The local linear approximation of a function $f(x)$ at a point $x_0$ is given by the formula:

$L(x) = f(x_0) + f'(x_0)(x - x_0)$

Step 1: Find $f(x)$ and $f'(x)$

Given the function $f(x) = \sqrt{1 + x}$, we first need to compute its derivative $f'(x)$.

$f(x) = (1 + x)^{1/2}$ $f'(x) = \frac{1}{2}(1 + x)^{-1/2} \cdot 1 = \frac{1}{2\sqrt{1 + x}}$

Step 2: Evaluate $f(x_0)$ and $f'(x_0)$ at $x_0 = 0$

$f(0) = \sqrt{1 + 0} = 1$ $f'(0) = \frac{1}{2\sqrt{1 + 0}} = \frac{1}{2}$

Step 3: Write the local linear approximation $L(x)$

Using the formula for the linear approximation:

$L(x) = f(0) + f'(0)(x - 0)$ $L(x) = 1 + \frac{1}{2}(x)$ $L(x) = 1 + \frac{x}{2}$

Step 4: Approximate $\sqrt{0.9}$ and $\sqrt{1.1}$

• To approximate $\sqrt{0.9}$, let $x = -0.1$: $L(-0.1) = 1 + \frac{-0.1}{2} = 1 - 0.05 = 0.95$

• To approximate $\sqrt{1.1}$, let $x = 0.1$: $L(0.1) = 1 + \frac{0.1}{2} = 1 + 0.05 = 1.05$

Final Answer:

• $\sqrt{0.9} \approx 0.95$
• $\sqrt{1.1} \approx 1.05$

Would you like more details or have any questions?

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Here are 5 related questions:

1. How does the local linear approximation differ from a Taylor series expansion?
2. What is the error between the approximation and the exact value for $\sqrt{0.9}$?
3. Can we apply the same method to approximate $\sqrt{0.8}$ or $\sqrt{1.2}$?
4. How does the choice of $x_0$ affect the accuracy of the approximation?
5. What are other applications of linear approximation in calculus?

Tip: Local linear approximations are particularly accurate when the point of interest is close to the point where the approximation is made.