Question

Given the function
`y=(m^2-1)x^2+2(m-1)x+2` , find the values of parameter m for which the function is always positive.

Answer 1

Answer:
`(-\infty, -1) \cup (1, \infty)`

Explanation:

The function is always positive when it has a positive leading coefficient (since that means the graph will open up), and when the discriminant is negative (meaning the graph will never cross the x-axis).

Condition I. Leading coefficient is positive

`m^2 -1 > 0 \implies m < -1 \text{ or } m > 1`

Condition II. Discriminant is negative

`(2(m-1))^2 -4(m^2 -1)(2) < 0\\\\4(m^2 -2m+1)-8(m^2 -1) < 0\\\\4m^2 -8m+4-8m^2 +8 < 0\\\\-4m^2 -8m+12 < 0\\\\m^2 +2m-3 > 0\\\\(m+3)(m-1) > 0\\\\m < -3 \text{ or } m > 1`

Taking the intersection of these intervals, we get
`m < -1` or
`m > 1`.