Answer:
4th answer option
Explanation:
the domain is easy : x must not be 1 to avoid 0 in the denominator (and hence a .../0 expression).
that eliminates the 1st and 3rd answer.
for the range :
x = 0 gives us y = -4/-1 = 4
x = 2 gives us y = 10/1 = 10
so, it is very likely that limit for y = 7 for x = 1, which is then the impossible y-value in the range for the impossible x- value.
let's verify with
l'Hopital's Rule
If lim x→a f(x) = 0
and
lim x→a g(x) = 0
then
lim x→a f(x)/g(x) = lim x→a f'(x)/g'(x)
f = 3x² + x - 4
f' = 6x + 1
g = x - 1
g' = 1
f'(x)/g'(x) = (6x + 1)/1 = 6x + 1
for a = 1, therefore x = 1
6×1 + 1 = 7
so, yes, as the limit for x = 1 is y = 7, for the invalid domain value x = 1, the invalid range value is y = 7.