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To solve the problem using differentials and the graph of \( f \), we will approximate the values of \( f(5.9) \) and \( f(6.04) \) using the linear approximation formula:

\[
f(x) \approx f(a) + f'(a)(x - a),
\]

where \( a \) is a point near \( x \), and \( f'(a) \) is the derivative of \( f \) at \( a \).

### Given:
- The graph shows that \( f(6) = 5 \), and we need to estimate the values of \( f(5.9) \) and \( f(6.04) \).
- We'll use \( a = 6 \) since it's close to both 5.9 and 6.04.
- We need to determine the slope \( f'(6) \) from the graph. The line tangent to \( f \) at \( (6, 5) \) indicates the slope.

From the graph, the slope of the tangent line \( f'(6) \) seems to be approximately 1 (as it rises about 1 unit for every unit it runs).

### (a) Approximating \( f(5.9) \):
Using the linear approximation formula:
\[
f(5.9) \approx f(6) + f'(6)(5.9 - 6).
\]
Substitute the values:
\[
f(5.9) \approx 5 + 1 \cdot (5.9 - 6) = 5 + 1 \cdot (-0.1) = 5 - 0.1 = 4.9.
\]

### (b) Approximating \( f(6.04) \):
Using the linear approximation formula:
\[
f(6.04) \approx f(6) + f'(6)(6.04 - 6).
\]
Substitute the values:
\[
f(6.04) \approx 5 + 1 \cdot (6.04 - 6) = 5 + 1 \cdot 0.04 = 5 + 0.04 = 5.04.
\]

### Final Answers:
- \( f(5.9) \approx 4.90 \)
- \( f(6.04) \approx 5.04 \)

Would you like a more detailed explanation, or do you have any questions?

### Related Questions:
1. How does the differential method help in approximating function values?
2. Why is it important to choose a point \( a \) close to \( x \) for linear approximation?
3. How do you determine the slope of the tangent line from a graph?
4. What are the limitations of using linear approximation?
5. Can linear approximation be used for functions with non-continuous derivatives?

### Tip:
Linear approximation is most accurate when \( x \) is very close to \( a \). For larger deviations, higher-order approximations like quadratic or cubic may be necessary for better accuracy.

Use differentials and the graph of f to approximate the following values: f(5.9) and f(6.04).

Learn how to use differentials and linear approximation to estimate the values of f(5.9) and f(6.04) based on the graph of f. This problem covers key calculus concepts such as tangent line approximation and differential equations. Suitable for students in Grades 10-12.

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