since f= f1.f2 so f will be equal to 1 iff f1=1 and f2 =1
f1 is 1 for the minterms 0,1,2,3,5,7 and f2 is 1 for the minterm 0,1,2,4,5,7
so f will be 1 for the minterm 0,1,2,5,7 where 7 is the dont care minterm .
so f = ∑m(1,5)+d(0,2,7)
According to me F is AnD operation of f1 and f2 so in f minterm will be common min-term of both function.
since it is AND operation true is only for f1=1 f2=1.
and for dont care first take common and
then check the f1 dont care is present in minterm of f2 or not if yes then take it in(e.g. let 2 in f1 be 1 and 2 is in f2 so it will be 1) similarly check for f2 vice versa. this is done because . 1 and Dn't care is a don't care because the value will now depend on what i take the don't care to be . so we just take intersections of don't care with minterms of f2.
f1.f2 = $\sum$m (1, 5) + $\oslash$( 0, 2, 7) using : 1 . $\oslash$ = $\oslash$
where $\oslash$ is dont care. 0. $\oslash$ = 0
$\oslash$ . $\oslash$ = $\oslash$