Answer
$(-\infty,4)\cup(4,+\infty)$
Work Step by Step
$(f+x)(x)=3x+4+\frac{5}{4-x}=\frac{(3x+4)(4-x)+5}{4-x}$ simplify
$\frac{12x+3x^2+16-4x+5}{4-x}=\frac{-3x^2+8x+21}{4-x}$ Since the denominator can't be equal to 0 $x\ne4$
the domain is $(-\infty,4)\cup(4,+\infty)$
