1. **State the problem:** We are given the function $f(x) = 2x^3 - 3x^2 - 12x + 4$ and we want to analyze it. 2. **Formula and rules:** For a cubic function $f(x) = ax^3 + bx^2 + cx + d$, we can find critical points by computing the derivative $f'(x)$ and setting it to zero. Critical points help find extrema (maxima and minima). 3. **Find the derivative:** $$f'(x) = \frac{d}{dx}(2x^3 - 3x^2 - 12x + 4) = 6x^2 - 6x - 12$$ 4. **Set derivative to zero to find critical points:** $$6x^2 - 6x - 12 = 0$$ Divide both sides by 6: $$x^2 - x - 2 = 0$$ 5. **Solve quadratic equation:** $$x^2 - x - 2 = 0$$ Factor: $$(x - 2)(x + 1) = 0$$ So, $$x = 2 \quad \text{or} \quad x = -1$$ 6. **Find function values at critical points:** $$f(2) = 2(2)^3 - 3(2)^2 - 12(2) + 4 = 16 - 12 - 24 + 4 = -16$$ $$f(-1) = 2(-1)^3 - 3(-1)^2 - 12(-1) + 4 = -2 - 3 + 12 + 4 = 11$$ 7. **Determine the nature of critical points using second derivative:** $$f''(x) = \frac{d}{dx}(6x^2 - 6x - 12) = 12x - 6$$ Evaluate at $x=2$: $$f''(2) = 12(2) - 6 = 24 - 6 = 18 > 0$$ So $x=2$ is a local minimum. Evaluate at $x=-1$: $$f''(-1) = 12(-1) - 6 = -12 - 6 = -18 < 0$$ So $x=-1$ is a local maximum. **Final answer:** The function $f(x) = 2x^3 - 3x^2 - 12x + 4$ has a local maximum at $x = -1$ with value $f(-1) = 11$ and a local minimum at $x = 2$ with value $f(2) = -16$.