problem: Find the domain of the real-valued function \[f(x)=\sqrt{-6x^2+11x-4}.\] Give the endpoints in your answer as common fractions (not mixed numbers or decimals).
solution: We need $-6x^2+11x-4\geq 0$.  The quadratic factors as \[(2x-1)(-3x+4) \ge 0.\] Thus the zeroes of the quadratic are at $\frac{1}{2}$ and $\frac{4}{3}$.  Since the quadratic faces downward, it is nonnegative between the zeroes.  So the domain is $x \in \boxed{\left[\frac{1}{2}, \frac{4}{3}\right]}$.