1. **Problem:** Find the critical points of $F(x) = x^3 - 3x^2$ and classify them as maxima or minima. 2. **Step 1: Find the first derivative** Use the power rule: $$\frac{d}{dx} x^n = n x^{n-1}$$ $$F'(x) = \frac{d}{dx}(x^3) - \frac{d}{dx}(3x^2) = 3x^2 - 6x$$ 3. **Step 2: Find critical points** Critical points occur where $F'(x) = 0$: $$3x^2 - 6x = 0$$ Factor out $3x$: $$3x(x - 2) = 0$$ So, $x = 0$ or $x = 2$. 4. **Step 3: Classify critical points using the second derivative test** Find the second derivative: $$F''(x) = \frac{d}{dx}(3x^2 - 6x) = 6x - 6$$ Evaluate at $x=0$: $$F''(0) = 6(0) - 6 = -6 < 0$$ Since $F''(0) < 0$, $x=0$ is a local maximum. Evaluate at $x=2$: $$F''(2) = 6(2) - 6 = 12 - 6 = 6 > 0$$ Since $F''(2) > 0$, $x=2$ is a local minimum. 5. **Step 4: Find coordinates of critical points** Calculate $F(0)$: $$F(0) = 0^3 - 3(0)^2 = 0$$ Calculate $F(2)$: $$F(2) = 2^3 - 3(2)^2 = 8 - 12 = -4$$ **Final answer:** - Critical points at $(0,0)$ (local maximum) - Critical points at $(2,-4)$ (local minimum)