1. **State the problem:** We need to find the local minimum and local maximum of the function $$f(x) = 2x^3 - 30x^2 + 54x + 9$$ by estimating from its graph. 2. **Find the derivative:** To find local extrema, we first find the critical points by setting the derivative equal to zero. $$f'(x) = \frac{d}{dx}(2x^3 - 30x^2 + 54x + 9) = 6x^2 - 60x + 54$$ 3. **Set derivative to zero:** $$6x^2 - 60x + 54 = 0$$ Divide both sides by 6: $$\cancel{6}x^2 - \cancel{6}0x + \cancel{6}9 = 0 \Rightarrow x^2 - 10x + 9 = 0$$ 4. **Solve quadratic equation:** $$x = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot 9}}{2} = \frac{10 \pm \sqrt{100 - 36}}{2} = \frac{10 \pm \sqrt{64}}{2} = \frac{10 \pm 8}{2}$$ So, $$x_1 = \frac{10 - 8}{2} = 1$$ $$x_2 = \frac{10 + 8}{2} = 9$$ 5. **Determine nature of critical points:** Find the second derivative: $$f''(x) = \frac{d}{dx}(6x^2 - 60x + 54) = 12x - 60$$ Evaluate at $x=1$: $$f''(1) = 12(1) - 60 = 12 - 60 = -48 < 0$$ Since $f''(1) < 0$, $x=1$ is a local maximum. Evaluate at $x=9$: $$f''(9) = 12(9) - 60 = 108 - 60 = 48 > 0$$ Since $f''(9) > 0$, $x=9$ is a local minimum. 6. **Find function values at critical points:** $$f(1) = 2(1)^3 - 30(1)^2 + 54(1) + 9 = 2 - 30 + 54 + 9 = 35$$ $$f(9) = 2(9)^3 - 30(9)^2 + 54(9) + 9 = 2(729) - 30(81) + 486 + 9 = 1458 - 2430 + 486 + 9 = -477$$ **Final answer:** - Local maximum at $x=1$ with output value $35$ - Local minimum at $x=9$ with output value $-477$