Showing a particular type of continuous function is uniformly bounded Let $I = [0,\infty)$ and $f:I \to I$ be continuous with f(0) = 0. Show that if 
\begin{equation}
f(t) \leq 1 + \frac{1}{10}f(t)^2, \text{ for all } t \in I
\end{equation}
then $f$ is uniformly bounded throughout $I$.
I've had much trouble with this one... I haven't been able to find any problems like this either.  Anyone have ideas?
 A: For $y$ real, the equation
$$y\leq 1+\frac{y^2}{10}$$
Is equivalent to $y^2-10y+10\geq 0$. The roots of the polynomial $y^2-10y+10$ are $\frac{10\pm\sqrt{60}}{2}=5\pm\sqrt{15}$, which are both positive.
Now we use this with $y=f(t)$. This means that $f(t)$ is never between $5-\sqrt{15}$ and $5+\sqrt{15}$.
We claim that $f(t)\leq 5+\sqrt{15}$ for all $t$. Indeed, if this was not the case, then there would be some $t_0$ for which $f(t_0)>5+\sqrt{15}$. But $f(0)=0$ and by the Intermediate Value Theorem, all values between $0$ and $f(t_0)$ are attained by $f$, in particular all values strictly between $5-\sqrt{15}$ and $5+\sqrt{15}$, which contradicts the above.
Therefore, $f$ is bounded in $I$ by $5+\sqrt{15}$. You can in fact use the same argument and show that $f$ is bounded by $5-\sqrt{15}$.
A: If a number $x$ satisfies $$x\leq1+\frac{x^2}{10},$$
then $x^2-10x+10\geq0$. Since this is a parabola, it will be positive outside the interval given by its roots. We have
$$
x_1=\frac{10+\sqrt{100-40}}2=5+\sqrt{15},\ \ x_2=5-\sqrt{15}.
$$
This means that if $5-\sqrt{15}<f(t)<5+\sqrt{15}$, then $f(t)>1+\frac1{10}\,f(t)^2$. 
As $f$ is continuous and $f(0)=0$, its values can never cross the interval $5-\sqrt{15},5+\sqrt{15}]$, so we have
$$
f(t)\leq 5-\sqrt{15}
$$
for all $t$. 
