1. **State the problem:** Find the maximum and minimum values of the function $$f(x) = 3x^4 - 2x^3 - 6x^2 + 6x + 1$$. 2. **Formula and rules:** To find maxima and minima, we first find the critical points by setting the first derivative $$f'(x)$$ equal to zero. Then, use the second derivative test to classify each critical point. 3. **Find the first derivative:** $$f'(x) = \frac{d}{dx}(3x^4 - 2x^3 - 6x^2 + 6x + 1) = 12x^3 - 6x^2 - 12x + 6$$ 4. **Set the first derivative equal to zero to find critical points:** $$12x^3 - 6x^2 - 12x + 6 = 0$$ Divide both sides by 6: $$2x^3 - x^2 - 2x + 1 = 0$$ 5. **Factor the cubic equation:** Try rational roots: test $x=1$: $$2(1)^3 - (1)^2 - 2(1) + 1 = 2 - 1 - 2 + 1 = 0$$ So, $x=1$ is a root. Divide the cubic by $(x-1)$: $$2x^3 - x^2 - 2x + 1 = (x-1)(2x^2 + x - 1)$$ 6. **Factor the quadratic:** $$2x^2 + x - 1 = 0$$ Use quadratic formula: $$x = \frac{-1 \pm \sqrt{1^2 - 4(2)(-1)}}{2 \cdot 2} = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4}$$ So, $$x = \frac{-1 + 3}{4} = \frac{2}{4} = 0.5$$ $$x = \frac{-1 - 3}{4} = \frac{-4}{4} = -1$$ 7. **Critical points are:** $$x = -1, 0.5, 1$$ 8. **Find the second derivative:** $$f''(x) = \frac{d}{dx}(12x^3 - 6x^2 - 12x + 6) = 36x^2 - 12x - 12$$ 9. **Evaluate the second derivative at each critical point:** - At $x = -1$: $$f''(-1) = 36(-1)^2 - 12(-1) - 12 = 36 + 12 - 12 = 36 > 0$$ So, $x = -1$ is a local minimum. - At $x = 0.5$: $$f''(0.5) = 36(0.5)^2 - 12(0.5) - 12 = 36(0.25) - 6 - 12 = 9 - 6 - 12 = -9 < 0$$ So, $x = 0.5$ is a local maximum. - At $x = 1$: $$f''(1) = 36(1)^2 - 12(1) - 12 = 36 - 12 - 12 = 12 > 0$$ So, $x = 1$ is a local minimum. 10. **Find the function values at critical points:** - At $x = -1$: $$f(-1) = 3(-1)^4 - 2(-1)^3 - 6(-1)^2 + 6(-1) + 1 = 3(1) + 2 - 6 - 6 + 1 = 3 + 2 - 6 - 6 + 1 = -6$$ - At $x = 0.5$: $$f(0.5) = 3(0.5)^4 - 2(0.5)^3 - 6(0.5)^2 + 6(0.5) + 1 = 3(0.0625) - 2(0.125) - 6(0.25) + 3 + 1 = 0.1875 - 0.25 - 1.5 + 3 + 1 = 2.4375$$ - At $x = 1$: $$f(1) = 3(1)^4 - 2(1)^3 - 6(1)^2 + 6(1) + 1 = 3 - 2 - 6 + 6 + 1 = 2$$ **Final answer:** - Local minima at $x = -1$ with value $$f(-1) = -6$$ - Local maximum at $x = 0.5$ with value $$f(0.5) = 2.4375$$ - Local minimum at $x = 1$ with value $$f(1) = 2$$