1. **Problem statement:** Use linear approximation to estimate (a) $\sqrt[3]{1001}$ and (b) $8.06^{\frac{2}{3}}$. 2. **Recall linear approximation formula:** For a function $f(x)$ near $x=a$, the linear approximation is $$f(x) \approx f(a) + f'(a)(x - a)$$ ### Part (a) $\sqrt[3]{1001}$ 3. Define $f(x) = x^{1/3}$. 4. Choose $a=1000$ because $1000$ is a perfect cube and close to $1001$. 5. Calculate $f(a) = 1000^{1/3} = 10$. 6. Calculate derivative: $$f'(x) = \frac{1}{3}x^{-2/3}$$ 7. Evaluate derivative at $a=1000$: $$f'(1000) = \frac{1}{3} \times 1000^{-2/3} = \frac{1}{3} \times \frac{1}{100^{2/3}} = \frac{1}{3} \times \frac{1}{100^{2/3}}$$ Since $1000^{1/3} = 10$, then $1000^{-2/3} = 10^{-2} = \frac{1}{100}$. So, $$f'(1000) = \frac{1}{3} \times \frac{1}{100} = \frac{1}{300}$$ 8. Use linear approximation: $$f(1001) \approx f(1000) + f'(1000)(1001 - 1000) = 10 + \frac{1}{300} \times 1 = 10 + \frac{1}{300} = 10.00333$$ ### Part (b) $8.06^{2/3}$ 9. Define $g(x) = x^{2/3}$. 10. Choose $a=8$ because $8$ is a perfect cube and close to $8.06$. 11. Calculate $g(a) = 8^{2/3} = (8^{1/3})^2 = 2^2 = 4$. 12. Calculate derivative: $$g'(x) = \frac{2}{3} x^{-1/3}$$ 13. Evaluate derivative at $a=8$: $$g'(8) = \frac{2}{3} \times 8^{-1/3} = \frac{2}{3} \times \frac{1}{2} = \frac{1}{3}$$ 14. Use linear approximation: $$g(8.06) \approx g(8) + g'(8)(8.06 - 8) = 4 + \frac{1}{3} \times 0.06 = 4 + 0.02 = 4.02$$ **Final answers:** - (a) $\sqrt[3]{1001} \approx 10.00333$ - (b) $8.06^{2/3} \approx 4.02$