Maximal Solutions Let $I\subseteq \boldsymbol{R}$ be a non-degenerated interval and $f:I\to \boldsymbol{R}$ a continuous function and $\forall x\in I$, $f(x)\neq 0$. Given an $a\in I$ let 
$$\begin{align}   F_a:&I\to\boldsymbol{R} \\ x&\mapsto \int_{a}^{x}\frac{ds}{f(s)}\end{align}.$$
If $\varphi_a$ is the inverse of $F_a$, how to prove that, for a $c\in\boldsymbol{R}$, $\varphi_{a,c}(t) = \varphi_a(t-c)$ is a maximal solution of the differential equation
$$\frac{dx}{dt} = f\big(x(t)\big)?$$
 A: First: using the derivative of the inverse function and the fundamental theorem of calculus, you get:
$$
\frac{\partial \varphi_a}{\partial t}(t)=\frac{1}{\frac{\partial F_a}{ \partial x} (\varphi_a(t))}=f(\varphi_a(t))
$$
Therefore $\varphi_a(t)$ is a solution of the differential equation.
I believe you did not ask for first part of the problem, but just in case.
Second: We can suppose without loss of generality that $f>0$ and say that $I=(A,B)$. Note that
$$
Dom(\varphi_a)=Im(F_a)=\left(-\int_A^a\frac{1}{f}\,,\,\int_a^B \frac{1}{f}\right),
$$
and then
$$
Dom (\,\varphi_{a,c}\,)=\left(c-\int_A^a\frac{1}{f}\,,\,c+\int_a^B \frac{1}{f}\right)
$$
Now consider the initial condition, say $x(t_0)=\alpha \in I $. Then we need that $t_0$ belong to the domain of $\varphi_{a,c}$. For any $a$ we always can choose $c$ such that $t_0 \in Dom (\varphi_{a,c})$:
$$
\varphi_{a,c}(t_0)=\varphi_a(t_0-c)=\alpha \;\;\mbox{ iff }\;\; t_0-F_{a}(\alpha)=c
$$
Finally,
$$
Dom (\,\varphi_{a,c}\,)=\left(t_0-\int_A^\alpha\frac{1}{f}\,,\,t_0+\int_\alpha^B \frac{1}{f}\right)
$$
