1. **Problem Statement:** Find the critical numbers, intervals of increase/decrease, and relative extrema values for the function $$f(x) = x^3 - 3x^2 + 1$$ on the interval $$[-3,3]$$. 2. **Step 1: Find the critical numbers** Critical numbers occur where the derivative $$f'(x)$$ is zero or undefined. Calculate the derivative: $$f'(x) = 3x^2 - 6x$$ Set the derivative equal to zero: $$3x^2 - 6x = 0$$ Factor out $$3x$$: $$3x(x - 2) = 0$$ So, $$x = 0$$ or $$x = 2$$ are critical numbers. 3. **Step 2: Determine intervals of increase and decrease** Test values in intervals determined by critical numbers: $$(-3,0)$$, $$(0,2)$$, and $$(2,3)$$. - For $$x < 0$$, pick $$x = -1$$: $$f'(-1) = 3(-1)^2 - 6(-1) = 3 + 6 = 9 > 0$$, so $$f$$ is increasing on $$(-3,0)$$. - For $$0 < x < 2$$, pick $$x = 1$$: $$f'(1) = 3(1)^2 - 6(1) = 3 - 6 = -3 < 0$$, so $$f$$ is decreasing on $$(0,2)$$. - For $$x > 2$$, pick $$x = 3$$: $$f'(3) = 3(3)^2 - 6(3) = 27 - 18 = 9 > 0$$, so $$f$$ is increasing on $$(2,3)$$. 4. **Step 3: Find the values of $$f(x)$$ at the relative extrema** Evaluate $$f(x)$$ at critical points: - At $$x=0$$: $$f(0) = 0^3 - 3(0)^2 + 1 = 1$$ - At $$x=2$$: $$f(2) = 2^3 - 3(2)^2 + 1 = 8 - 12 + 1 = -3$$ 5. **Summary:** - Critical numbers: $$x=0$$ and $$x=2$$. - Increasing on $$(-3,0)$$ and $$(2,3)$$. - Decreasing on $$(0,2)$$. - Relative maximum at $$x=0$$ with $$f(0)=1$$. - Relative minimum at $$x=2$$ with $$f(2)=-3$$.