1. **State the problem:** We want to estimate the value of $\sqrt{8.9}$ using local linear approximation. 2. **Formula and concept:** Local linear approximation uses the tangent line at a point close to the value we want to estimate. The formula is: $$f(x) \approx f(a) + f'(a)(x - a)$$ where $a$ is a point near $x$ where $f(a)$ and $f'(a)$ are easy to calculate. 3. **Choose a function and point:** Let $f(x) = \sqrt{x} = x^{1/2}$. Choose $a = 9$ because $\sqrt{9} = 3$ is exact and close to 8.9. 4. **Calculate $f(a)$:** $$f(9) = \sqrt{9} = 3$$ 5. **Calculate the derivative $f'(x)$:** $$f'(x) = \frac{1}{2\sqrt{x}} = \frac{1}{2x^{1/2}}$$ 6. **Calculate $f'(a)$:** $$f'(9) = \frac{1}{2\sqrt{9}} = \frac{1}{2 \times 3} = \frac{1}{6}$$ 7. **Apply the linear approximation formula:** $$f(8.9) \approx f(9) + f'(9)(8.9 - 9) = 3 + \frac{1}{6} \times (-0.1) = 3 - \frac{0.1}{6} = 3 - 0.0167 = 2.9833$$ 8. **Interpretation:** The estimated value of $\sqrt{8.9}$ using local linear approximation is approximately $2.9833$. This method works well because 8.9 is close to 9, where the function and its derivative are easy to compute.