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To approximate \( \frac{1}{0.254} \) using linear approximation, we will follow these steps:

1. Let \( f(x) = \frac{1}{x} \).
2. Use a "nice" point \( x_0 \) near 0.254 to find the equation of the tangent line to \( f(x) \) at that point. A convenient point close to 0.254 is \( x_0 = 0.25 \), since \( \frac{1}{0.25} = 4 \).

### Steps to Solve:
1. **Find the derivative of \( f(x) \)**:  
   \( f'(x) = -\frac{1}{x^2} \).
   
2. **Evaluate \( f(x_0) \) and \( f'(x_0) \) at \( x_0 = 0.25 \)**:  
   - \( f(0.25) = \frac{1}{0.25} = 4 \).  
   - \( f'(0.25) = -\frac{1}{(0.25)^2} = -16 \).

3. **Tangent line equation at \( x_0 = 0.25 \)**:  
   Using the point-slope form of the equation of a line, the tangent line is given by:  
   \( L(x) = f(x_0) + f'(x_0)(x - x_0) \),  
   which simplifies to:  
   \[
   L(x) = 4 + (-16)(x - 0.25) = 4 - 16(x - 0.25).
   \]

4. **Approximate \( \frac{1}{0.254} \)** using the tangent line \( L(x) \) at \( x = 0.254 \):  
   \[
   L(0.254) = 4 - 16(0.254 - 0.25) = 4 - 16(0.004) = 4 - 0.064 = 3.936.
   \]

Thus, the linear approximation of \( \frac{1}{0.254} \) is approximately **3.936**.

Would you like any further details or clarifications on this solution?

Here are 5 related questions to expand on this:

1. Why is linear approximation useful for functions like \( f(x) = \frac{1}{x} \)?
2. How accurate is this approximation compared to the exact value of \( \frac{1}{0.254} \)?
3. Could we use another nearby value of \( x \) for a better approximation?
4. What other functions are well-approximated by their tangent lines?
5. How does the error in linear approximation depend on the distance from the tangent point?

**Tip**: When using linear approximation, always choose a point \( x_0 \) where the function and its derivative are easy to compute!

Use linear approximation, i.e. the tangent line, to approximate 1/0.254. Let f(x) = 1/x and find the equation of the tangent line to f(x) at a 'nice' point near 0.254. Then use this to approximate 1/0.254.

Learn how to use linear approximation to estimate 1/0.254 by finding the equation of the tangent line to f(x) = 1/x near a point close to 0.254. This solution covers key concepts such as differentiation, tangent lines, and linear approximation. Suitable for students in Grades 11-12.

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