1. **State the problem:** We have the function $$f(x) = 2x^3 - 12x^2 + 18x - 8$$ and need to find where it is increasing, decreasing, and its local extrema. 2. **Find the derivative:** To analyze increasing/decreasing intervals and local extrema, compute $$f'(x)$$: $$f'(x) = \frac{d}{dx}(2x^3 - 12x^2 + 18x - 8) = 6x^2 - 24x + 18$$ 3. **Find critical points:** Set $$f'(x) = 0$$: $$6x^2 - 24x + 18 = 0$$ Divide both sides by 6: $$x^2 - 4x + 3 = 0$$ Factor: $$(x - 3)(x - 1) = 0$$ So critical points are at $$x = 1$$ and $$x = 3$$. 4. **Determine intervals of increase/decrease:** Test values in intervals determined by critical points: - For $$x < 1$$, pick $$x=0$$: $$f'(0) = 6(0)^2 - 24(0) + 18 = 18 > 0$$, so $$f$$ is increasing on $$(-\infty, 1)$$. - For $$1 < x < 3$$, pick $$x=2$$: $$f'(2) = 6(4) - 24(2) + 18 = 24 - 48 + 18 = -6 < 0$$, so $$f$$ is decreasing on $$(1, 3)$$. - For $$x > 3$$, pick $$x=4$$: $$f'(4) = 6(16) - 24(4) + 18 = 96 - 96 + 18 = 18 > 0$$, so $$f$$ is increasing on $$(3, \infty)$$. 5. **Find local extrema values:** Evaluate $$f(x)$$ at critical points: - At $$x=1$$: $$f(1) = 2(1)^3 - 12(1)^2 + 18(1) - 8 = 2 - 12 + 18 - 8 = 0$$ - At $$x=3$$: $$f(3) = 2(27) - 12(9) + 18(3) - 8 = 54 - 108 + 54 - 8 = -8$$ 6. **Classify extrema:** Since $$f'$$ changes from positive to negative at $$x=1$$, $$f(1)=0$$ is a local maximum. Since $$f'$$ changes from negative to positive at $$x=3$$, $$f(3)=-8$$ is a local minimum. **Final answers:** (a) Increasing on $$(-\infty, 1) \cup (3, \infty)$$ (b) Decreasing on $$(1, 3)$$ (c) Local maximum value: $$0$$ at $$x=1$$ Local minimum value: $$-8$$ at $$x=3$$