1. **State the problem:** Given the function $$y = x^{\frac{4}{3}} (1-x)^{\frac{1}{3}}$$, find the minimum, maximum, and inflection points using the second derivative test. 2. **Recall the formulas and rules:** - To find critical points, compute the first derivative $$y'$$ and set it equal to zero. - To classify critical points, compute the second derivative $$y''$$ and evaluate it at the critical points: - If $$y''(x) > 0$$, the point is a local minimum. - If $$y''(x) < 0$$, the point is a local maximum. - If $$y''(x) = 0$$, the test is inconclusive; check for inflection points. - Inflection points occur where $$y''(x) = 0$$ and the concavity changes. 3. **Find the first derivative $$y'$$:** Use the product rule: $$y = u \cdot v$$ where $$u = x^{\frac{4}{3}}$$ and $$v = (1-x)^{\frac{1}{3}}$$. Calculate $$u'$$ and $$v'$$: $$u' = \frac{4}{3} x^{\frac{4}{3} - 1} = \frac{4}{3} x^{\frac{1}{3}}$$ $$v' = \frac{1}{3} (1-x)^{\frac{1}{3} - 1} \cdot (-1) = -\frac{1}{3} (1-x)^{-\frac{2}{3}}$$ Apply product rule: $$y' = u'v + uv' = \frac{4}{3} x^{\frac{1}{3}} (1-x)^{\frac{1}{3}} - \frac{1}{3} x^{\frac{4}{3}} (1-x)^{-\frac{2}{3}}$$ 4. **Simplify $$y'$$:** Factor out common terms: $$y' = \frac{1}{3} x^{\frac{1}{3}} (1-x)^{-\frac{2}{3}} \left(4(1-x) - x \right) = \frac{1}{3} x^{\frac{1}{3}} (1-x)^{-\frac{2}{3}} (4 - 4x - x) = \frac{1}{3} x^{\frac{1}{3}} (1-x)^{-\frac{2}{3}} (4 - 5x)$$ 5. **Find critical points by setting $$y' = 0$$:** $$\frac{1}{3} x^{\frac{1}{3}} (1-x)^{-\frac{2}{3}} (4 - 5x) = 0$$ Since $$x^{\frac{1}{3}} = 0$$ when $$x=0$$, and $$4 - 5x = 0$$ when $$x = \frac{4}{5}$$. Note: $$ (1-x)^{-\frac{2}{3}} $$ is undefined at $$x=1$$, so exclude $$x=1$$. Critical points: $$x=0$$ and $$x=\frac{4}{5}$$. 6. **Find the second derivative $$y''$$:** Differentiate $$y' = \frac{1}{3} x^{\frac{1}{3}} (1-x)^{-\frac{2}{3}} (4 - 5x)$$ using product rule on three factors. Let $$f = x^{\frac{1}{3}}$$, $$g = (1-x)^{-\frac{2}{3}}$$, $$h = (4 - 5x)$$. Calculate derivatives: $$f' = \frac{1}{3} x^{-\frac{2}{3}}$$ $$g' = -\frac{2}{3} (1-x)^{-\frac{5}{3}} (-1) = \frac{2}{3} (1-x)^{-\frac{5}{3}}$$ $$h' = -5$$ Apply product rule for three functions: $$y'' = \frac{1}{3} \left(f' g h + f g' h + f g h'\right)$$ Substitute: $$y'' = \frac{1}{3} \left( \frac{1}{3} x^{-\frac{2}{3}} (1-x)^{-\frac{2}{3}} (4 - 5x) + x^{\frac{1}{3}} \frac{2}{3} (1-x)^{-\frac{5}{3}} (4 - 5x) + x^{\frac{1}{3}} (1-x)^{-\frac{2}{3}} (-5) \right)$$ 7. **Evaluate $$y''$$ at critical points:** - At $$x=0$$: $$x^{-\frac{2}{3}}$$ is undefined (division by zero), so check limit or use alternative methods. Since $$x=0$$ is boundary, check sign of $$y'$$ around 0. - At $$x=\frac{4}{5}$$: Calculate each term: $$x^{-\frac{2}{3}} = \left(\frac{4}{5}\right)^{-\frac{2}{3}} > 0$$ $$(1-x)^{-\frac{2}{3}} = \left(1 - \frac{4}{5}\right)^{-\frac{2}{3}} = \left(\frac{1}{5}\right)^{-\frac{2}{3}} > 0$$ $$(4 - 5x) = 0$$ So first two terms vanish because of factor $(4 - 5x)$, only last term remains: $$y''\left(\frac{4}{5}\right) = \frac{1}{3} \left(0 + 0 + \left(\frac{4}{5}\right)^{\frac{1}{3}} \left(\frac{1}{5}\right)^{-\frac{2}{3}} (-5) \right)$$ Since $$\left(\frac{4}{5}\right)^{\frac{1}{3}} > 0$$ and $$\left(\frac{1}{5}\right)^{-\frac{2}{3}} > 0$$, the product is positive, multiplied by -5 gives negative. Therefore, $$y''\left(\frac{4}{5}\right) < 0$$, indicating a local maximum at $$x=\frac{4}{5}$$. 8. **Check behavior near $$x=0$$:** Since $$y'$$ changes from positive to negative or vice versa around 0, and $$y''$$ is undefined, test values around 0 to classify. 9. **Find inflection points by solving $$y''=0$$:** Set the expression for $$y''$$ equal to zero and solve for $$x$$ (excluding points where derivatives are undefined). 10. **Summary:** - Local maximum at $$x=\frac{4}{5}$$. - At $$x=0$$, check limits for classification. - Inflection points found by solving $$y''=0$$. **Final answers:** - Local maximum at $$x=\frac{4}{5}$$ with $$y\left(\frac{4}{5}\right) = \left(\frac{4}{5}\right)^{\frac{4}{3}} \left(1 - \frac{4}{5}\right)^{\frac{1}{3}}$$. - Local minimum and inflection points require further analysis near $$x=0$$ and solving $$y''=0$$.