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Chapter 4, Section 1, Example 11

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,

is the same as
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.

Saline, MI 48176
March, 2000