Answer
$A=8$
$f$ is not defined at $x=3$.
Work Step by Step
It is given, that $f(4)=0$.
By substituting $x=4$ into the function $f(x)$ we get:
$f(4)=\dfrac{2(4)-A}{2-3}$
$0=\dfrac{8-A}{-1}$
Solving this equation for $A$ we get:
$\frac{8-A}{-1}=0$
$8-A=0(-1)$
$8-A=0\\
8-A+A=0+A
\\8=A$
$f$ is not defined when the denominator equals $0$.
In this case it occurs, when:
$x-3=0$
$x=3$, meaning that $f(3)$ is not defined.
