### Instruction: 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).### Response: To find the domain of this function, I need to identify the values of x that make the expression under the square root non-negative.
That is, I want to solve the inequality \[-6x^2+11x-4\geq 0.\]
One way to do this is to factor the quadratic expression, if possible.
I can try to find two numbers that multiply to -24 (the product of the leading and constant coefficients) and add to 11 (the middle coefficient).
One such pair of numbers is -3 and 8, so I can write \[-6x^2+11x-4=(-3x+4)(2x-1).\]
Now, to make this product non-negative, either both factors have to be non-negative, or both factors have to be non-positive.
This leads to two cases.
Case 1: Both factors are non-negative.
This means \[-3x+4\geq 0\] and \[2x-1\geq 0.\]
Solving for x, I get \[x\leq \frac{4}{3}\] and \[x\geq \frac{1}{2}.\]
Combining these inequalities, I get \[\frac{1}{2}\leq x\leq \frac{4}{3}.\]
This is one part of the domain.
Case 2: Both factors are non-positive.
This means \[-3x+4\leq 0\] and \[2x-1\leq 0.\]
Solving for x, I get \[x\geq \frac{4}{3}\] and \[x\leq \frac{1}{2}.\]
However, these inequalities are contradictory, since no value of x can satisfy both of them at the same time.
This means there is no solution in this case.
Therefore, the domain of the function is only the interval \[\frac{1}{2}\leq x\leq \frac{4}{3}.\]
The endpoints of this interval are \[\frac{1}{2}\] and \[\frac{4}{3}\] in common fractions, as requested.