91Ó°ÊÓ

The function is given as \(f(x) = mx^2 - 9mx + 5m + 1\).

The discriminant for a quadratic function \(ax^2+bx+c\) is given by \(D = b^2 - 4ac\). For our function, we have \(a = m\), \(b = -9m\), and \(c = 5m + 1\). Now, let's find the discriminant: \(D = (-9m)^2 - 4(m)(5m + 1)\) \(D = 81m^2 - 20m^2 - 4m\) \(D = 61m^2 - 4m\) For the function to have no real roots, the discriminant must be less than zero, i.e., \(D < 0\).

We need to solve the inequality \(61m^2 - 4m < 0\). First, let's factor out the common factor of 'm': \(m(61m - 4) < 0\) Now, we can use the properties of inequalities to find the values of m that make the function positive for all x. We have two factors being multiplied together (m and 61m - 4), so one of them must be positive and the other must be negative: - If m > 0, then \(61m - 4 > 0\). Solving this, we get \(m > \frac{4}{61}\). - If m < 0, then \(61m - 4 < 0\). Solving this, we get \(m < \frac{4}{61}\). By considering the leading coefficient, which is 'm', we can determine that the function will be positive when m > 0. Thus, combining these conditions, we get: For the function \(f(x) = mx^2 - 9mx + 5m + 1\) to be positive for all \(x\), the values of m must satisfy \(0 < m < \frac{4}{61}\).