To find the linear approximation $L(x)$ of the function $f(x) = \frac{1}{x}$ at $a = 2$, we will determine the equation of the tangent line at $x = 2$ and use it to estimate $f(2.1)$ by substituting $x = 2.1$ into $L(x)$.

Steps

1. Find $f(a)$: $f(2) = \frac{1}{2} = 0.5$

2. Compute $f'(x)$: The derivative of $f(x) = \frac{1}{x}$ is $f'(x) = -\frac{1}{x^2}$

3. Evaluate $f'(a)$ at $x = 2$: $f'(2) = -\frac{1}{2^2} = -\frac{1}{4} = -0.25$

4. Write the equation of the tangent line (linear approximation $L(x)$): The equation of the tangent line at $x = 2$ is given by $L(x) = f(2) + f'(2)(x - 2)$ Substituting the values we found: $L(x) = 0.5 - 0.25(x - 2)$ Simplifying, we get $L(x) = 0.5 - 0.25x + 0.5 = 1 - 0.25x$

5. Estimate $f(2.1)$ using $L(x)$: Substitute $x = 2.1$ into $L(x)$: $L(2.1) = 1 - 0.25 \cdot 2.1 = 1 - 0.525 = 0.475$

Answer

Thus, the linear approximation $L(x)$ of $f(x) = \frac{1}{x}$ at $x = 2$ is $L(x) = 1 - 0.25x$ Using this, the estimated value of $\frac{1}{2.1}$ is approximately $0.475$.

Would you like a more detailed breakdown or have any questions?

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Here are five related questions to expand your understanding:

1. What would the linear approximation be for $f(x) = \frac{1}{x}$ at $a = 3$?
2. How does the accuracy of a linear approximation change as $x$ moves further from $a$?
3. Could we use a similar approach to estimate $f(2.05)$? How close would it be?
4. What other methods could provide better approximations for values close to $a$ than the linear one?
5. How does the slope $f'(a)$ influence the approximation's accuracy?

Tip: Linear approximations work best for values of $x$ near the point $a$ where the approximation is calculated.