If \( f(x)=\frac{\sqrt{x+2}}{3-3 x^{2}} \), for which values of \( x \) is 1.2.2 \( \quad f(x) \) non real.

To determine when the function \( f(x) = \frac{\sqrt{x + 2}}{3 - 3x^2} \) is **non-real**, we need to examine both the numerator and the denominator. ### 1. **Numerator Analysis:** The numerator involves a square root, \( \sqrt{x + 2} \). For this expression to be real: \[ x + 2 \geq 0 \implies x \geq -2 \] - **When \( x < -2 \):** The square root becomes the square root of a negative number, which is **non-real**. ### 2. **Denominator Analysis:** The denominator is \( 3 - 3x^2 \). For the function to be defined (and real), the denominator must not be zero: \[ 3 - 3x^2 \neq 0 \implies 1 - x^2 \neq 0 \implies x^2 \neq 1 \implies x \neq \pm 1 \] - **When \( x = 1 \) or \( x = -1 \):** The denominator becomes zero, making the function **undefined**. In many contexts, undefined values are considered as points where the function does not produce a real number. ### **Conclusion:** Combining both analyses: - **\( f(x) \) is non-real when:** - **\( x < -2 \)** - **\( x = 1 \) or \( x = -1 \)** ### **Final Answer:** All real numbers below –2 or equal to 1 or –1. In other words, f (x ) is non real for x < −2 or x = ±1.