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).