1. **State the problem:** We need to find the intervals where the function $$f(x) = x - \frac{1}{6}x^2 - \frac{1}{3}\ln x + \frac{1}{6}$$ is concave up and concave down, and also find its inflection points. 2. **Recall the formula and rules:** - The concavity of a function is determined by the sign of its second derivative $$f''(x)$$. - If $$f''(x) > 0$$ on an interval, $$f$$ is concave up there. - If $$f''(x) < 0$$ on an interval, $$f$$ is concave down there. - Inflection points occur where $$f''(x) = 0$$ and the concavity changes. 3. **Find the first derivative $$f'(x)$$:** $$f'(x) = \frac{d}{dx}\left(x - \frac{1}{6}x^2 - \frac{1}{3}\ln x + \frac{1}{6}\right) = 1 - \frac{1}{3}x - \frac{1}{3}\cdot \frac{1}{x} + 0 = 1 - \frac{1}{3}x - \frac{1}{3x}$$ 4. **Find the second derivative $$f''(x)$$:** $$f''(x) = \frac{d}{dx}\left(1 - \frac{1}{3}x - \frac{1}{3x}\right) = 0 - \frac{1}{3} - \frac{1}{3} \cdot \frac{d}{dx}\left(x^{-1}\right) = -\frac{1}{3} + \frac{1}{3x^2}$$ 5. **Simplify $$f''(x)$$:** $$f''(x) = -\frac{1}{3} + \frac{1}{3x^2} = \frac{1}{3x^2} - \frac{1}{3} = \frac{1 - x^2}{3x^2}$$ 6. **Find where $$f''(x) = 0$$ to locate possible inflection points:** $$\frac{1 - x^2}{3x^2} = 0 \implies 1 - x^2 = 0 \implies x^2 = 1 \implies x = \pm 1$$ 7. **Check domain:** Since $$f(x)$$ contains $$\ln x$$, the domain is $$x > 0$$, so only $$x = 1$$ is valid. 8. **Determine concavity intervals by testing values around $$x=1$$:** - For $$0 < x < 1$$, pick $$x=0.5$$: $$f''(0.5) = \frac{1 - (0.5)^2}{3(0.5)^2} = \frac{1 - 0.25}{3 \times 0.25} = \frac{0.75}{0.75} = 1 > 0$$ so concave up. - For $$x > 1$$, pick $$x=2$$: $$f''(2) = \frac{1 - 4}{3 \times 4} = \frac{-3}{12} = -\frac{1}{4} < 0$$ so concave down. 9. **Conclusion:** - $$f$$ is concave up on $$\left(0, 1\right)$$. - $$f$$ is concave down on $$\left(1, \infty\right)$$. - There is an inflection point at $$x=1$$. 10. **Find the inflection point coordinates:** $$f(1) = 1 - \frac{1}{6} \times 1^2 - \frac{1}{3} \ln 1 + \frac{1}{6} = 1 - \frac{1}{6} - 0 + \frac{1}{6} = 1$$ **Final answer:** - Concave up on $$\left(0, 1\right)$$ - Concave down on $$\left(1, \infty\right)$$ - Inflection point at $$(1, 1)$$