1. **State the problem:** Determine which statement is true about the extreme points of the function $$f(x) = x^3 - 3x$$. 2. **Find the first derivative:** To find extreme points, calculate $$f'(x)$$: $$f'(x) = \frac{d}{dx}(x^3 - 3x) = 3x^2 - 3$$ 3. **Find critical points:** Set $$f'(x) = 0$$ to find critical points: $$3x^2 - 3 = 0 \implies x^2 = 1 \implies x = \pm 1$$ 4. **Determine the nature of critical points:** Use the second derivative test: $$f''(x) = \frac{d}{dx}(3x^2 - 3) = 6x$$ Evaluate at $$x=1$$: $$f''(1) = 6(1) = 6 > 0$$, so $$x=1$$ is a local minimum. Evaluate at $$x=-1$$: $$f''(-1) = 6(-1) = -6 < 0$$, so $$x=-1$$ is a local maximum. 5. **Conclusion:** The point $$x=1$$ is a minimum and $$x=-1$$ is a maximum. **Answer:** (B) $$x=-1$$ & $$1$$ are points of maxima & minima respectively.