This problem involves estimating $f(4.8)$ using the local linear approximation (or tangent line approximation) for the function $f(x)$ at $x = 5$.

The formula for the local linear approximation is: $f(x) \approx f(a) + f'(a)(x - a),$ where:

• $a = 5$,
• $f(a) = f(5) = 3$,
• $f'(a) = f'(5) = 4$,
• $x = 4.8$.

Step-by-step solution:

1. Substitute values into the formula: $f(4.8) \approx f(5) + f'(5)(4.8 - 5).$ Substituting $f(5) = 3$, $f'(5) = 4$, and $4.8 - 5 = -0.2$, we get: $f(4.8) \approx 3 + 4(-0.2).$

2. Simplify the expression: $f(4.8) \approx 3 - 0.8 = 2.2.$

Final Answer:

The estimate for $f(4.8)$ is $\mathbf{2.2}$ (Option A).

Would you like further explanation or details on linear approximations? Here are related questions to explore:

1. How is the linear approximation derived from the tangent line equation?
2. Why is $f'(x)$ (the derivative) crucial for making these approximations?
3. How does this method differ from other numerical estimation techniques?
4. What is the accuracy of local linear approximations, and when do they fail?
5. Can the linear approximation formula be used for larger deviations from $a$?

Tip: The closer $x$ is to $a$, the more accurate the linear approximation will be!