Expecting a Discontinuity in Piecewise Function on TI-89 but finding Unexpected Results

When I look at the following function $f(x)$, it would seem to me that because there are no $\le$ or $\ge$ inequalities included where the function's formula changes (for the parts relating to 7,) the integer 7 should (most likely) not be included in the line of this graphed function for $x$.

$$f(x)=\begin {cases} \frac{5}{x-10}& x \le 5 \\\frac{5}{5x-30}& 5 \lt x \lt 7\\\frac{x-2}{12-x}& x \gt 7 \end {cases}$$

But when I go to graph the line (and subsequently check the function's data table for values) I find that the line of this function actually appears to continue through the integer 7 on my TI-89 calculator (and I am getting a defined value of 1. for $y$ when $x$ equals 7!)

Here is the nested when() functions I am using to graph this on the calculator:

when(x<7, when(x<=5, ((5)/(x-10)), ((5)/(5x-30))), ((x-2)/(12-x)))


How is it possible for $f(x)$ to be defined at the integer 7 if the integer 7 is being skipped in the rules for the function itself?

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It is probably an artifact of how your calculator handles piecewise defined functions. The limit of $f$ as $x$ goes to $7$ is $1$, though, so (unless there's some context) there should not be any problem defining $f(7) =1$. – Arthur Apr 22 '13 at 18:58
@Arthur Interesting point about the artifacts possibility. As a follow-up to the last part of your comment: if I were to use $f(7)$, though, how would the function know what formula to use? (Which reminds me of case statements in programming languages... and how case statements often contain a "default" or "else" statement to catch extra cases...) – summea Apr 22 '13 at 19:03

You are correct that the function is not defined at $x=7$. The gap is just the one point, however, and the limit from both sides is $1$. I am not surprised the graph looks like it goes through $(7,1)$ but it shouldn't be reporting a value at $7$.

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Thanks for your answer! Unfortunately, it may be the way I've entered the function into my calculator; when I use the when() command, it looks like it might be using the last part of the formula as a "catch-all" that includes the 7... – summea Apr 22 '13 at 19:38
Sorry for wasting everyone's time on this question; I'll try to find a better way to enter the function into my calculator to avoid the "catch-all" else statement that is happening; – summea Apr 22 '13 at 19:53
Would this particular question be "too localized" for me to write down what I used to solve this issue? (i.e. how to define a piecewise function that includes a gap between $x$ values on a TI-89) If so, I can delete it; – summea Apr 22 '13 at 20:20
@summea: I think it would be, as it relates to the use of the specific calculator. The mathematical point was that there was a gap in the domain of the function. But I would leave what is here, and it probably doesn't hurt to put an answer to show what you found. – Ross Millikan Apr 22 '13 at 20:22

Unfortunately, this ended up being an issue with the nested when() functions within the overall $f(x)$ function on a TI-89 calculator. Nested when() statements are great for when there are no gaps between areas where the formula changes in a piecewise function. But when there are gaps, nested when() functions can get pretty complicated.

Instead of using nested when() functions (when there is a gap present in the line being graphed,) it is possible to just define this piecewise function as an entire user-defined function (using the Func command on the calculator,) like this:

Func: If x<=5 Then:Return (5)/(x-10):ElseIf 5<x<7 Then:Return (5)/(5x-30):ElseIf x>7 Then:Return (x-2)/(12-x):Else:Return -1:EndIf:EndFunc


After using this newly-created function on the calculator, when checking the $f(7)$ in the table of the graphed function, 7 is now undefined* (as expected.)

*Note: "undefined" in this case is displayed as -1 until I can find a better way to represent that the value does not exist at 7.

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