1. **State the problem:** We are given the function $f(x) = 6x^5 + 3x^3 - 2x - 8$ and we want to analyze or work with it as needed. 2. **Identify the function type:** This is a polynomial function of degree 5. 3. **Find the derivative:** To find critical points or analyze behavior, we use the power rule: $$f'(x) = \frac{d}{dx}(6x^5) + \frac{d}{dx}(3x^3) - \frac{d}{dx}(2x) - \frac{d}{dx}(8)$$ 4. **Apply the power rule:** $$f'(x) = 6 \cdot 5x^{5-1} + 3 \cdot 3x^{3-1} - 2 \cdot 1x^{1-1} - 0$$ 5. **Simplify:** $$f'(x) = 30x^4 + 9x^2 - 2$$ 6. **Set derivative equal to zero to find critical points:** $$30x^4 + 9x^2 - 2 = 0$$ 7. **Let $y = x^2$ to solve quadratic in $y$:** $$30y^2 + 9y - 2 = 0$$ 8. **Use quadratic formula:** $$y = \frac{-9 \pm \sqrt{9^2 - 4 \cdot 30 \cdot (-2)}}{2 \cdot 30} = \frac{-9 \pm \sqrt{81 + 240}}{60} = \frac{-9 \pm \sqrt{321}}{60}$$ 9. **Calculate approximate roots:** $$\sqrt{321} \approx 17.9165$$ $$y_1 = \frac{-9 + 17.9165}{60} = \frac{8.9165}{60} \approx 0.1486$$ $$y_2 = \frac{-9 - 17.9165}{60} = \frac{-26.9165}{60} \approx -0.4486$$ 10. **Since $y = x^2$ and $x^2$ cannot be negative, discard $y_2$.** 11. **Find $x$ values:** $$x = \pm \sqrt{0.1486} \approx \pm 0.3855$$ **Final answer:** The critical points of $f(x)$ occur at approximately $x = 0.3855$ and $x = -0.3855$.