Divergence of the solution. I have two functions $f,g:\mathbb{R}\longrightarrow{\mathbb{R}}$, both continuous and with $g$ bounded. I also have the following Cauchy problem 
$$
\begin{cases}x'=f(t)g(x)\\x(t_0)=x_0
\end{cases}
$$ 
If $\phi$ is the solution of the system defined in $(-\infty, b)$, I have 
$$
\lim_{t \to{+}b}{\phi'(t)=f(t)g(\phi(t))=f(b)k}
$$ 
with $k \geq g(\phi(t))$ 
since g is bounded. And now the question: 
in my notes it claimed that $\phi(t)$ goes to $\infty$ in $b$ since is not defined there. Is it true?
I don't see why $\phi \to \infty$ in $b$. I think it is simply not defined but I don't see why it should go to infinity.
Showing this I would get a contradiction since $\phi(t)'$ is bounded, hence proving that $\phi$ is defined in $\mathbb{R}$
Regards.
Edit -------
Original problem:
Let $f,g:\mathbb{R}\longrightarrow{\mathbb{R}}$, both continuous and with $g$ bounded. Prove that for all $(t_0,x_0)\in \mathbb{R^2}$, the maximal solutions of the Cauchy problem
$$
\begin{cases}x'=f(t)g(x)\\x(t_0)=x_0
\end{cases}
$$
are defined for all $\mathbb{R}$.
The attempt of solution is done by contradiction; supposing $\phi : (-\infty,b) \to \mathbb{R}$ is a solution, we want to show the derivative of $\phi$ is bounded (as I noted at the begining of this post) while $\phi$ goes to $\infty$ in $b$. This last part is what I don't get. I don't see why is going to $\infty$.
 A: A direct argument, not a proof by contradiction, for proving the existence of a global in time solution for every initial data is the following one. Assuming $k$ is an upper bound for $g:\mathbb{R}\to\mathbb{R}$, i. e. $|g(x)|\le k$ for all $x\in\mathbb{R}$, we also know that $f:\mathbb{R}\to\mathbb{R}$ is continuous: this implies that, in any (time) interval we have
$$
|f(x)|\le \max_{x\in I} |f(x)|<\infty.
$$
Considering $I=[t_0,t_1]$ or $[t_1,t_0]$  (we must also consider the behavior of the solution backward in time) and defining $\max_{t\in I} |f(t)|\triangleq M^{t_1}_{t_0}<\infty$, 
we have that
$$
|\phi^\prime(t)|\le k M^{t_1}_{t_0}<\infty\quad\forall t\in I\label{1}\tag{1}
$$
Note that $M^{t_1}_{t_0}$ depends in general on both $t_1$ and $t_2$.
Equation \eqref{1} implies
$$
|x(t)-x_0|=\Bigg|\int\limits_{t_0}^{t}\phi^\prime(s)\mathrm{d}s\Bigg|\le
\begin{cases}
\displaystyle\int^{t_0}_tk M^{t_1}_{t_0} \mathrm{d}s &t_1<t_0\\
\\
\displaystyle\int_{t_0}^tk M^{t_1}_{t_0} \mathrm{d}s &t_1>t_0
\end{cases}
\le k M^{t_1}_{t_0} |t_1-t_0|
$$
i. e.
$$
|x(t)|\le k M^{t_1}_{t_0} |t_1-t_0|+|x_0|<\infty\quad \forall t\in I,\;\forall(t_0,x_0)\in\mathbb{R}\label{2}\tag{2}
$$
The arbitrariness of $t_1$ and formula \eqref{2} imply that the solution $x(t)$ of the posed Cauchy problem exists finite for each $(t_0,x_0)\in\mathbb{R}$ and each time $t$.
