Considero
$f:x in ZZ to 4x+12 in 2ZZ$ e $g:x in 2ZZ to |x/2|+1 in NN$
$f({0,2,-4})={y in 2ZZ: EE x in X={0,2,-4}:y=f(x)}={12,20,-4}$,
$g({0,2,-2,-4})={y in NN:EE x in S={0,2,-2,-4}:y=g(x)}={1,2,2,3}$,
$f^(-1)({-4,-6,2,4})={x in ZZ:f(x) in Y={-4,-6,2,4}},$
Per cui si ha:
$4x+12=-4 <=>4x=-4-12=-16 <=> x=-4 in ZZ$,
$4x+12=-6 <=> 4x=-12-6=-18 <=> x=-9/2 notin ZZ$,
$4x+12=2 <=> 4x=-12+2=-10 <=> x=-5/2 notin ZZ$,
$4x+12=4<=> 4x=-12+4=-8 <=> x=-2 in ZZ $,
infine $f^(-1)({-4,-6,2,4})={-2-4}$,
$f^(-1)(2ZZ)={x in ZZ:f(x) in 2ZZ}={x in ZZ: y=4x+12 in 2ZZ}=ZZ$,
$f(ZZ)={y in 2ZZ:EEx in ZZ : y=f(x)}={y in 2ZZ:EE x in ZZ : y=4x+12}={y in 2ZZ:EE x in ZZ : y=4(x+3)}=4ZZ$,
$forall y in 2ZZ$ $f^(-1)({y})={x in ZZ : f(x)=y}={x in ZZ : y=4x+12}={x in ZZ : x=(y-12)/4}$
gli elementi $y in 2ZZ$ possono essere del tipo $y=2k$ oppure $y=4k$ con $k in ZZ$ per cui:
se $y=2k$ in tal caso risulta $x=(2k-12)/4=(k-6)/2 notin ZZ to f^(-1)({y}) = emptyset $,
se $y=4k$ in tal caso risulta $x=(4k-12)/4=(k-3) in ZZ to f^(-1)({y}) ={x in ZZ : x=k-3}$,
L'esercizio non è terminato, ma vi chiedo se quanto scritto è corretto.
Grazie