SOLUTION: I can not figure out this.... solve the inequality (x-3)(2-x)/(x-1)^2 >0 then I am to graph the solution set.

Question 579368: I can not figure out this.... solve the inequality (x-3)(2-x)/(x-1)^2 >0 then I am to graph the solution set.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
the inequality equation is:
(x-3)(2-x) / (x-1)^2 > 0
set this equation equal to 0 to find the 0 points of the equation if there are any.
also note that the equation is undefined when x = 1 because the denominator then becomes equal to 0.
when you graph the equation, you'll see that there is a vertical asymptote at x = 1.
anyway, set the equation equal to 0 to get:
(x-3)(2-x) / (x-1)^2 = 0
multiply both sides of the equation by (x-1)^2 to get:
(x-3)(2-x) = 0
this equation will be true when either (x-3) = 0 or when (2-x) = 0 or when both of them are equal to 0.
set x-3 = 0 and solve for x to get x = 3.
set 2-x = 0 and solve for x to get x = 2.
the equation crosses or touches the x-axis when x = 3 or x = 2.
it is either above the x-axis or below the x-axis in between.
when x is < 2, the equation becomes negative.
when x is > 2 and < 3 the equation becomes positive.
when x is > 3 the equation becomes negative again.
for x is < 2, pick a value for x like 1.5.
when x = 1.5, the value of the equation becomes:
(1.5-3)(2-1.5) = -1.5 * .5 which is negative.
for x is > 2 and < 3, pick a value for x like x = 2.5
when x = 2.5, the value of the equation becomes:
(2.5-3)*(2-2.5) = -.5 * -.5 which is positive.
for x is > 3, pick a value for x like x = 3.5
(3.5-3)(2-3.5) = .5 * -1.5 which is negative.
the equation is positive between 2 and 3 and is negative otherwise.
this tells you that the solution to the equation is x < 2 and x > 3.
in interval notation, this would be (-infinity,2) union (3,+infinity)
a graph of the equation looks like this:

a more distant view looks like this:

an even more distant view looks like this:

you can see that the equation only goes positive between x = 2 and x = 3 and stays negative otherwise.

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