1. **State the problem:** Find the point where the function $$f(x) = x(x - 3)(x + 3)$$ attains its local maximum. Round the answer to 2 decimal places. 2. **Rewrite the function:** Expand the function for easier differentiation. $$f(x) = x(x^2 - 9) = x^3 - 9x$$ 3. **Find the derivative:** To find local maxima or minima, compute the first derivative. $$f'(x) = \frac{d}{dx}(x^3 - 9x) = 3x^2 - 9$$ 4. **Find critical points:** Set the derivative equal to zero and solve for $$x$$. $$3x^2 - 9 = 0$$ $$3x^2 = 9$$ $$x^2 = 3$$ $$x = \pm \sqrt{3}$$ 5. **Determine which critical point is a local maximum:** Use the second derivative test. $$f''(x) = \frac{d}{dx}(3x^2 - 9) = 6x$$ Evaluate at $$x = \sqrt{3}$$: $$f''(\sqrt{3}) = 6\sqrt{3} > 0$$ (local minimum) Evaluate at $$x = -\sqrt{3}$$: $$f''(-\sqrt{3}) = -6\sqrt{3} < 0$$ (local maximum) 6. **Find the local maximum value:** Substitute $$x = -\sqrt{3}$$ into $$f(x)$$. $$f(-\sqrt{3}) = (-\sqrt{3})^3 - 9(-\sqrt{3}) = -3\sqrt{3} + 9\sqrt{3} = 6\sqrt{3} \approx 10.39$$ **Final answer:** The function attains its local maximum at $$x = -1.73$$ (rounded to 2 decimal places).