### 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 this function, I need to figure out what values of x make the expression under the square root non-negative, since I can't take the square root of a negative number.
To do this, I'll first try to factor the quadratic expression -10x^2-11x+6, which might make it easier to see the zeros and the sign changes of the expression.
To factor a quadratic expression of the form ax^2+bx+c, I need to find two numbers that add up to b and multiply to ac.
In this case, a=-10, b=-11, and c=6, so I need to find two numbers that add up to -11 and multiply to -60.
One pair of numbers that works is -15 and 4, since -15+4=-11 and -15*4=-60.
So I can write -10x^2-11x+6 as -10x^2-15x+4x+6.
Next, I'll use a technique called grouping, where I split the expression into two groups and factor out the greatest common factor from each group.
The first group is -10x^2-15x, and the greatest common factor is -5x.
The second group is 4x+6, and the greatest common factor is 2.
So I get -5x(2x+3)+2(2x+3).
Now, I notice that both groups have a common factor of 2x+3, so I can factor that out as well, and get (2x+3)(-5x+2).
So, the expression under the square root is equivalent to (2x+3)(-5x+2), and I can use this to find the domain of f(x).
The expression (2x+3)(-5x+2) is zero when either 2x+3=0 or -5x+2=0, which means when x=-3/2 or x=2/5.
These are the zeros of the expression, and they divide the number line into three intervals: (-infinity, -3/2), (-3/2, 2/5), and (2/5, infinity).
I want to find out which of these intervals make the expression non-negative, so I'll pick a test point in each interval and plug it into the expression.
For the first interval, I'll pick x=-2, which is less than -3/2.
Plugging in x=-2, I get (2(-2)+3)(-5(-2)+2) = (-1)(12) = -12, which is negative.
So this interval is not part of the domain of f(x).
For the second interval, I'll pick x=0, which is between -3/2 and 2/5.
Plugging in x=0, I get (2(0)+3)(-5(0)+2) = (3)(2) = 6, which is positive.
So this interval is part of the domain of f(x).
For the third interval, I'll pick x=1, which is greater than 2/5.
Plugging in x=1, I get (2(1)+3)(-5(1)+2) = (5)(-3) = -15, which is negative.
So this interval is not part of the domain of f(x).
Therefore, the domain of f(x) is the interval (-3/2, 2/5), and I can write the endpoints as common fractions: -3/2 and 2/5.