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!