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.

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To determine for which values of $$ x $$ the function 
$$ f(x) = \frac{\sqrt{x+2}}{3 - 3x^{2}} $$
is non-real, we need to examine two main aspects:

1. **The Square Root in the Numerator:**
   - The expression inside the square root, $$ x + 2 $$, must be non-negative for the square root to yield a real number.
   - Therefore, $$ x + 2 \geq 0 $$ which simplifies to:
     $$
     x \geq -2
     $$
   - If $$ x < -2 $$, the square root becomes the square root of a negative number, which is not real.

2. **The Denominator:**
   - The denominator, $$ 3 - 3x^{2} $$, must not be zero because division by zero is undefined.
   - Setting the denominator to zero:
     $$
     3 - 3x^{2} = 0 \quad \Rightarrow \quad x^{2} = 1 \quad \Rightarrow \quad x = \pm 1
     $$
   - At $$ x = 1 $$ and $$ x = -1 $$, the function is undefined.

**Conclusion:**
- The function $$ f(x) $$ is **non-real** in two scenarios:
  1. **When $$ x < -2 $$:** The square root becomes non-real.
  2. **When $$ x = 1 $$ or $$ x = -1 $$:** The function is undefined due to division by zero.

However, typically when discussing the realness of a function, the primary concern is whether outputs are real numbers. Thus, the most significant condition making $$ f(x) $$ non-real is when $$ x < -2 $$.

### Final Answer:
All real numbers below –2. The function f (x ) is not real for every x with x < −2.
$$ f(x) $$ is non-real for all $$ x < -2 $$.

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.

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