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...

Question

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...

loser66 · Answer

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?

loser66 · Answer

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

ybarrap · Answer

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

ybarrap · Answer

http://www.wolframalpha.com/input/?i=x%28x%2B3%2F2%29%2F%28x-1%29

loser66 · Answer

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

ybarrap · Answer

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

geerky42 · Answer

@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!

ybarrap · Answer

Yep, that's it!

geerky42 · Answer

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

geerky42 · Answer

But I got what I needed, thanks!

loser66 · Answer

ok