https://file.aiquickdraw.com/mathbot/file/1727232971080w45lsvyn.jpg

Let's go through the steps to solve the problem visible in the image. The task asks for using linear approximation to estimate \(\sqrt{16.1}\), using the function \(f(x) = \sqrt{x}\) and the tangent line at \(x = 16\).

### Step 1: Find the equation of the tangent line at \(x = 16\)

The formula for linear approximation (tangent line approximation) is:

\[
L(x) = f(a) + f'(a)(x - a)
\]

where \(a\) is the point where the tangent line is computed (in this case \(a = 16\)).

1. **Calculate \(f(16)\):**
   \[
   f(16) = \sqrt{16} = 4
   \]
   
2. **Find \(f'(x)\):**
   The derivative of \(f(x) = \sqrt{x}\) is:
   \[
   f'(x) = \frac{1}{2\sqrt{x}}
   \]
   
   Now, calculate \(f'(16)\):
   \[
   f'(16) = \frac{1}{2\sqrt{16}} = \frac{1}{8}
   \]
   
   So, the slope \(m = f'(16) = \frac{1}{8}\).

### Step 2: Write the tangent line equation

The equation of the tangent line at \(x = 16\) is:
\[
y = f(16) + f'(16)(x - 16)
\]
Substitute the values:
\[
y = 4 + \frac{1}{8}(x - 16)
\]
This is the equation of the tangent line.

### Step 3: Approximate \(\sqrt{16.1}\)

Now, use this tangent line equation to approximate \(\sqrt{16.1}\), which corresponds to \(x = 16.1\).

Substitute \(x = 16.1\) into the tangent line equation:
\[
y = 4 + \frac{1}{8}(16.1 - 16)
\]
\[
y = 4 + \frac{1}{8}(0.1) = 4 + 0.0125 = 4.0125
\]

Therefore, the approximation for \(\sqrt{16.1}\) is **4.0125**.

### Final Answers:
- Slope \(m = \frac{1}{8} = 0.125\)
- Intercept \(b = 4\)
- Approximation of \(\sqrt{16.1} = 4.0125\)

---

Would you like further details or have any questions? 

Here are five follow-up questions related to this concept:
1. How does linear approximation differ from other methods of approximation?
2. Why is the tangent line used to approximate values of the function near a specific point?
3. How does the accuracy of the approximation change as we move further from the point of tangency?
4. Could we use linear approximation for non-continuous functions? Why or why not?
5. How would the approximation change if we used a different point, say \(x = 9\), instead of \(x = 16\)?

**Tip**: Linear approximation is generally more accurate the closer \(x\) is to the point of tangency.

Use linear approximation to estimate \(\sqrt{16.1}\) using the function \(f(x) = \sqrt{x}\) and the tangent line at \(x = 16\). Find the slope (m) and intercept (b) of the tangent line, and use this to approximate \(\sqrt{16.1}\).

Learn how to estimate the square root of 16.1 using linear approximation and the tangent line method. This solution walks through finding the slope and intercept at x = 16, using key calculus concepts like derivatives and linear approximation.

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