1. **Problem Statement:** Find the local maximum and minimum values of the function $$y = x^5 - 5x^4 + 5x^3 - 10$$. 2. **Formula and Rules:** To find local maxima and minima, we use the first derivative test. - Find the first derivative $$y' = \frac{dy}{dx}$$. - Solve $$y' = 0$$ to find critical points. - Use the second derivative $$y''$$ to determine the nature of each critical point: - If $$y''(x) > 0$$, local minimum. - If $$y''(x) < 0$$, local maximum. 3. **Find the first derivative:** $$y = x^5 - 5x^4 + 5x^3 - 10$$ $$y' = 5x^4 - 20x^3 + 15x^2$$ 4. **Factor the first derivative:** $$y' = 5x^2(x^2 - 4x + 3)$$ Factor quadratic: $$x^2 - 4x + 3 = (x - 1)(x - 3)$$ So, $$y' = 5x^2 (x - 1)(x - 3)$$ 5. **Find critical points by setting $$y' = 0$$:** $$5x^2 (x - 1)(x - 3) = 0$$ Critical points are $$x = 0, 1, 3$$. 6. **Find the second derivative:** $$y' = 5x^4 - 20x^3 + 15x^2$$ $$y'' = 20x^3 - 60x^2 + 30x$$ 7. **Evaluate $$y''$$ at each critical point:** - At $$x=0$$: $$y''(0) = 0$$ (inconclusive, test with first derivative sign change) - At $$x=1$$: $$y''(1) = 20(1)^3 - 60(1)^2 + 30(1) = 20 - 60 + 30 = -10 < 0$$ (local maximum) - At $$x=3$$: $$y''(3) = 20(27) - 60(9) + 30(3) = 540 - 540 + 90 = 90 > 0$$ (local minimum) 8. **Check behavior at $$x=0$$:** Test values around 0 in $$y'$$: - For $$x = -0.1$$, $$y' > 0$$ - For $$x = 0.1$$, $$y' > 0$$ No sign change, so no local extremum at $$x=0$$. 9. **Find function values at critical points:** - $$y(1) = 1 - 5 + 5 - 10 = -9$$ - $$y(3) = 243 - 405 + 135 - 10 = (243 - 405) + 125 = -162 + 125 = -37$$ 10. **Final answer:** - Local maximum at $$x=1$$ with value $$y = -9$$. - Local minimum at $$x=3$$ with value $$y = -37$$.