What can we say about $f$ when we know its derivative at few points? Let $b>0$ and $f: \mathbb R \to \mathbb R$ is a differentiable function  with the following properties:
 (1) $f(0)=0=f(b)$ (2) $f'(0)=1$
 and $f'(b)<0$ (3) $f> 0$ on $(0,b).$

My Question: Can we find a constant $C> 0$ such that $f(x)\geq C  \min \{ x, b-x \}$ for all $x\in (0, b)$?

 A: The answer is yes. First we use the condition $f(0) =0$ and $f'(1)>0$. By definition of derivative, for all $\epsilon >0$, there is $\delta>0$ so that 
$$\left| \frac{f(x)-f(0)}{x-0} - f'(0)\right| < \epsilon$$
for all $x$ so that $|x-0| <\delta$. We choose $\epsilon = f'(0)/2$, so 
$$ |f(x)/x - f'(0)| <f'(0)/2$$
for $x< \delta_1$. This implies 
$$ f(x) \ge \frac{1}{2} f'(0) x \ge C \min\{x, b-x\}$$
whenever $C\le \frac 12 f'(0)$ and $x<\delta_1$. 
Similarly, using $f(b) =0$ and $f'(b)<0$, there is $\delta_2$ so that 
$$ f(x) \ge \frac{1}{2} |f'(b)| (b-x) \ge C \min\{x, b-x\}$$
whenever $C \le \frac 12 |f'(b)| $ and $x\in (b-\delta_2, b)$. 
Now consider the interval $[\delta_1, b-\delta_2]$. Note $f$ and $\min\{x, b-x\}$ is is positive on this interval. Let $m_1$ be the minimum of $f$ and $M_1$ the maximum of $\{x, b-x\}$ in this interval. $m_1>0$ since $[\delta_1, b-\delta_2]$ is closed and bounded. Then 
$$ f(x) \ge m_1 = \frac{m_1}{M_1} M_1 \ge \frac{m_1}{M_1} \{x, b-x\}$$
on $[\delta_1, b-\delta_2]$. Grouping all three information, we have 
$$f(x) \ge C \{x, b-x\} \ \ \ \ \text{for all }x\in (0,b),$$
where $C = \min\{ m_1/M_1, \frac 12 f'(0) , \frac 12 |f'(b)|\}$.
