Question

Find the domain:
[1 / sqroot of 2x+3] + (sqroot of 6-4x)

Answer 1

so we have a fraction with a square root on the bottom
so what ever is in the square root must be greater than 0
and for the square root on the right, we need the stuff on the inside to be >=0
so we need the intersection of
2x+3>0 and 6-4x>=0
x>-3/2 and x<=3/2
so
we need
-3/2<x<=3/2

Answer 2

does this make sense?

Answer 3

Yes! Thank you so much!!!

Answer 4

np