1. **Problem statement:** Find the coordinates of all absolute maximums and minimums of the function $$f(x) = (x^2 - 3)^{\frac{2}{3}}$$ on the interval $$[-8, 2]$$. 2. **Formula and rules:** To find absolute extrema on a closed interval, evaluate the function at critical points (where $$f'(x) = 0$$ or undefined) and at the endpoints. 3. **Find the derivative:** $$f(x) = (x^2 - 3)^{\frac{2}{3}}$$ Using the chain rule: $$f'(x) = \frac{2}{3}(x^2 - 3)^{-\frac{1}{3}} \cdot 2x = \frac{4x}{3(x^2 - 3)^{\frac{1}{3}}}$$ 4. **Find critical points:** Set numerator and denominator conditions: - Numerator zero: $$4x = 0 \Rightarrow x = 0$$ - Denominator zero: $$x^2 - 3 = 0 \Rightarrow x = \pm \sqrt{3}$$ (derivative undefined here) 5. **Evaluate $$f(x)$$ at critical points and endpoints:** - At $$x = -8$$: $$f(-8) = ((-8)^2 - 3)^{\frac{2}{3}} = (64 - 3)^{\frac{2}{3}} = 61^{\frac{2}{3}}$$ - At $$x = -\sqrt{3}$$: $$f(-\sqrt{3}) = (3 - 3)^{\frac{2}{3}} = 0$$ - At $$x = 0$$: $$f(0) = (-3)^{\frac{2}{3}} = ((-3)^2)^{\frac{1}{3}} = 9^{\frac{1}{3}} = \sqrt[3]{9}$$ - At $$x = \sqrt{3}$$: $$f(\sqrt{3}) = 0$$ - At $$x = 2$$: $$f(2) = (4 - 3)^{\frac{2}{3}} = 1^{\frac{2}{3}} = 1$$ 6. **Compare values:** - $$f(-8) = 61^{\frac{2}{3}} \approx 15.53$$ - $$f(-\sqrt{3}) = 0$$ - $$f(0) = \sqrt[3]{9} \approx 2.08$$ - $$f(\sqrt{3}) = 0$$ - $$f(2) = 1$$ 7. **Conclusion:** - Absolute maximum at $$x = -8$$ with $$f(-8) \approx 15.53$$ - Absolute minimums at $$x = -\sqrt{3}$$ and $$x = \sqrt{3}$$ with $$f = 0$$ **Final answer:** - Absolute maximum: $$(-8, 61^{\frac{2}{3}})$$ - Absolute minimums: $$(-\sqrt{3}, 0)$$ and $$(\sqrt{3}, 0)$$