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

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

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Let's analyze the function $$ f(x) = \frac{\sqrt{x + 2}}{3 - 3x^2} $$ to determine the conditions for each part of your question. ### 1. **When is $$ f(x) $$ Non-Real?** The function $$ f(x) $$ involves a square root in the numerator, $$ \sqrt{x + 2} $$. For this expression to be real: $$ x + 2 \geq 0 \implies x \geq -2 $$ Therefore, $$ f(x) $$ is **non-real** when: $$ x < -2 $$ ### 2. **When is $$ f(x) $$ Undefined?** The function $$ f(x) $$ is undefined in two scenarios: 1. **When the denominator is zero:** $$ 3 - 3x^2 = 0 \implies 1 - x^2 = 0 \implies x^2 = 1 \implies x = \pm 1 $$ 2. **When the square root is not real:** As determined above, this occurs when $$ x < -2 $$. However, being undefined typically refers to points where the function does not have a valid expression due to division by zero or other issues, not necessarily where it's non-real. Therefore, $$ f(x) $$ is **undefined** at: $$ x = -1 \quad \text{and} \quad x = 1 $$ ### 3. **When is $$ f(x) > 0 $$?** For $$ f(x) $$ to be positive, both the numerator and the denominator must be positive (since the numerator, being a square root, is always non-negative). 1. **Numerator ($$ \sqrt{x + 2} > 0 $$):** $$ x + 2 > 0 \implies x > -2 $$ 2. **Denominator ($$ 3 - 3x^2 > 0 $$):** $$ 3 - 3x^2 > 0 \implies 1 - x^2 > 0 \implies -1 < x < 1 $$ Combining these two conditions: $$ -1 < x < 1 $$ Therefore, $$ f(x) > 0 $$ when: $$ -1 < x < 1 $$ ### **Summary** 1. **Non-real:** $$ x < -2 $$ 2. **Undefined:** $$ x = -1 $$ and $$ x = 1 $$ 3. **Positive:** $$ -1 < x < 1 $$ --- **Final Answer:** 1. All real numbers below –2; 2. x = –1 and x = 1; 3. All real x with –1 < x < 1. - $$ f(x) $$ is non-real when $$ x < -2 $$. - $$ f(x) $$ is undefined at $$ x = -1 $$ and $$ x = 1 $$. - $$ f(x) > 0 $$ when $$ -1 < x < 1 $$.

If , for which values of is

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