The question in the uploaded image involves using a local linear approximation (or tangent line approximation) to estimate $f(4.8)$. Here's the breakdown and solution:

Problem Details:

• Given:
  • $f(5) = 3$
  • $f'(5) = 4$
• Estimate $f(4.8)$ using the local linear approximation.

Solution:

The local linear approximation formula is: $f(x) \approx f(a) + f'(a)(x - a)$ where $a$ is the point of tangency, and $x$ is the value we want to approximate.

Here:

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

Substitute these values into the formula: $f(4.8) \approx f(5) + f'(5)(4.8 - 5)$ $f(4.8) \approx 3 + 4(-0.2)$ $f(4.8) \approx 3 - 0.8$ $f(4.8) \approx 2.2$

Thus, the estimate for $f(4.8)$ is $2.2$, which matches choice A.

Do you want further clarification or more details?

Here are 5 follow-up questions to expand your understanding:

1. What is the geometric interpretation of a tangent line approximation?
2. How would the approximation change if $f'(5)$ were negative?
3. What is the error involved in using linear approximation versus the actual value?
4. How could you use a quadratic approximation to improve accuracy?
5. Can you derive the formula for local linear approximation from the definition of a derivative?

Tip:

For small changes near $a$, the tangent line is a very close estimate of the function. However, as $x$ moves further away, the approximation may deviate significantly!