Find the domain of the real-valued function
Give the endpoints in your answer as common fractions (not mixed numbers or decimals).
The stuff under square root needs to be bigger than 0.
We factor to get: -6(x^2-2x-3)>= 0, thus -6(x-3)(x+1) >= 0.
We can get rid of -6, but we need to change the sign. (x-3)(x+1) <= 0.
Two pairs of possibilities: x-3>= 0, x+1 <=0
and x-3 <= 0, x+1 >=0.
Simplify to get x>=3, x<=-1(which is impossible so we abandon it)
and x<=3, x>=-1(which makes sense)
Finally, the domain is [-1, 3]