1. The problem asks us to find where the function $k(x) = \frac{1}{x^2 + 3}$ is convex downward. 2. Recall that a function is convex downward where its second derivative is negative: $k''(x) < 0$. 3. Start by finding the first derivative $k'(x)$ using the quotient or chain rule: $$k'(x) = \frac{d}{dx} \left( (x^2 + 3)^{-1} \right) = -1 \cdot (x^2 + 3)^{-2} \cdot 2x = \frac{-2x}{(x^2 + 3)^2}$$ 4. Now find the second derivative $k''(x)$ using the quotient/product rule: $$k''(x) = \frac{d}{dx} \left( \frac{-2x}{(x^2 + 3)^2} \right)$$ Using the quotient rule: $$\frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{f'(x) g(x) - f(x) g'(x)}{g(x)^2}$$ where $f(x) = -2x$ and $g(x) = (x^2 + 3)^2$. Calculate $f'(x) = -2$ and $g'(x) = 2 (x^2 + 3) \cdot 2x = 4x (x^2 + 3)$. So, $$k''(x) = \frac{-2 \cdot (x^2 + 3)^2 - (-2x) \cdot 4x (x^2 + 3)}{(x^2 + 3)^4} = \frac{-2 (x^2 + 3)^2 + 8x^2 (x^2 + 3)}{(x^2 + 3)^4}$$ Factor numerator: $$= \frac{-2 (x^2 + 3)^2 + 8x^2 (x^2 + 3)}{(x^2 + 3)^4} = \frac{(x^2 + 3) (-2 (x^2 + 3) + 8x^2)}{(x^2 + 3)^4}$$ Simplify inside parentheses: $$-2 (x^2 + 3) + 8x^2 = -2 x^2 - 6 + 8 x^2 = 6 x^2 - 6 = 6 (x^2 - 1)$$ Thus, $$k''(x) = \frac{(x^2 + 3) \cdot 6 (x^2 - 1)}{(x^2 + 3)^4} = \frac{6 (x^2 - 1)}{(x^2 + 3)^3}$$ 5. Since $(x^2 + 3)^3 > 0$ for all real $x$, the sign of $k''(x)$ depends on $6 (x^2 - 1)$. 6. Solve inequality for convex downward where $k''(x) < 0$: $$6 (x^2 - 1) < 0 \implies x^2 - 1 < 0 \implies x^2 < 1 \implies -1 < x < 1$$ 7. Therefore, the function is convex downward on the interval $]-1, 1[$. Final answer: D) $] -1, 1[$