To solve the problem, we need to approximate $f(2.1)$ using the tangent line to the graph of $f$ at $x = 2$. Here's the step-by-step process:

Step 1: The equation of the tangent line

The general equation of a tangent line is: $y - f(a) = f'(a)(x - a)$ Here:

• $a = 2$,
• $f(2) = 4$,
• $f'(2) = -\frac{1}{2}$.

Substitute these values into the equation: $y - 4 = -\frac{1}{2}(x - 2)$ Simplify: $y = -\frac{1}{2}(x - 2) + 4$ $y = -\frac{1}{2}x + 1 + 4$ $y = -\frac{1}{2}x + 5$

Step 2: Approximate $f(2.1)$ using the tangent line

Plug $x = 2.1$ into the tangent line equation: $y = -\frac{1}{2}(2.1) + 5$ $y = -1.05 + 5$ $y = 3.95$

Step 3: Conclusion

The approximate value of $f(2.1)$ is 3.95, which corresponds to option B.

Would you like further explanation or related examples? Here are some related questions for practice:

1. What is the significance of the tangent line in approximating function values?
2. How does the derivative at a point affect the slope of the tangent line?
3. If $f'(2)$ were positive, how would it affect the approximation of $f(2.1)$?
4. Can the tangent line provide an exact value of $f(x)$ for all $x$? Why or why not?
5. How does the approximation improve as $x$ gets closer to $a$ (e.g., $x = 2$)?

Tip: When approximating using tangent lines, always check that the $x$-value you're using is close to the point of tangency for better accuracy.