Show that $x(t)<0$ for all $t>0$ Show that if a function $x(t)$ satisfies $0\leq \frac{dx}{
dt}\leq x^2$ for all $t$, and $x(0) = -1$, then $x(t) < 0$
for all $t \in [0,\infty)$.
I tried integrating by separation of variables but I'm not sure if I am allowed to do this since there's a discontinuity at $0$. 
 A: By the mean value theorem, we have $x(t)-x(0)=x'(r)(t-0)$ for some $0<r<t$. Substituting values, we get  $x(t)+1=tx'(r)$. Applying the hypothesis, we have now that $x(t)+1\le tx(r)^2$. But $x'$ is positive everywhere, so $x$ is increasing, which means that $x(t)+1\le tx(r)^2\le tx(t)^2$ and so $tx(t)^2-x(t)-1\ge 0.$ Noting that this inequality implies that there is no $t\ge 0$ for which $x(t)=0$, and since $x(0)=-1$, we may invoke the intermediate value theorem to conclude that $x(t)<0$ on $[0,\infty).$
A: If $\frac{dx}{dt}$ exists for all $t$, then $x(t)$ is continuous. Locally (near $t=0$), the function $f(t)=\frac 1 x,f(0)=-1$ is defined (as long as $x\neq 0$), and one has $$\frac d{dt}\left(\frac 1x\right)=-\frac{x'}{x^2}\geq -1,$$ by the given assumption, hence one has $$f(t)=\frac 1x\geq -t-1\qquad\qquad (1)$$ on any connected interval containing $0$ in $[0,\infty)$ where $x\neq 0$.
Now if $x(t)$ is ever $0$ or positive, then by the intermediate value theorem, there exists $t_1>0$ such that $x(t_1)=0$. Let $$t_0=\inf \{t_1>0~|~x(t_1)=0\}.$$ Then one has by continuity $x(t_0)=0$ and $x(t)<0$ on $[0,t_0)$. But this is a contradiction to (1), since $\lim_{t\rightarrow t_0^{-}}\frac 1 x=-\infty$ but $\lim_{t\rightarrow t_0}-t-1=-t_0-1.$ Hence $x(t)<0$ for all $t\geq 0$. QED
