To approximate $f(-4.1)$ using the tangent line to the graph of $f$ at $x = -4$, we use the linear approximation formula based on the point-slope form of the tangent line. Since $f$ is differentiable at $x = -4$, the tangent line at this point can approximate values of $f$ near $x = -4$.

Step 1: Write the equation of the tangent line

The equation of the tangent line at $x = -4$ is: $y = f(-4) + f'(-4)(x + 4)$

From the problem statement:

• $f(-4) = -5$
• $f'(-4) = 3$

Thus, the equation of the tangent line becomes: $y = -5 + 3(x + 4)$

Step 2: Substitute $x = -4.1$ into the tangent line equation

Now, we substitute $x = -4.1$ to approximate $f(-4.1)$: $f(-4.1) \approx -5 + 3(-4.1 + 4)$

Calculating inside the parentheses: $-4.1 + 4 = -0.1$

So: $f(-4.1) \approx -5 + 3(-0.1)$

Then: $f(-4.1) \approx -5 - 0.3 = -5.3$

Conclusion

The approximation for $f(-4.1)$ using the tangent line is: $f(-4.1) \approx -5.3$

Would you like more details or have any questions?

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Here are some follow-up questions you might find helpful:

1. How does linear approximation relate to differentiability?
2. Why does the tangent line provide a good approximation near $x = -4$?
3. What would change if $f'(-4)$ were negative instead of positive?
4. How do we improve this approximation using higher-order terms?
5. How can we use this approximation method to estimate values at $x$ farther from -4?

Tip: The linear approximation is more accurate for values close to the point of tangency.