$$ \begin{array}{ll} ext { If } f(x)=\frac{\sqrt{x+2}}{3-3 x^{2}}, ext { for which values of } x ext { is } \ ext { 1.2.2 } & f(x) ext { non real. } \ ext { 1.2.3 } & f(x) ext { undefined } \ ext { 1.2.4 } & f(x)0\end{array} $$

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$$ \begin{array}{ll}	ext { If } f(x)=\frac{\sqrt{x+2}}{3-3 x^{2}}, 	ext { for which values of } x 	ext { is } \ 	ext { 1.2.2 } & f(x) 	ext { non real. } \ 	ext { 1.2.3 } & f(x) 	ext { undefined } \ 	ext { 1.2.4 } & f(x)>0\end{array} $$

UpStudy AI · Accepted Answer

Let's analyze the function $$ f(x) = \frac{\sqrt{x+2}}{3 - 3x^2} $$ to determine the conditions for each part of the question. ### 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: $$ x + 2 \geq 0 \implies x \geq -2 $$ If $$ x < -2 $$, the square root becomes the square root of a negative number, which is not real. Therefore: - **$$ f(x) $$ is non-real for all $$ x < -2 $$.** ### 1.2.3. When is $$ f(x) $$ **undefined**? The function $$ f(x) $$ is undefined when the denominator is zero: $$ 3 - 3x^2 = 0 \implies 3x^2 = 3 \implies x^2 = 1 \implies x = \pm 1 $$ Thus: - **$$ f(x) $$ is undefined at $$ x = 1 $$ and $$ x = -1 $$.** ### 1.2.4. When is $$ f(x) > 0 $$? For $$ f(x) $$ to be positive: 1. The numerator must be positive: $$ \sqrt{x + 2} > 0 \implies x + 2 > 0 \implies x > -2 $$ 2. The denominator must also be positive: $$ 3 - 3x^2 > 0 \implies 1 - x^2 > 0 \implies -1 < x < 1 $$ Combining these two conditions: $$ -2 < x < 1 $$ Therefore: - **$$ f(x) > 0 $$ for all $$ x $$ such that $$ -2 < x < 1 $$.** ### Summary 1. **Non-real:** $$ x < -2 $$ 2. **Undefined:** $$ x = 1 $$ and $$ x = -1 $$ 3. **Positive:** $$ -2 < x < 1 $$ - $$ f(x) $$ is non-real when $$ x < -2 $$. - $$ f(x) $$ is undefined at $$ x = 1 $$ and $$ x = -1 $$. - $$ f(x) > 0 $$ when $$ -2 < x < 1 $$.

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