I'm having a bit of trouble understanding how to approach this question:
The probability density function (PDF) of a continuous random variable $X$ is given by
$$f_X\left(x\right)= \begin{cases} \frac{\left(2x+3\right)}{4} & \text{0 $\leq$ x $\leq$ 1,} \\ 0 & \text{otherwise} \\ \end{cases}$$
Find the conditional PDF of $X$ given event $A = \{0.6 < X < 1.2\}$, $f_{X}\left(x \mid A\right)$.
Using the formula
$$f_{A}\left(x \mid A\right) = \frac{d}{dx} F_X \left(x \mid A \right)$$
and
$$F_X \left(x \mid A \right)=\frac{P\left[\{X \le x\} \cap A\right]}{P\left[A\right]}$$
I determined $F_X \left(x\right)$ as
$$F_X\left(x\right)= \begin{cases} 0 & \text{$x < 0$,} \\ \frac{x^2 + 3x}{4} & \text{$0 \le x \le 1$,} \\ 1 & x>1 \end{cases}$$
but this is where I got stuck. I'm not sure how to incorporate the given restrictions of $A$ and the restrinctions of $F_X \left(x\right)$ at the same time to get the full piecewise equation for $F_{X}\left(x \mid A\right)$. Any help would be greatly appreciated.