Question
Find the minimum and maximum of the function
y=\dfrac{x^2+x^2+2}{2x^2-x+1}
Answer
Rearrange the function
y(2x^2-x+1) = x^2+x^2+2
(1-2y)x^2+(1+y)x+2-y = 0
For the quadratic equation to have real solutions,
the discriminant must be greater than or equal to 0, that is,
(1+y)^2-4(1-2y)(2-y)\geq 0
Simplifying the inequality gives
7y^2-22y+7\leq 0
(1)
Divide the equation by 7 to make the coefficient of leading term 1
y^2-\dfrac{22}{7}y+1\leq 0
(2)
On the other hand,
suppose a,b are the minimum and maximum of the function y, that is
a< y < b
Then
(y-a)(y-b)<0
Expand the equation
y^2-(a+b)+ab<0
(3)
Comparing coefficients of each terms of (3) with those of (1) results in
a+b = \dfrac{22}{7}
ab = 1
Solving the system of equations for a,b yields
y_{min} = \dfrac{11-6\sqrt{2} }{7}
and
y_{man} = \dfrac{11+6\sqrt{2} }{7}