Find interval of defintion

Question

Let $f$ be defined on $(a,b)$. If $$f'+f^2+1 \ge 0,$$ $$\lim_{x\to a}f(x)=\infty$$ and $$\lim_{x\to b}f(x)=-\infty$$ Find an interval where $b-a$ might be.

Answer 1

We have

$$\frac{df}{dx} + f^2 + 1 \ge 0 \implies \frac{df}{f^2+1} \ge -dx$$

Integrating this on $[0,x]$ gives $f(x) \ge -\tan x$.

Now we have

$$-\infty = \lim_{x\to b-}f(x) \ge -\lim_{x\to b-} \tan x$$

so $\lim_{x\to b-} \tan x = +\infty$ which implies $b = \frac{(2k+1)\pi}2$ for some $k \in \mathbb{Z}$.

Now, if $b-a > \pi$, there would exist $t \in (a,b)$ such that $\lim_{x\to t+} \tan x = -\infty$ so $f(t) \ge -\lim_{x\to t+} \tan x = +\infty$ which is a contradiction because $f(t)$ is a well-defined real number.

Hence $b-a \le \pi$.

I believe this is the most we can say about $b-a$.

Answer 2

Hint

$$f'+f^2+1\geq 0\iff \frac{f'}{f^2+1}\geq -1.$$