1. The problem involves finding the equation of the tangent line to the function $f(x) = \sqrt[3]{x}$ at $x=8$ and using it to approximate $\sqrt[3]{11}$. 2. Recall the formula for the derivative of $f(x) = x^{1/3}$: $$f'(x) = \frac{1}{3}x^{-2/3} = \frac{1}{3\sqrt[3]{x^2}}.$$ This derivative gives the slope of the tangent line at any point $x$. 3. Calculate the slope at $x=8$: $$f'(8) = \frac{1}{3\sqrt[3]{8^2}} = \frac{1}{3\sqrt[3]{64}} = \frac{1}{3 \times 4} = \frac{1}{12}.$$ 4. Find the point on the curve at $x=8$: $$f(8) = \sqrt[3]{8} = 2.$$ So the point is $(8, 2)$. 5. Use the point-slope form of the line: $$y - y_1 = m(x - x_1),$$ where $m = \frac{1}{12}$ and $(x_1, y_1) = (8, 2)$. So, $$y - 2 = \frac{1}{12}(x - 8).$$ 6. Simplify the tangent line equation: $$y = 2 + \frac{1}{12}(x - 8) = 2 + \frac{x}{12} - \frac{8}{12} = 2 + \frac{x}{12} - \frac{2}{3} = \frac{6}{3} - \frac{2}{3} + \frac{x}{12} = \frac{4}{3} + \frac{x}{12}.$$ 7. To approximate $\sqrt[3]{11}$, substitute $x=11$ into the tangent line: $$y \approx \frac{4}{3} + \frac{11}{12} = \frac{16}{12} + \frac{11}{12} = \frac{27}{12} = \frac{9}{4} = 2.25.$$ 8. The exact cube root of 11 is about 2.224, so the approximation $2.25$ is close. Final answer: The equation of the tangent line is $$y = \frac{4}{3} + \frac{x}{12}$$ and the approximation for $\sqrt[3]{11}$ is $$\frac{9}{4} = 2.25.$$