1.2 If \( f(x)=\frac{\sqrt{x+2}}{3-3 x^{2}} \), for which values of \( x \) is \( \begin{array}{ll}1.2 .2 & f(x) \text { non real. } \\ 1.2 .3 & f(x) \text { undefined } \\ 1.2 .4 & f(x)>0\end{array} \)

To determine when \( f(x) = \frac{\sqrt{x+2}}{3-3x^2} \) is non-real, we need to focus on the square root in the numerator. The expression under the square root, \( x + 2 \), must be non-negative. Thus, \( x + 2 \geq 0 \) leads to \( x \geq -2 \). If \( x < -2 \), \( f(x) \) will be non-real since the square root of a negative number isn’t defined in the reals. For \( f(x) \) to be undefined, we look at the denominator \( 3 - 3x^2 \). Setting this equal to zero gives \( 3 - 3x^2 = 0 \), which simplifies to \( x^2 = 1 \), leading to possible undefined points at \( x = 1 \) and \( x = -1 \). Therefore, \( f(x) \) is undefined at these two values. To find when \( f(x) > 0 \), both the numerator and the denominator must be positive. The numerator \( \sqrt{x+2} > 0 \) when \( x + 2 > 0 \) (i.e., \( x > -2 \)). The denominator \( 3 - 3x^2 > 0 \) simplifies to \( 1 > x^2 \) or \( -1 < x < 1 \). Combining these conditions, \( -2 < x < 1 \) is where \( f(x) > 0 \). Thus, all intervals where \( f(x) \) is positive are between \(-2\) and \( 1\) excluding \(-1\) where it becomes undefined.