SOLUTION: Show that, if x is real, the values of the expression {{{(x^2-3x+1)/(2x^2-3x+2)}}} can only lie between -1 and 5/7.

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Show that, if x is real, the values of the expression {{{(x^2-3x+1)/(2x^2-3x+2)}}} can only lie between -1 and 5/7.      Log On


   

Question 1041377: Show that, if x is real, the values of the expression %28x%5E2-3x%2B1%29%2F%282x%5E2-3x%2B2%29 can only lie between -1 and 5/7.
Found 2 solutions by Edwin McCravy, robertb:
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
Let y = the expression

y=%28x%5E2-3x%2B1%29%2F%282x%5E2-3x%2B2%29

Show that the denominator is never 0.

Show that it has derivative

%28dy%29%2F%28dx%29=%283%28x%5E2-1%29%29%2F%282x%5E2-3x%2B2%29%5E2

Set that = 0 and show that it has relative 
and absolute max at (-1,5/7) and an absolute
minimum at (1,-1).

Edwin

Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
The domain of the expression is the set of all real numbers. (The discriminant b%5E2-4ac+=+-17 of the denominator implies that it itself is never zero.)
Let y+=+%28x%5E2-3x%2B1%29%2F%282x%5E2-3x%2B2%29.
==> y%282x%5E2-3x%2B2%29+=+x%5E2-3x%2B1 (after multiplication; note that the denominator is never zero)
==> 2yx%5E2-3yx%2B2y+=+x%5E2-3x%2B1 <==> %282y-1%29x%5E2%2B%283-3y%29x%2B%282y-1%29+=+0.
For x to have real values, the discriminant of this equation (with x as variable) must be greater than or equal to 0.
==> %283-3y%29%5E2-4%282y-1%29%5E2+%3E=+0 <==> -7y%5E2-2y%2B5+%3E=0, or
7y%5E2%2B2y-5+%3C=0,
<==> %287y-5%29%28y%2B1%29+%3C=+0
The inequality is only true when y ∈ [-1, 5/7].