1. We want to find where the function \( k(x) = \frac{1}{x^2 + 3} \) is convex downward. \n2. Recall that a function is convex downward where its second derivative \( k''(x) < 0 \). \n3. First, find the first derivative using the quotient or chain rule: \n\[ k'(x) = -\frac{2x}{(x^2 + 3)^2} \] \n4. Now find the second derivative: \n\[ k''(x) = -\frac{d}{dx} \left( \frac{2x}{(x^2 + 3)^2} \right) = - \left( \frac{2(x^2 + 3)^2 - 2x \cdot 2 (x^2 + 3)(2x)}{(x^2 + 3)^4} \right) \] \n Simplify numerator: \n\[ 2(x^2 + 3)^2 - 8x^2 (x^2 + 3) = 2(x^2 + 3) [ (x^2 + 3) - 4x^2 ] = 2(x^2 + 3)(-3x^2 + 3) = 6(x^2 + 3)(1 - x^2) \] \n So: \n\[ k''(x) = - \frac{6 (x^2 + 3)(1 - x^2)}{(x^2 + 3)^4} = - \frac{6 (1 - x^2)}{(x^2 + 3)^3} \] \n5. Since \((x^2 + 3)^3 > 0\) for all real \(x\), the sign of \( k''(x) \) depends on \( -6(1 - x^2) \). \n6. Set \( k''(x) < 0 \) for convex downward: \n\[ -6(1 - x^2) < 0 \implies 1 - x^2 > 0 \implies -1 < x < 1 \] \n7. Therefore, the function is convex downward on the interval \( ] -1, 1 [ \). \n\nFinal answer: **Option D) ] -1, 1 [**