## geerky42 one year ago I need a quick favor: I want strictly increasing elementary function $$f(x)$$ such that $$f(0)=0$$, $$f\left(\frac12\right)=1$$, and $$\displaystyle \lim_{x\to1^-}f(x) = \infty$$. I just want one example of function that satisfies given conditions. Cannot figure this out...

1. Loser66

How about this $$f(x) =\dfrac{x-x^2}{1-x}$$ f(0) =0 $$f(1/2) =\dfrac{1/2-1/4}{1-1/2}=1$$ as x approach 1 from the left , lim --> infinitive Does it work?

2. Loser66

and if we simplify, we get f(x) =x, it is increasing, right?

3. ybarrap

You think this works? $$f(x)=\frac{x(x+\frac{3}{2})}{x-1}\\$$

4. ybarrap
5. Loser66

@ybarrap but your graph is decreasing from x =-1

6. ybarrap

Or maybe... $$f(x)=\frac{x(x+\frac{3}{2})}{1-x}\\$$ http://www.wolframalpha.com/input/?i=x%28x%2B3%2F2%29%2F%281-x%29

7. geerky42

@ybarrap Not exactly what I needed, but I modified it to $$f(x) = -\dfrac{x(x+\frac32)}{2(x-1)}$$, which works for me. Thanks!

8. ybarrap

Yep, that's it!

9. geerky42

@Loser66 Well, I want $$\lim_{x\to1^-}f(x)$$ to be $$\infty$$,

10. geerky42

But I got what I needed, thanks!

11. Loser66

ok