1. **Problem:** Find the critical points and analyze the function $f(x) = \frac{x^3}{9} - 3x$. 2. **Formula and rules:** To find critical points, we use the derivative $f'(x)$ and set it equal to zero: $$f'(x) = 0.$$ Critical points occur where the derivative is zero or undefined. 3. **Calculate the derivative:** $$f(x) = \frac{x^3}{9} - 3x = \frac{1}{9}x^3 - 3x$$ $$f'(x) = \frac{1}{9} \cdot 3x^2 - 3 = \frac{1}{3}x^2 - 3$$ 4. **Find critical points:** Set derivative equal to zero: $$\frac{1}{3}x^2 - 3 = 0$$ Multiply both sides by 3: $$x^2 - 9 = 0$$ $$x^2 = 9$$ $$x = \pm 3$$ 5. **Second derivative test:** Calculate second derivative: $$f''(x) = \frac{d}{dx} \left( \frac{1}{3}x^2 - 3 \right) = \frac{2}{3}x$$ Evaluate at critical points: - At $x=3$: $$f''(3) = \frac{2}{3} \times 3 = 2 > 0$$ (local minimum) - At $x=-3$: $$f''(-3) = \frac{2}{3} \times (-3) = -2 < 0$$ (local maximum) 6. **Find function values at critical points:** - At $x=3$: $$f(3) = \frac{3^3}{9} - 3 \times 3 = \frac{27}{9} - 9 = 3 - 9 = -6$$ - At $x=-3$: $$f(-3) = \frac{(-3)^3}{9} - 3 \times (-3) = \frac{-27}{9} + 9 = -3 + 9 = 6$$ 7. **Summary:** - Local maximum at $(-3, 6)$ - Local minimum at $(3, -6)$ Final answer: The function $f(x) = \frac{x^3}{9} - 3x$ has a local maximum at $x=-3$ with value 6 and a local minimum at $x=3$ with value -6.