To approximate $f(1.9)$ and $f(2.08)$ using differentials and the graph of $f$, we can apply linear approximation. The formula for linear approximation at $x = a$ is:

$f(x) \approx f(a) + f'(a)(x - a)$

From the graph, we can see that the point $(2, 1)$ is given, meaning:

$f(2) = 1$

Next, we need to estimate the slope $f'(2)$, which represents the derivative of $f$ at $x = 2$. The slope of the tangent line through $(2, 1)$ appears to be negative, and it seems to be approximately $-2$ based on the graph's visual representation.

(a) Approximating $f(1.9)$:

We use $a = 2$, $x = 1.9$, and $f'(2) \approx -2$:

$f(1.9) \approx f(2) + f'(2)(1.9 - 2) = 1 + (-2)(1.9 - 2)$ $f(1.9) \approx 1 + (-2)(-0.1) = 1 + 0.2 = 1.2$

(b) Approximating $f(2.08)$:

Similarly, we use $a = 2$, $x = 2.08$, and $f'(2) \approx -2$:

$f(2.08) \approx f(2) + f'(2)(2.08 - 2) = 1 + (-2)(2.08 - 2)$ $f(2.08) \approx 1 + (-2)(0.08) = 1 - 0.16 = 0.84$

Final Answers:

• $f(1.9) \approx 1.200$
• $f(2.08) \approx 0.840$

Would you like further details on linear approximation or any specific steps explained?

Here are some related questions to consider:

1. How do you calculate the derivative of a function graphically?
2. What is the concept of local linearity in calculus?
3. How does the accuracy of the linear approximation depend on the slope?
4. What are the limitations of using differentials for approximation?
5. Can you apply differentials to functions with higher degrees of complexity?

Tip: When approximating using differentials, the closer $x$ is to the point $a$, the more accurate your linear approximation will be!