### Instruction: Find the domain of the real-valued function $$f(x)=\sqrt{-10x^2-11x+6}.$$ Give the endpoints in your answer as common fractions, not mixed numbers or decimals.### Response: To find the domain of the function, I need to determine the values of x for which the expression under the square root is non-negative, since the square root of a negative number is not defined for real numbers.
I can start by factoring the quadratic expression -10x^2-11x+6 using the AC method.
I need to find two numbers that multiply to -60 (the product of the leading coefficient and the constant term) and add to -11 (the coefficient of the linear term).
After some trial and error, I find that -15 and 4 work.
So I can rewrite the expression as -10x^2-15x+4x+6.
Then I can group the terms and factor by grouping.
I get -10x^2-15x+4x+6=-5x(2x+3)+2(2x+3)=(2x+3)(-5x+2).
So the function can be rewritten as $$f(x)=\sqrt{(2x+3)(-5x+2)}.$$
Now I can see that the expression under the square root is zero when x=-3/2 or x=2/5.
These are the endpoints of the interval where the expression is non-negative, since the parabola opens downward (the leading coefficient is negative) and these are the x-intercepts.
To check that the expression is indeed non-negative on this interval, I can plug in a test value between -3/2 and 2/5, say x=0.
Then I get $$f(0)=\sqrt{(2(0)+3)(-5(0)+2)}=\sqrt{6}>0,$$ which confirms that the expression is positive on the interior of the interval.