Step 1: Use Linear Approximation to Estimate $f(3.1)$

The function is given as $f(x) = \sqrt{x + 1}$.

Linear Approximation Formula:

$f(x) \approx f(a) + f'(a)(x - a)$ Here, $a = 3$, and $x = 3.1$.

Calculate $f(a)$ at $a = 3$:

$f(3) = \sqrt{3 + 1} = \sqrt{4} = 2$

Find the derivative $f'(x)$:

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

Evaluate $f'(a)$ at $a = 3$:

$f'(3) = \frac{1}{2\sqrt{3 + 1}} = \frac{1}{2\sqrt{4}} = \frac{1}{4}$

Use the Linear Approximation Formula:

$f(3.1) \approx f(3) + f'(3)(3.1 - 3)$ $f(3.1) \approx 2 + \frac{1}{4}(3.1 - 3)$ $f(3.1) \approx 2 + \frac{1}{4}(0.1) = 2 + 0.025 = 2.025$

Thus, the linear approximation estimate is: $f(3.1) \approx 2.025$

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Step 2: Calculate the Exact Value of $f(3.1)$

$f(3.1) = \sqrt{3.1 + 1} = \sqrt{4.1}$

To calculate $\sqrt{4.1}$, we approximate: $\sqrt{4.1} \approx 2.024845673$

Thus, the exact value is approximately: $f(3.1) \approx 2.0248$

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Step 3: Compare Linear Approximation and Exact Value

The linear approximation yielded: $f(3.1) \approx 2.025$ The exact value is: $f(3.1) \approx 2.0248$

The two results are extremely close, differing only by: $\text{Difference} = |2.025 - 2.0248| = 0.0002$

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Step 4: Explanation of Why Linear Approximation is Reasonable

The linear approximation provides a good estimate for $f(x)$ near $x = 3$ because:

1. The function $f(x) = \sqrt{x + 1}$ is smooth and differentiable around $x = 3$.
2. For values close to $x = 3$, the tangent line at $x = 3$ closely matches the curve of $f(x)$.
3. The derivative $f'(x)$ does not change drastically near $x = 3$, so the linear model remains valid for small changes in $x$.

Thus, linear approximation is effective for small deviations from $x = 3$.

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Do you have any questions or need further clarification? Here are related questions to explore:

1. How do we calculate higher-order approximations like quadratic or cubic?
2. What are the limitations of linear approximations for functions with rapid changes?
3. How would the error increase if $x$ were farther from $3$?
4. Can we generalize this process for other functions $f(x)$?
5. How does the second derivative influence the accuracy of linear approximation?

Tip: Linear approximations work best for functions with small curvature (low second derivatives) near the point of interest.