\( f(x)=\frac{\sqrt{x+2}}{3-3 x^{2}} \); for whic valus of \( x \) is \( f(x)>0 \)

To determine the values of \( x \) for which \( f(x) = \frac{\sqrt{x+2}}{3 - 3x^2} > 0 \), let's analyze both the numerator and the denominator. ### 1. **Domain of \( f(x) \)** - **Numerator (\( \sqrt{x+2} \))**: - The expression under the square root must be non-negative. - \( x + 2 \geq 0 \Rightarrow x \geq -2 \). - **Denominator (\( 3 - 3x^2 \))**: - The denominator cannot be zero to avoid undefined expressions. - \( 3 - 3x^2 \neq 0 \Rightarrow 1 - x^2 \neq 0 \Rightarrow x \neq \pm 1 \). Thus, the domain of \( f(x) \) is: \[ x \geq -2 \quad \text{and} \quad x \neq 1, -1 \] ### 2. **When is \( f(x) > 0 \)?** For \( f(x) \) to be positive: - The **numerator** (\( \sqrt{x+2} \)) must be positive. - The **denominator** (\( 3 - 3x^2 \)) must also be positive. #### **Numerator Analysis:** - \( \sqrt{x+2} > 0 \) implies \( x > -2 \). #### **Denominator Analysis:** - \( 3 - 3x^2 > 0 \Rightarrow 1 - x^2 > 0 \). - Solving \( 1 - x^2 > 0 \): - \( x^2 < 1 \). - \( -1 < x < 1 \). ### 3. **Combining the Conditions** - From the numerator: \( x > -2 \). - From the denominator: \( -1 < x < 1 \). The combined conditions require: \[ -1 < x < 1 \] ### 4. **Excluding Undefined Points** - \( x = -1 \) and \( x = 1 \) are excluded because they make the denominator zero. ### **Conclusion** \( f(x) > 0 \) for all real numbers \( x \) such that: \[ -1 < x < 1 \] **Answer:** All real numbers x with x between –1 and 1. In other words, f (x ) > 0 for –1 < x < 1