1. **State the problem:** Find the local and absolute maxima and minima of the function $$f(x) = 3x - 6 \cos(x)$$ on the interval $$[-\pi, \pi]$$. 2. **Find the derivative:** To find critical points, compute $$f'(x)$$: $$f'(x) = 3 - 6(-\sin(x)) = 3 + 6\sin(x)$$. 3. **Set derivative to zero to find critical points:** $$3 + 6\sin(x) = 0 \implies \sin(x) = -\frac{1}{2}$$. 4. **Solve for $$x$$ in $$[-\pi, \pi]$$ where $$\sin(x) = -\frac{1}{2}$$:** The solutions are $$x = -\frac{\pi}{6}$$ and $$x = -\frac{5\pi}{6}$$. 5. **Determine the nature of critical points using the second derivative:** $$f''(x) = 6\cos(x)$$. Evaluate at critical points: - At $$x = -\frac{\pi}{6}$$, $$f''\left(-\frac{\pi}{6}\right) = 6\cos\left(-\frac{\pi}{6}\right) = 6 \times \frac{\sqrt{3}}{2} = 3\sqrt{3} > 0$$, so local minimum. - At $$x = -\frac{5\pi}{6}$$, $$f''\left(-\frac{5\pi}{6}\right) = 6\cos\left(-\frac{5\pi}{6}\right) = 6 \times \left(-\frac{\sqrt{3}}{2}\right) = -3\sqrt{3} < 0$$, so local maximum. 6. **Evaluate $$f(x)$$ at critical points and endpoints to find absolute extrema:** - $$f\left(-\frac{\pi}{6}\right) = 3\left(-\frac{\pi}{6}\right) - 6\cos\left(-\frac{\pi}{6}\right) = -\frac{\pi}{2} - 6 \times \frac{\sqrt{3}}{2} = -\frac{\pi}{2} - 3\sqrt{3}$$ - $$f\left(-\frac{5\pi}{6}\right) = 3\left(-\frac{5\pi}{6}\right) - 6\cos\left(-\frac{5\pi}{6}\right) = -\frac{5\pi}{2} - 6 \times \left(-\frac{\sqrt{3}}{2}\right) = -\frac{5\pi}{2} + 3\sqrt{3}$$ - $$f(-\pi) = 3(-\pi) - 6\cos(-\pi) = -3\pi - 6(-1) = -3\pi + 6$$ - $$f(\pi) = 3\pi - 6\cos(\pi) = 3\pi - 6(-1) = 3\pi + 6$$ 7. **Compare values:** - $$f(-\pi) = -3\pi + 6 \approx -9.424 + 6 = -3.424$$ - $$f\left(-\frac{5\pi}{6}\right) \approx -\frac{5 \times 3.1416}{2} + 3 \times 1.732 = -7.854 + 5.196 = -2.658$$ - $$f\left(-\frac{\pi}{6}\right) \approx -\frac{3.1416}{2} - 3 \times 1.732 = -1.571 - 5.196 = -6.767$$ - $$f(\pi) = 3\pi + 6 \approx 9.424 + 6 = 15.424$$ 8. **Conclusions:** - Local maxima at $$x = -\frac{5\pi}{6}$$ with value $$f\left(-\frac{5\pi}{6}\right)$$. - Local minima at $$x = -\frac{\pi}{6}$$ with value $$f\left(-\frac{\pi}{6}\right)$$. - Absolute maximum at $$x = \pi$$ with value $$f(\pi)$$. - Absolute minimum at $$x = -\frac{\pi}{6}$$ with value $$f\left(-\frac{\pi}{6}\right)$$. **Final answers:** - Local maxima: $$-\frac{5\pi}{6}$$ - Local minima: $$-\frac{\pi}{6}$$ - Absolute maxima: $$\pi$$ - Absolute minima: $$-\frac{\pi}{6}$$