Hey I've got this question, and I really don't even know where to start, here it goes:
PROVE or DISPROVE: There exists a function $f:\Bbb{R}\rightarrow \Bbb{R}\space$ such that $f'(x)=-f^2(x)$
for every $x\in\Bbb{R}$ and such that $f(x)\ne0\space$ for every $x\in\Bbb{R}$.
My feeling is that there isn't one, and so I thought of assuming contradiction and then calculating some integral to get a contradiction (maybe since $\int \frac{f'(x)}{-f^2(x)}dx=1$ and $f(x)\ne0$), but this is based only on a sort of similar exercise we had in class..
any help much appreciated!
Best Answer
$-\dfrac{f'(x)}{f^2(x)}=1\Rightarrow \bigg(\dfrac{1}{f(x)}\bigg)'=(x)'$ so $\dfrac{1}{f(x)}=x+c,\ \forall x\in\mathbb{R}$ so $\dfrac{1}{f(-c)}=0$ contradiction