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

Let's analyze the function \( f(x) = \frac{\sqrt{x+2}}{3 - 3x^{2}} \) to determine the specified conditions. ### 1.2.2. When is \( f(x) \) Non-Real? For \( f(x) \) to be real, the expression inside the square root must be non-negative: \[ x + 2 \geq 0 \quad \Rightarrow \quad x \geq -2 \] **Conclusion:** \( f(x) \) is **non-real** for all \( x < -2 \). --- ### 1.2.3. When is \( f(x) \) Undefined? A function is undefined when the denominator is zero or when the expression inside the square root is negative. However, from 1.2.2, we've already handled the case when the square root is negative. Here, we focus on the denominator: \[ 3 - 3x^{2} = 0 \quad \Rightarrow \quad 1 - x^{2} = 0 \quad \Rightarrow \quad x = \pm 1 \] **Conclusion:** \( f(x) \) is **undefined** at \( x = 1 \) and \( x = -1 \). --- ### 1.2.4. When is \( f(x) > 0 \)? For \( f(x) \) to be positive: 1. **Numerator:** \( \sqrt{x+2} \) is always non-negative and positive when \( x > -2 \). 2. **Denominator:** \( 3 - 3x^{2} > 0 \Rightarrow 1 - x^{2} > 0 \Rightarrow -1 < x < 1 \). Combining these conditions: \[ x > -2 \quad \text{and} \quad -1 < x < 1 \quad \Rightarrow \quad -1 < x < 1 \] **Conclusion:** \( f(x) > 0 \) for all \( x \) in the interval \( (-1, 1) \). --- ### Summary - **1.2.2.** \( f(x) \) is **non-real** for all \( x < -2 \). - **1.2.3.** \( f(x) \) is **undefined** at \( x = 1 \) and \( x = -1 \). - **1.2.4.** \( f(x) > 0 \) for all \( x \) in \( (-1, 1) \).