1. **State the problem:** Find the relative maxima of the function $$y=\frac{1}{3}x^3 + x^2 - 15x + 15$$. 2. **Formula and rules:** To find relative maxima, we first find the critical points by setting the first derivative $$y'$$ equal to zero. Then, use the second derivative test to determine if those points are maxima or minima. 3. **Find the first derivative:** $$y' = \frac{d}{dx}\left(\frac{1}{3}x^3 + x^2 - 15x + 15\right) = x^2 + 2x - 15$$ 4. **Set the first derivative equal to zero to find critical points:** $$x^2 + 2x - 15 = 0$$ 5. **Factor the quadratic:** $$ (x + 5)(x - 3) = 0 $$ 6. **Solve for $$x$$:** $$x = -5 \quad \text{or} \quad x = 3$$ 7. **Find the second derivative:** $$y'' = \frac{d}{dx}(x^2 + 2x - 15) = 2x + 2$$ 8. **Evaluate the second derivative at each critical point:** - At $$x = -5$$: $$y''(-5) = 2(-5) + 2 = -10 + 2 = -8 < 0$$, so $$x = -5$$ is a relative maximum. - At $$x = 3$$: $$y''(3) = 2(3) + 2 = 6 + 2 = 8 > 0$$, so $$x = 3$$ is a relative minimum. 9. **Find the corresponding $$y$$ values:** - At $$x = -5$$: $$y = \frac{1}{3}(-5)^3 + (-5)^2 - 15(-5) + 15 = \frac{1}{3}(-125) + 25 + 75 + 15 = -\frac{125}{3} + 115 = \frac{-125 + 345}{3} = \frac{220}{3} \approx 73.33$$ - At $$x = 3$$ (for completeness): $$y = \frac{1}{3}(27) + 9 - 45 + 15 = 9 + 9 - 45 + 15 = -12$$ **Final answer:** The function has a relative maximum at $$\left(-5, \frac{220}{3}\right)$$.