Chapter 4, Section 1, Example 11 presents a graph of the functions
f(x) = ( 1 + 1/x)x g(x) = e1
along with a paragraph related to that graph.
If we restrict ourselves to the f(x) function, then the graph given in the book is
Figure 1
It is interesting, and important, to
note that the graph was produced on a TI-85 using the ZSTD
zoom settings. However,
if we attempt the same graph on a TI-86, we get the graph
Figure 2
Something is wrong. We can not have two different graphs for the same function.
The problem gets worse if we change the WINDOW settings on our TI-86. For example,
we could shift to ZDECM and we will get the graph
Figure 3
This graph is more similar to the TI-85 version, however there are some extra
points plotted in the interval (– 1,0). We can change the
yMin and yMax values to be – 2 and 6
respectively, and then produce
a new version of the graph. (The image below has been altered to circle the
extraneous dots in red.)
Figure 4
And, if we choose our own values for xMin and xMax, as in
Figure 5
We can get a graph such as
Figure 6
With respect to the TI-86, if we set the WINDOW values as
Figure 7
we will get a graph that does look just like the one in the text, namely,
Figure 8
All of these are all graphs of the same function f(x) = ( 1 + 1/x)x.
all that we have done is to shift to the TI-86 and then we have changed the domain
by altering the WINDOW settings.
Why are we getting such different graphs?
We need to divide the problem into five pieces. First, when x>0, there is
no difference in the graphs. When x>0, the function
f(x) = ( 1 + 1/x)x
resolves to being a base that is greater than 1 raised to a positive power.
This is always defined for real numbers. [It is interesting to consider that as
the value of x gets close to 0, the value of 1/x becomes larger, which means that
the value of the base, (1+1/x) is large. However, at the same time,
the exponent is close to 0 so we are taking higher roots of the base.
For example, when x=0.01, 1/x=100, and the base is 101,
however we need to raise this to the 0.01 power, which gives the
approximate functional value 1.0472.]
Second, when x=0 the function is undefined. According to the book, in the
second sentence of the bottom paragraph of page 266
Further, since f(0) is not defined we graph an open point at (0,0) indicating
that the point (0,0) is not part of the graph.
There are two problems with this statement. The first is that we can not see
an open point at (0,0) on the graph. The axes go through (0,0) and, therefore,
the point (0,0) is turned opaque. The second problem is a bit more subtle. It
is true that the function is not defined when x=0, but that means that we
do not graph any point where x=0. It is not the case that we have an open
point at (0,0), we have an entire open column. Again, we can not see the "openness" of
that column because it is covered by the y-axis. However, if we turn off the axes
and if we regenerate the function graph on a TI-86 with the ZSTD settings, we
get the graph
Figure 9
which has been modified to include the red arrow. That arrow points to the open
spot in the graph caused by the function not being defined when x=0.
However, if we compare this graph with the graph in Figure 2 (identical except that
the earlier graph included the axes), we note that the spot indicated by the arrow
is really (0,1), not (0,0). As noted above, when x>0 and as
x gets closer and closer to 0, the value of the function gets closer to the
value 1. Nonetheless, we should say that when x=0 we have an open column
in the graph.
Third, we need to consider the values of x in the
open interval (– 1,0).
For values in that interval, the expression 1/x is a negative number
less than – 1. This means that for values in the
interval (– 1,0), the base (1+1/x) is a
negative number. We are raising that base to the x power, which again
is a negative number between – 1 and 0.
Therefore, we will be taking a root of a negative number. Although
the cube root of a negative value is defined, the square root of a negative
value is not defined for real numbers. Thus, the function
f(x) = ( 1 + 1/x)x
will be defined at some points of the interval and it will be undefined at other
points of the interval. Remember that the calculator does not plot each value
of the domain of the function. Rather, using the xMin and xMax WINDOW settings
the calculator determines a number of values of the domain to evaluate, and
for each of those values the calculator attempts to obtain the functional value.
Then, the calculator plots those successive points, and if the calculator is in the
DrawLine mode, the calculator will "connect" adjacent defined functional values.
In the case of Figures 2 and 9 above, the selected values in the interval
(– 1,0) happen to be defined values for the function.
Therefore, the calculator plotted them and connected adjacent points.
In Figures 3 and 4, the WINDOW settings have changed, which means that the calculator
selects different values to use. It turns out, for those figures, that only four
points in the interval are defined. Furthermore, there are
values between those "defined points" where the calculator has tested the value
and determined that the function is "undefined" for that value. As a result,
in Figures 3 and 4, the calculator has plotted isolated points
in the interval (– 1,0).
In Figure 6, with new xMin and xMax values, the calculator has determined
that there twelve adjacent defined points, and then another eight undefined points
in our interval. The twelve points are connected, while the eight undefined points
leave that section of the graph blank. The graph in Figure 6, for most of the twelve defined
points in our interval, seems to oscilate between positive and negative values.
Although we can not see just how the calculator is computing the functional values,
we do know that if x=– 5/7 then we should
get a negative functional value, and if
x=– 4/7 we should get a positive functional value.
Although this makes sense, it
does not explain the calculator operations.
Figure 10 shows the graph of Figure 3 in TRACE mode, where we have
selected one of the twelve defined points.
Figure 10
Fourth, we need to look at the special case
where x=– 1.
In that case the base is 0.
This produces
0-1 which
is 1/0 which is undefined.
Therefore, our graph should have a missing column corresponding to
x=– 1.
Fifth, we need to look at the remaining values of x,
namely x<– 1. For such values 1/x is a negative number
between – 1 and 0. Therefore, the base is a positive
number between 0 and 1. Because the value of x is negative
the expression we are evaluating,
(1+1/x)x
is the same as
1/(1+1/x)|x|
which is the same as
1/( (x+1)/x )|x|
which is the same as
( x/(1+x) )|x|
This expression is defined for all values of x<– 1. Therefore,
the graphs that we have seen above have been consistent over this interval.
The work that was done above was based on the TI-86. The TI-85 does not show these
issues because, on the TI-85, raising a negative number to a negative power
seems to be undefined, even though we know that in some cases it is quite well defined.
The TI-89 behaves more like the TI-86, however, the TI-89 produces some even more
dramatic aberations in the (– 1,0) interval.
Graphing the function on the TI-89 with the ZoomStd settings produces:
Figure 11
which is again different from Figure 2. In fact, Figure 11 combines some of the
aspects of Figure 2 and Figure 6. One small distinction between the standard operation of
the TI-89 and that of the TI-86 is the default value of the xres setting.
Figure 11 uses that default, namely xres=2.
If we change the default to xres=1 then the graph changes ever so slightly to
Figure 12
where we note that there is a new "open" column near the
x=– 0.5 value.
A complete change of the WINDOW settings to
Figure 13
produces another version of the graph,
Figure 14
and, then changing the xres value to be 2 and redrawing the graph produces
Figure 15
Interestingly, using the ZoomDec setting on the TI-89 produces
Figure 16
However, ZoomDec causes the xres value to be 2.
Without changing the other variables,
if we change xres to be 1, the graph changes to
Figure 17
again showing some of the defined points in our interval of interest.
Another setting,
Figure 18
produces yet another version of the graph
Figure 19
and changing xres to be 2 produces quite a different graph,
Figure 20
The lesson to be learned here is that having the calculator produce a graph
does not mean that the graph is really correct. At times it is essential
to look at the graph, interpret the results, ask questions about things that
may seem strange, and to use the options on the calculator to examine the
graph to try to arrive at some explanations and further understandings.
