1. **State the problem:** Find the x-coordinate(s) of the inflection point(s) of the function $$g(x) = \frac{1}{x^2 + 1}$$. 2. **Recall inflection points:** Inflection points occur where the second derivative changes sign, i.e., where $$g''(x) = 0$$ and changes concavity. 3. **Find the first derivative $$g'(x)$$:** $$g(x) = (x^2 + 1)^{-1}$$ Using the chain rule, $$g'(x) = -1 \cdot (x^2 + 1)^{-2} \cdot 2x = -\frac{2x}{(x^2 + 1)^2}$$ 4. **Find the second derivative $$g''(x)$$:** Using quotient/product rule on $$g'(x) = -2x(x^2 + 1)^{-2}$$: $$g''(x) = -2 \cdot \left[(x^2 + 1)^{-2} + x \cdot \frac{d}{dx}(x^2 + 1)^{-2} \right]$$ The derivative inside: $$\frac{d}{dx}(x^2 + 1)^{-2} = -2 (x^2 + 1)^{-3} \cdot 2x = -\frac{4x}{(x^2 + 1)^3}$$ So, $$g''(x) = -2 \left[(x^2 + 1)^{-2} - 4x^2 (x^2 + 1)^{-3} \right] = -2 \left[ \frac{1}{(x^2 + 1)^2} - \frac{4x^2}{(x^2 + 1)^3} \right]$$ 5. **Combine terms with common denominator:** $$g''(x) = -2 \cdot \frac{(x^2 + 1) - 4x^2}{(x^2 + 1)^3} = -2 \cdot \frac{1 - 3x^2}{(x^2 + 1)^3}$$ 6. **Set $$g''(x) = 0$$ to find candidate inflection points:** Since denominator is always positive, set numerator equal to zero: $$1 - 3x^2 = 0 \implies 3x^2 = 1 \implies x^2 = \frac{1}{3} \implies x = \pm \frac{1}{\sqrt{3}}$$ 7. **Verify concavity change:** Check values around $$x=\pm \frac{1}{\sqrt{3}}$$ to confirm change of sign of $$g''(x)$$, indicating inflection points. **Final answer:** The inflection points occur at $$x = \pm \frac{1}{\sqrt{3}}$$. Answer choice: **D) x = 1 / \sqrt{3} and x = -1 / \sqrt{3}**.