Question 896344
you want to find out the intervals where the square root is positive.
those intervals will be your domain.


start with 2x^3 - x^2 - 18x + 9 = 0


fortunately this factors cleanly.


if not, then you probably would need to graph the equation to see where the intervals lie.


graphing is the easiest way if you have a graphing calculator handy.


i did both.


first the factoring method.


your equation to factor is:


2x^3 - x^2 - 18x + 9 = 0


break this up into 2 separate parts as shown below:


(2x^3 - x^2) - (18x - 9) = 0


factor the first part to get:


x^2 * (2x - 1) - (18x - 9) = 0


factor the second part so the expression in the parentheses will be equal to 2x - 1.


you get:


x^2 * (2x - 1) - 9 * (2x - 1) = 0


factor out the common term of (2x - 1) to get:


(x^2 - 9) * (2x - 1) = 0


x^2 - 9 can further be factored into (x - 3) * (x + 3)


your equation becomes:


(x - 3) * (x + 3) * (2x - 1) = 0


the roots of this equation are:


x = 3
x = -3
x = 1/2


those are the values of x where the value of the equation is equal to 0.


in other words, the graph of the equations crosses the x-axis at those values of x.


you now need to test each interval to see whether the graph is negative or positive in those intervals.


first interval is x < -3
graph is negative there.


second interval is x > -3 and < 1/2
graph is positive there.


third interval is x > 1/2 and < 3
graph is negative there.


fourth interval is x > 3
graph is positive there.


your domain is the intervals where the square root of the equation is not negative.


those intervals will be:


-3 <= x <= 1/2


x >= 3


here's a graph of the equation within the square root sign where you wanted to find the intervals where it was positive and where it was negative.


the equation used is y = 2x^3 - x^2 - 18x + 9


<img src = "http://theo.x10hosting.com/2014/083004.jpg" alt="$$$" </>


here's a graph of the equation you wanted to find the domain for.


that equation is y = sqrt(2x^3 - x^2 - 18x + 9)


<img src = "http://theo.x10hosting.com/2014/083005.jpg" alt="$$$" </>



since the intervals where the square root is negative are invalid because they don't result in any real values, the graph is missing in those intervals.