1. **Problem Statement:** Find the slope of the tangent line to the function $$y = (2\sqrt{x} + 1)(x^3 - 6)$$ at $$x = 0$$. 2. **Formula and Rules:** The slope of the tangent line at a point is the derivative $$\frac{dy}{dx}$$ evaluated at that point. 3. **Step 1: Rewrite the function for easier differentiation.** $$y = (2x^{\frac{1}{2}} + 1)(x^3 - 6)$$ 4. **Step 2: Use the product rule for differentiation:** If $$y = u \cdot v$$, then $$\frac{dy}{dx} = u'v + uv'$$. Here, $$u = 2x^{\frac{1}{2}} + 1$$ and $$v = x^3 - 6$$. 5. **Step 3: Find derivatives of $$u$$ and $$v$$:** $$u' = 2 \cdot \frac{1}{2} x^{-\frac{1}{2}} = x^{-\frac{1}{2}} = \frac{1}{\sqrt{x}}$$ $$v' = 3x^2$$ 6. **Step 4: Apply the product rule:** $$\frac{dy}{dx} = u'v + uv' = \frac{1}{\sqrt{x}}(x^3 - 6) + (2\sqrt{x} + 1)(3x^2)$$ 7. **Step 5: Simplify each term:** First term: $$\frac{1}{\sqrt{x}}(x^3 - 6) = \frac{x^3}{\sqrt{x}} - \frac{6}{\sqrt{x}} = x^{3 - \frac{1}{2}} - 6x^{-\frac{1}{2}} = x^{\frac{5}{2}} - 6x^{-\frac{1}{2}}$$ Second term: $$ (2\sqrt{x} + 1)(3x^2) = 3x^2 \cdot 2x^{\frac{1}{2}} + 3x^2 \cdot 1 = 6x^{2 + \frac{1}{2}} + 3x^2 = 6x^{\frac{5}{2}} + 3x^2$$ 8. **Step 6: Combine terms:** $$\frac{dy}{dx} = x^{\frac{5}{2}} - 6x^{-\frac{1}{2}} + 6x^{\frac{5}{2}} + 3x^2 = (1 + 6)x^{\frac{5}{2}} + 3x^2 - 6x^{-\frac{1}{2}} = 7x^{\frac{5}{2}} + 3x^2 - 6x^{-\frac{1}{2}}$$ 9. **Step 7: Evaluate the derivative at $$x=0$$:** Note that $$x^{-\frac{1}{2}} = \frac{1}{\sqrt{x}}$$ is undefined at $$x=0$$, so the derivative does not exist at $$x=0$$. 10. **Step 8: Check the original function's behavior at $$x=0$$:** $$y(0) = (2\sqrt{0} + 1)(0^3 - 6) = (0 + 1)(-6) = -6$$ 11. **Step 9: Use the definition of derivative (limit) to find slope at $$x=0$$:** $$m = \lim_{h \to 0} \frac{y(h) - y(0)}{h}$$ Calculate $$y(h)$$: $$y(h) = (2\sqrt{h} + 1)(h^3 - 6) = (2h^{\frac{1}{2}} + 1)(h^3 - 6)$$ Expand: $$= (2h^{\frac{1}{2}})(h^3 - 6) + 1 \cdot (h^3 - 6) = 2h^{\frac{7}{2}} - 12h^{\frac{1}{2}} + h^3 - 6$$ So, $$\frac{y(h) - y(0)}{h} = \frac{2h^{\frac{7}{2}} - 12h^{\frac{1}{2}} + h^3 - 6 + 6}{h} = \frac{2h^{\frac{7}{2}} - 12h^{\frac{1}{2}} + h^3}{h}$$ Simplify powers: $$= 2h^{\frac{7}{2} - 1} - 12h^{\frac{1}{2} - 1} + h^{3 - 1} = 2h^{\frac{5}{2}} - 12h^{-\frac{1}{2}} + h^2$$ As $$h \to 0$$, $$h^{-\frac{1}{2}} \to \infty$$, so the limit does not exist. 12. **Conclusion:** The slope of the tangent line at $$x=0$$ does not exist because the derivative is undefined and the limit diverges. **Final answer:** The slope of the tangent line to the function at $$x=0$$ does not exist.