1. **State the problem:** Find the inflection point(s) of the function $$g(x) = \frac{x^2}{2} (9 - 2 \ln(x))$$ and analyze its concavity. 2. **Recall formulas:** - First derivative: $$g'(x) = \frac{d}{dx} \left( \frac{x^2}{2} (9 - 2 \ln(x)) \right)$$ - Second derivative: $$g''(x) = \frac{d}{dx} g'(x)$$ - Inflection points occur where $$g''(x) = 0$$ and the concavity changes. 3. **Calculate the first derivative using product rule:** $$g'(x) = \frac{d}{dx} \left( \frac{x^2}{2} \right) (9 - 2 \ln(x)) + \frac{x^2}{2} \frac{d}{dx} (9 - 2 \ln(x))$$ 4. **Compute derivatives:** - $$\frac{d}{dx} \left( \frac{x^2}{2} \right) = x$$ - $$\frac{d}{dx} (9 - 2 \ln(x)) = - \frac{2}{x}$$ 5. **Substitute:** $$g'(x) = x (9 - 2 \ln(x)) + \frac{x^2}{2} \left(- \frac{2}{x} \right) = x (9 - 2 \ln(x)) - x$$ 6. **Simplify:** $$g'(x) = x (9 - 2 \ln(x) - 1) = x (8 - 2 \ln(x))$$ 7. **Calculate the second derivative:** $$g''(x) = \frac{d}{dx} \left( x (8 - 2 \ln(x)) \right) = \frac{d}{dx} (8x) - \frac{d}{dx} (2x \ln(x))$$ 8. **Derivatives:** - $$\frac{d}{dx} (8x) = 8$$ - Use product rule for $$2x \ln(x)$$: $$\frac{d}{dx} (2x \ln(x)) = 2 \left( \ln(x) + 1 \right)$$ 9. **Substitute:** $$g''(x) = 8 - 2 (\ln(x) + 1) = 8 - 2 \ln(x) - 2 = 6 - 2 \ln(x)$$ 10. **Find inflection points by setting $$g''(x) = 0$$:** $$6 - 2 \ln(x) = 0 \implies 2 \ln(x) = 6 \implies \ln(x) = 3 \implies x = e^3$$ 11. **Evaluate concavity:** - For $$x < e^3$$, say at $$x=1$$: $$g''(1) = 6 - 2 \ln(1) = 6 - 0 = 6 > 0$$ (concave up) - For $$x > e^3$$, say at $$x=25$$: $$g''(25) = 6 - 2 \ln(25) < 0$$ (concave down) 12. **Calculate $$g(e^3)$$ for the inflection point coordinates:** $$g(e^3) = \frac{(e^3)^2}{2} (9 - 2 \ln(e^3)) = \frac{e^6}{2} (9 - 2 \times 3) = \frac{e^6}{2} (9 - 6) = \frac{e^6}{2} \times 3 = \frac{3 e^6}{2}$$ 13. **Final answer:** - Inflection point at $$\left(e^3, \frac{3 e^6}{2} \right)$$ - Concave up on $$ (0, e^3) $$ - Concave down on $$ (e^3, \infty) $$ This completes the curvature and inflection point analysis for the function $$g(x)$$.