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.