1. Problem: Analyze the function $$y = \frac{x^3}{3} - \frac{x^2}{2} - 2x + \frac{1}{3}$$ to find local maxima, minima, inflection points, and intervals of concavity. 2. Find the first derivative: $$y' = x^2 - x - 2$$ 3. Solve $$y' = 0$$ for critical points: $$x^2 - x - 2 = 0 \implies (x-2)(x+1) = 0 \implies x=2, x=-1$$ 4. Find the second derivative: $$y'' = 2x - 1$$ 5. Evaluate $$y''$$ at critical points: - At $$x=2$$: $$y''(2) = 3 > 0$$, local minimum. - At $$x=-1$$: $$y''(-1) = -3 < 0$$, local maximum. 6. Find inflection points by solving $$y''=0$$: $$2x - 1 = 0 \implies x=\frac{1}{2}$$ 7. Determine concavity intervals: - Concave up where $$y'' > 0$$: $$x > \frac{1}{2}$$ - Concave down where $$y'' < 0$$: $$x < \frac{1}{2}$$ 8. Calculate function values at critical and inflection points: - $$y(2) = \frac{8}{3} - 2 - 4 + \frac{1}{3} = -\frac{1}{3}$$ - $$y(-1) = -\frac{1}{3} - \frac{1}{2} + 2 + \frac{1}{3} = \frac{3}{2}$$ - $$y\left(\frac{1}{2}\right) = \frac{1}{24} - \frac{1}{8} - 1 + \frac{1}{3} = -\frac{7}{24}$$ Final answer: - Local maximum at $$(-1, \frac{3}{2})$$ - Local minimum at $$(2, -\frac{1}{3})$$ - Inflection point at $$\left(\frac{1}{2}, -\frac{7}{24}\right)$$ - Concave down on $$(-\infty, \frac{1}{2})$$ - Concave up on $$\left(\frac{1}{2}, \infty\right)$$