I am looking for a function that satisfies the following criteria:
$\ f(0) = 0$
$\ f'(0) = 0 $
$\ f'(x) > 0, 0 < x < 1$
$\ f''(x) > 0, 0 < x < 1$
$\ \lim_{x\to 1} f'(x) = \infty$
Additionally, I need:
$\ f'(0.01) \approx 0.24$
$\ f'(0.5) \approx 0.25$
$\ f'(0.98) \approx 0.26$
Basically, this means the derivative starts off at 0, quickly grows to around 1/4, increases slowly between 0.01 and 0.98, and then tends to $\infty$ between 0.98 and 1.
Currently, I achieve this with the following piecewise function:
$\ f(x) = \begin{cases} Ax^{2}, & \text{0 $\le$ x < 0.01} \\ B+ Cx + Dx^{2}, & \text{0.01 $\le$ x < 0.98 } \\ \frac{1}{1-x} + E + Fx, & \text{0.98 $\le$ x < 1 } \\ \end{cases} $
With appropriate values for A-F. However, I am trying to find a simple function with a continuous $2^{nd}$ derivative that also has all these properties. I am also looking to find cases where the derivative of the function has more than 1 'near plateau' (e.g. ranges where it is increasing slowly). For a function that satisfied the original 5 criteria above and:
$\ f'(0.01) \approx 0.24$
$\ f'(0.5) \approx 0.25$
$\ f'(0.6) \approx 0.5$
$\ f'(0.99) \approx 0.51$
You may try the model $$ f(x)=\begin{cases} Ax^3+Bx^2+Cx+D & x\in[0,0.01]\\ Ex^2+Fx+G & x\in(0.01,0.98)\\ H\frac1{1-x}+Ix+J & x\in[0.98,1] \end{cases} $$. Then you have enough variables to satisfy $$ f(0)=0\\ f'(0)=0\\ f'(0.01)=0.24\\ f'(0.98)=0.26 $$ You will get automatically $f'(0.5)\approx 0.25$ and $\lim_{x\nearrow 1}f'(x)=\infty$. But to make $f'$ differentiable you have to check $$ \lim_{x\nearrow 0.01} f'(x)=\lim_{x\searrow 0.01}f'(x)\text{ and } \lim_{x\nearrow 0.98} f'(x)=\lim_{x\searrow 0.98}f'(x). $$ But this is possible in the model above.
Edit:
Consider $$ f_n(x)=0.26\left(\sqrt{x^2+\frac1n}-\frac1{\sqrt{n}}\right)-(1-x)^{\frac1n}-\frac{x}n+1. $$ Then we have $f_n(0)=0$ for all $n\in\mathbb{N}$. Further $$ f_n'(x)=0.26\cdot\frac{x}{\sqrt{x^2+\frac1n}}+\frac1{n}(1-x)^{-\frac1n}-\frac1n $$ which yields $f_n'(0)=0$ and $\lim_{x\to 1}f_n'(x)=\infty$ for all $n\in\mathbb{N}$ and $f_n'(x)>0$ for all $0<x<1$ and all $n\in\mathbb{N}$. For $0<x<1$ we get $$ \lim_{n\to\infty} \frac{x}{\sqrt{x^2+\frac1n}}=1\text{ and }\lim_{n\to\infty}\frac1n(1-x)^{-\frac1n}-\frac1n=0. $$ This yields $\lim_{n\to\infty} f_n'(x)=0.26$. Therefore you can choose $n$ large enough such that $f_n'(x)\approx 0.26$ for all $x\in[0.01,0.98]$.
Finally $$ f_n''(x)=0.26\cdot\frac1{n(x^2+\frac1n)^{\frac32}}+\frac1{n^2}(1-x)^{-1-\frac1n}>0 $$ for all $n\in\mathbb{N}$ and all $0<x<1$.
If you choose $n$ sufficient large, you get all the properties you supposed. But unfortunately the convergence isn't very fast. $$ \begin{array}{|c|c|c|c|} \hline n & f_n'(0.01) & f_n'(0.5) & f_n'(0.98)\\ \hline 10,000 & \approx 0.1838 & \approx 0.2599 & \approx 0.25999\\ \hline 20,000 & \approx 0.2123 & \approx 0.26 & \approx 0.26\\ \hline 30,000 & \approx 0.2252 & \approx 0.26 & \approx 0.26\\ \hline 40,000 & \approx 0.2326 & \approx 0.26 & \approx 0.26\\ \hline 50,000 & \approx 0.2373 & \approx 0.26 & \approx 0.26\\ \hline 60,000 & \approx 0.2407 & \approx 0.26 & \approx 0.26\\ \hline \end{array} $$
The function you are looking for could be $f_{50,000}$ or $f_{60,000}$.