Question 899350
{{{f(x)=sqrt(3x^2-x-2)}}}


The restriction starts with {{{3x^2-x-2>=0}}};
which leads to {{{(3x+2)(x-1)>=0}}}
whose critical points are at -2/3 and 1.


You can test the signs around the critical values for x and find that the acceptable values for x are {{{x<=-2/3}}} and {{{1<=x}}}.  No values of x between the critical points are acceptable in the function because they force square root of a negative number  (which makes for complex solutions containing imaginary numbers).