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How do you call a (linear) function (or the point), which breaks down to 0 on ordinate (axis y), as soon as you breach a certain x1 value?

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Marc van Leeuwen
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feder
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  • Do you mean you want the graph to have the vertical line above $x_1$, or you just mean that the graph quits at $x_1$? – rschwieb Sep 10 '13 at 13:40
  • From what you say @rschwieb, I imply it's not the same. I want to indicate, that if value x is greater than x1, y always results 0. – feder Sep 10 '13 at 14:41
  • OK, yes, $0$ after $x_1$. But you did not actually mean to draw the vertical line then? – rschwieb Sep 10 '13 at 15:21
  • True. That is my bad. It entails that having x1 could result in any y covered by that vertical slope section. total non-sense. sorry for that, folks. – feder Sep 10 '13 at 19:44

2 Answers2

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You could describe it as piecewise linear, and write it e.g. $f:\mathbb{R}^+ \rightarrow \mathbb{R}$ defined by $$f(x)=\begin{cases}mx+y_1-mx_1 & \text{if } 0 \leq x < x_1 \\ 0 & \text{if } x \geq x_1 \end{cases}$$ where $m$ is the slope (in this case, it'll be negative).

If "break down" means we want the function undefined (instead of $0$), we can define the domain of the function as $[0,x_1]$.

Rebecca J. Stones
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    I think the phrase "piecewise linear" is usually used to mean that the function is not only linear on "pieces" but also continuous, even at the endpoints of the pieces. – Andreas Blass Sep 10 '13 at 13:01
  • Dear @AndreasBlass I don't believe the inclusion of continuity is necessarily standard. I recall seeing several textbooks ask the student to "pick values for these constants to make this piecewise linear function continuous" etc. – rschwieb Sep 10 '13 at 13:39
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Assumming you meant a vertical segment at $x=x_1$, this is not (the graph of) a function at all, and even less (if that were possible) a linear function. A (non-linear) function could make a jump at $x=x_1$, but it would need to have a unique value at $x_1$ (not a whole range of values). Your description also suggests it would have to have (unique) values beyond $x_1$ (it wouldn't have to if you defined a function with domain $[0,x_1]$ but I don't think you intended that).

Marc van Leeuwen
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  • I wouldn't say "not the graph of a function at all" because one could feasibly consider it as a function of $y$, but of course we're all agreed it's not a function of $x$. I guess some readers might not accept any other variable than the one on the horizontal $x$ axis as a candidate for input, and that would be OK too: this is just for the rest of us. – rschwieb Sep 10 '13 at 13:44