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I need a function that does the following:

  • x can not be greater than 1
  • y can not be greater then 2
  • As x approaches the value 0, then y approaches, but does not reach the value 2.
  • As y approaches the 0, then x appproaches, but does not reach the value 1.

The graph of the function should look something like the following image, except for the part where the x value exceeds the value of 1. I need x to only approach the value of 1.

This is kinda what I'm looking for

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Does the function have to qualify as a nonlinear function also, or will a linear function do? There exist an uncountable infinity of nonlinear functions which satisfy your conditions. –  Doug Spoonwood Sep 14 '11 at 17:43
    
"As x approaches the value 0, then y approaches, but does not reach the value 2." This is ambiguous. What about $y=2*(x-1)^2$? When $x$ "approaches" (but does not reach) 0, $y$ approaches (but does not reach) 1. Or perhaps you meant "when $x$ tends to 0, $y$ tends to some $y_0<1$ –  leonbloy Sep 14 '11 at 19:00

1 Answer 1

Why do you tag this conic sections? Does $y=2(1-x^p)$ , where $0 \lt p \lt 1$ and the smaller $p$ gets, the closer the bend is to the origin meet your needs? As Arturo Magidin suggests, you should state $x \in (0,1)$

Added: if you really want a conic section, you could use $y=\frac{2}{1-x}$ restricted to $(0,1).$ If that doesn't go deep enough into the corner, $y=\frac{2}{(1-x)^p}$, again with $0 \lt p \lt 1,$ is available, but is no longer a conic section.

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Of course, you want to restrict the domain explicitly; or throw in some "dummy factor" that makes the function undefined for $x\geq 1$ and for $x\leq 0$; perhaps $\frac{\sqrt{x(1-x)}}{\sqrt{x(1-x)}}$? –  Arturo Magidin Sep 14 '11 at 17:29
    
@Arturo: that factor cancels very nicely, no? Probably you intended something like $\sqrt{\mathrm{sign}(x(1-x))}$... –  J. M. Sep 14 '11 at 17:32
    
@J. M., if you pretend you don't know about imaginary numbers, then Arturo's factor cannot cancel outside $[0,1]$, because neither numerator nor denominator even exist. –  Henning Makholm Sep 14 '11 at 17:46
    
@Henning: True, but if this one's going to be for programming purposes, I've seen compilers aggressively optimize out anything of the form x-x and x/x in expressions... –  J. M. Sep 14 '11 at 17:53
    
which is why one should always turn on the "strict floating-point" option in one's compiler. –  Henning Makholm Sep 14 '11 at 18:00

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