Determine the values of m for the inequality (m-1)x^2-(m+1)x+(m+1)>0 holds.
For the inequality to hold, we'll have to impose the following:
(m-1) > 0, for the function to be convex and m - 1 different from 0, for the function to keeps it's order.
If the quadratic function is positive for any real value of x, then the graph of the function is entirely above x axis. That means that the quadratic function is not cancelling for any real value of x. For this reason, the discriminant of the function is positive.
delta = b^2 - 4ac
a,b,c are the coefficients of the quadratic.
a = (m-1) , b = -(m+1) , c = (m+1)
delta = (m+1)^2 - 4(m-1)(m+1)
delta = m^2 + 2m + 1 - 4(m^2 - 1)
delta = m^2 + 2m + 1 - 4m^2 + 4
delta = -3m^2 + 2m + 5
We'll impose the constraint for delta to be positive:
-3m^2 + 2m + 5 > 0
3m^2 - 2m - 5 < 0
m1 = [2+sqrt(4 + 60)]/6
m1 = (2+8)/6
m1 = 10/6
m1 = 5/3
m2 = -1
Delta is negative if m is in the interval (-1 , 10/6).
But, to respect the constraint for the function to be convex (y has to be above x axis), m-1>0 => m>1.
The common interval of possible values of m is: (1 , 10/6).