ■ GRAPHING CALCULATORS AND COMPUTERS
Thomson Brooks-Cole copyright 200719. We see that the graphs of f(x) = x2 and g(x) = sinx intersect twice. One
solution is x = 0. The other solution of f = g is the x-coordinate of the
point of intersection in the first quadrant. Using an intersect feature or
zooming in, we find this value to be approximately 0.88. Alternatively, we
could find that value by finding the positive zero of h(x) = x2 − sinx.
Note: After producing the graph on a TI-83 Plus, we can find the approximate value 0.88 by using the following keystrokes:
. The “1” is just a guess for 0.88.
21. g(x) = x3/10 is larger than f(x) = 10x2 whenever x > 100.
23.
We see from the graphs of y = |sinx− x| and y = 0.1 that there are
two solutions to the equation |sinx− x| = 0.1: x ≈ −0.85 and
x ≈ 0.85. The condition |sinx− x| < 0.1 holds for any x lying
between these two values.
25. (a) The root functions y =
√
x,
y = 4
√
x and y = 6
√
x
(b) The root functions y = x,
y = 3
√
x and y = 5
√
x
(c) The root functions y =
√
x,
y = 3
√
x, y = 4
√
x and y = 5
√
x
(d) • For any n, the nth root of 0 is 0 and the nth root of 1 is 1; that is, all nth root functions pass through the points (0, 0)
and (1, 1).
• For odd n, the domain of the nth root function is R, while for even n, it is {x ∈ R | x ≥ 0}.
• Graphs of even root functions look similar to that of√x, while those of odd root functions resemble that of 3√x.
• As n increases, the graph of n√x becomes steeper near 0 and flatter for x > 1.
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GRAPHING CALCULATORS AND COMPUTERS ■ 11
Thomson Brooks-Cole copyright 200727. f(x) = x4 + cx2 + x. If c < −1.5, there are three humps: two
minimum points and a maximum point. These humps get flatter as c
increases, until at c = −1.5 two of the humps disappear and there is
only one minimum point. This single hump then moves to the right
and approaches the origin as c increases.
29. y = xn2−x. As n increases, the
maximum of the function moves further
from the origin, and gets larger. Note,
however, that regardless of n, the
function approaches 0 as x→∞.
31. y2 = cx3 + x2
If c < 0, the loop is to the right of the origin, and if c is positive, it is to the
left. In both cases, the closer c is to 0, the larger the loop is. (In the limiting
case, c = 0, the loop is “infinite,” that is, it doesn’t close.) Also, the larger
|c| is, the steeper the slope is on the loopless side of the origin.
33. The graphing window is 95 pixels wide and we want to start with x = 0 and end with x = 2π. Since there are 94 “gaps”
between pixels, the distance between pixels is 2π−0
94
. Thus, the x-values that the calculator actually plots are x = 0 + 2π
94
· n,
where n = 0, 1, 2, .. . , 93, 94. For y = sin 2x, the actual points plotted by the calculator are

2π
94 · n, sin

2 · 2π
94 · n

for
n = 0, 1, ... , 94. For y = sin 96x, the points plotted are

2π
94
· n, sin96 · 2π
94
· n for n = 0, 1, ... , 94. But
sin

96 · 2π
94
· n = sin94 · 2π
94
· n+ 2 · 2π
94
· n = sin2πn+ 2 · 2π
94
· n
= sin

2 · 2π
94
· n
[by periodicity of sine],
n = 0, 1, .. . , 94
So the y-values, and hence the points, plotted for y = sin 96x are identical to those plotted for y = sin 2x.
Note: Try graphing y = sin 94x. Can you see why all the y-values are zero?
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