### 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 find the values of x that make the expression inside the square root non-negative, since I cannot take the square root of a negative number.
So I need to solve the inequality \[-6x^2+11x-4 \geq 0.\]
This is a quadratic inequality, and one way to solve it is to find the zeros of the corresponding quadratic equation \[-6x^2+11x-4 = 0\] and then test the intervals between them.
To find the zeros, I can either use the quadratic formula or factor the expression.
I notice that the expression can be factored as \[-6x^2+11x-4 = -(2x-1)(3x-4)\] by looking for two numbers that multiply to -24 and add to 11.
So the zeros are x = 1/2 and x = 4/3, and I can write them as fractions since the problem asks for that.
Now I need to test the intervals \[(-\infty, 1/2), (1/2, 4/3),\] and \[(4/3, \infty)\] to see where the inequality is satisfied.
I can choose any value in each interval and plug it into the expression.
For example, I can choose x = 0, x = 1, and x = 2.
When x = 0, the expression is -4, which is negative, so the interval \[(-\infty, 1/2)\] does not work.
When x = 1, the expression is 1, which is positive, so the interval \[(1/2, 4/3)\] works.
When x = 2, the expression is -20, which is negative, so the interval \[(4/3, \infty)\] does not work.
So the domain of the function is the interval \[(1/2, 4/3)\] where the expression is non-negative.