2


The expected curve looks like below.
^　　　　　　　　　　　　　　　　　　　
|2____________________　
|                                      o              x　　　　　　
|                 o           x　　　　　　　　　　　　
|       o       x　　　　　　　　　　　　　　　
|   o  x　　　　　　　　　　　　　　　　　
| ox　　　　　　　　　　　　　　　　　　
|1____________________　
|　　　　　　　　　　　　　　　　　　　
|　　　　　　　　　　　　　　　　　　　
|　　　　　　　　　　　　　　　　　　　
|　　　　　　　　　　　　　　　　　　　
|_____________________>　

Rick Chin
  • 21

2 Answers2

1

Take a function which you already know has a similar behaviour and then try to adapt it. For example take $g(x)=-\frac{1}{x}$, we know $\lim_{x\rightarrow\infty} g(x)=0$. It is now just a matter of "shifting this upwards and to the left", so that $\lim_{x\rightarrow\infty} f(x)=2$ and $f(0)=1$. For this example you can try using $f(x)=-\frac{1}{x+1}+2$ and see that this function has the desired property.

Note that here $f(x)=g(x+1)+2$, where $g(x+1)$ shifts the graph of $g$ one unit to the left and the $+2$ shifts it 2 units upwards.

fidbc
  • 909
  • Thanks for guiding. It works. But, it would be ideal to have a less inward-bended curve, so Y value steps up more obvious when getting close to Y=2. If we allow the curve to hit and go through Y=2, which would only happen in a too large X value, is there a chance to make a less inward-bended curve? Thanks. – Rick Chin Feb 24 '13 at 02:24
  • You have to be a bit more precise about what you mean by "inward-bended". One way of bending the curve would be by taking $g(x)=-\frac{k}{x}$ for some constant $k$. Depending on the type of "bending" you want you should choose an appropriate $k$.

    For this function, if you shift it up by $\epsilon>0$ then it will take the value 2 for some $x$, but it will take values greater than 2 afterwards since it is an increasing function on the non negative real numbers.

     – fidbc Feb 24 '13 at 04:19
0

Another function is $y=1+(2/\pi)\arctan x$. Another is $y=2-e^{-x}$. Another is $$y={2\root3\of x+1\over\root3\of x+1}$$ If you can translate the phrase "inward-bended" into something precise and mathematical, maybe we'd come up with an answer better suited to your needs.

Gerry Myerson
  • 179,216
  • Thanks. I added an additional "o" curve into above graph. The o curve is more inward-bended than the x curve. I guess g(x)=-k/x is less inward bended than f(x)=-1/x. What I wanted was something similar to the x curve. So, as X value increases, the Y value would not step up so steeply and can slowly approach an unreachable limit of Y=2. – Rick Chin Feb 25 '13 at 05:14
  • So, the three examples I suggested --- how do they look to you? – Gerry Myerson Feb 25 '13 at 05:31