To estimate $\sqrt{100.6}$ using a linear approximation, we can use the concept of differentials.

1. Identify a function: Let $f(x) = \sqrt{x}$.

2. Select a point close to 100.6 where the square root is easy to compute. The closest such value is $x = 100$, because $\sqrt{100} = 10$.

3. Find $f'(x)$: We need the derivative of $f(x) = \sqrt{x}$. $f'(x) = \frac{1}{2\sqrt{x}}$

4. Compute $f'(100)$: $f'(100) = \frac{1}{2 \cdot 10} = \frac{1}{20} = 0.05$

5. Set up the linear approximation formula: The linear approximation at $x = 100$ is given by: $f(x) \approx f(100) + f'(100) \cdot (x - 100)$

6. Substitute values: $f(100.6) \approx 10 + 0.05 \cdot (100.6 - 100)$ $= 10 + 0.05 \cdot 0.6$ $= 10 + 0.03 = 10.03$

Therefore, the linear approximation estimate for $\sqrt{100.6}$ is 10.03.

Would you like further explanation on this process or have any questions?

Related Questions:

1. How would you apply linear approximation to estimate $\sqrt{99.4}$?
2. What is the general formula for linear approximation?
3. Can linear approximation be used for functions other than square roots?
4. How accurate is the linear approximation for values close to the point of approximation?
5. What happens to the accuracy of the estimate if the value is far from 100?

Tip:

Using a point close to the target value makes the linear approximation more accurate.