Given that the domain of the function f (x) = √ MX2 + 4x + 4 is a set of real numbers, the value range of the real number m is obtained

Given that the domain of the function f (x) = √ MX2 + 4x + 4 is a set of real numbers, the value range of the real number m is obtained

To define a field as a set of real numbers, MX2 + 4x + 4 > = 0 holds
When m = 0, it is obviously not in accordance with the meaning of the question
If M is not equal to 0, Let f (x) = MX2 + 4x + 4
Obviously, when M0, if the value of the function is always greater than or equal to 0, then there is no or only one intersection point between the function and the x-axis
Discriminant: 16-16m = 1