Find the domain of the real-valued function
\[f(x)=\sqrt{-6x^2+12x+18}.\]
Give the endpoints in your answer as common fractions (not mixed numbers or decimals).
+191 The stuff under square root needs to be bigger than 0.
-6x^2+12x+18>= 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]
