Consider the function f(x)=x+1−−−−√f(x)=x+1.

Use linear approximation at x=3x=3 to estimate f(3.1)f(3.1).

Calculate the exact value of f(3.1)f(3.1) and compare it to your estimate from part (1).

Explain why the linear approximation provides a reasonable estimate for values close to x=3x=3.

### 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:
\[
f(3.1) \approx 2.025
\]

---

### 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:
\[
f(3.1) \approx 2.0248
\]

---

### 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:
\[
\text{Difference} = |2.025 - 2.0248| = 0.0002
\]

---

### 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 \( x = 3 \).

---

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.

Consider the function f(x) = √(x + 1). Use linear approximation at x = 3 to estimate f(3.1). Calculate the exact value of f(3.1) and compare it to your estimate. Explain why the linear approximation provides a reasonable estimate for values close to x = 3.

Understand how to use linear approximation to estimate the value of f(3.1) for the function √(x + 1) at x = 3. Explore the step-by-step solution, exact value comparison, and reasoning behind the accuracy of linear approximations. Ideal for Grades 11-12 calculus students.

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