Assume $F:{\rm I\!R}\times[0,\infty)\rightarrow {\rm I\!R}$ is non-decreasing function for all fixed $t$, $F(x,t)\leq F(y,t)$, $ \forall t\geq 0, x \in {\rm I\!R}, x\geq y,$ and $\phi : [0,\infty)\rightarrow (0,1]$ is continuous, with property $\phi(0)=1 \geq \phi(s) \geq \phi(t)>0, \forall t>s\geq 0.$
Under this assumptions, is the following function continuous? $$f(t):=\inf\lbrace x : F(x,t)<\phi(t) \rbrace, \quad \forall t>0 $$
As Danny Park-Keung Chan said in the comment, $F$ should be non-increasing, otherwise $f(t)$ will always be $-\infty$ as long as $\{x:F(x,t)<\phi(t)\}$ is not empty.
$f(t)$ is not necessarily continuous.
Consider the counterexample $$ F(x,t)= \begin{cases} \mathrm{e}^{-tx},&\text{if }t\text{ is rational,}\\ \mathrm{e}^{-2tx},&\text{otherwise,} \end{cases} $$ and $\phi(t)=\mathrm{e}^{-t}$, then for $t>0$, $$ F(x,t)<\phi(t) \Leftrightarrow \begin{cases} x>1,&\text{if } t\text{ is rational,}\\ x>\frac{1}{2},&\text{otherwise.} \end{cases} $$
Therefore $$ f(t)=\inf\{x:F(x,t)<\phi(t)\}=\begin{cases} 1,&\text{if }t\text{ is rational,}\\ \frac{1}{2},&\text{otherwise,} \end{cases} $$ which is not continuous.