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The image provides an exercise about **Errors of Linear Approximations**, focusing on the function \( f(x) = \ln(x) \) and its linearization at \( x = 1 \). The main goal is to calculate and analyze the error terms.

Here's a breakdown of the exercise:

### Part (a)
You are asked to compute the following expressions for different values of \( x \):

\[
\frac{R_1(1.02)}{(1.02 - 1)^2}, \quad \frac{R_1(1.002)}{(1.002 - 1)^2}, \quad \frac{R_1(1.0002)}{(1.0002 - 1)^2}
\]
where \( R_1(x) = f(x) - L(x) \) is the error between \( f(x) = \ln(x) \) and its linearization \( L(x) \) at \( x = 1 \).

### Linearization:
For \( f(x) = \ln(x) \), we compute the linear approximation \( L(x) \) at \( x = 1 \) using:
\[
L(x) = f(1) + f'(1)(x - 1)
\]
Since:
\[
f(1) = \ln(1) = 0 \quad \text{and} \quad f'(x) = \frac{1}{x} \quad \text{so} \quad f'(1) = 1,
\]
we get the linearization:
\[
L(x) = 0 + 1 \cdot (x - 1) = x - 1.
\]
Thus, \( R_1(x) = \ln(x) - (x - 1) \).

Now, we need to compute these error terms and divide them by \( (x - 1)^2 \) for \( x = 1.02, 1.002, \) and \( 1.0002 \).

### Part (b)
The question asks to analyze Formula 1, which is:
\[
R_1(x) = \frac{f''(c)}{2} (x - a)^2
\]
and determine:
- The factors that affect the magnitude of \( R_1(x) \),
- The factor affecting the sign of \( R_1(x) \).

Would you like me to calculate these values for part (a) and provide more detailed explanations for part (b)?

Let f(x) = ln(x). Define L(x) as the linearization of f(x) at x = 1, and R1(x) = f(x) - L(x). (a) Compute R1(1.02)/(1.02 - 1)^2, R1(1.002)/(1.002 - 1)^2, and R1(1.0002)/(1.0002 - 1)^2. What do you observe? (b) Based on Formula 1, what factors determine the magnitude of R1(x) in the linear approximation? What factor affects the sign of R1(x)?

Explore how to calculate the error in the linear approximation of ln(x) at x=1. This includes detailed computations of R1(x) for values like 1.02, 1.002, and 1.0002, along with an analysis of the factors affecting the error's magnitude and sign based on Cauchy's Mean Value Theorem.

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