To find the approximate value of $\sqrt{(2.99)^2} + \sqrt{(4.03)^2}$, we can use differentials to approximate the values of the square roots at points close to 3 and 4, respectively.

Step 1: Define functions for approximation

Let: $f(x) = \sqrt{x}$ We want to approximate $f(2.99^2)$ and $f(4.03^2)$ using differentials around the points $x = 3$ and $x = 4$, respectively.

Step 2: Use differentials for approximation

The differential approximation formula is: $f(x + \Delta x) \approx f(x) + f'(x) \cdot \Delta x$ where $f'(x)$ is the derivative of $f(x)$.

The derivative of $f(x) = \sqrt{x}$ is: $f'(x) = \frac{1}{2\sqrt{x}}$

Now, we approximate $\sqrt{2.99^2}$ and $\sqrt{4.03^2}$.

Step 3: Approximate $\sqrt{2.99^2}$

At $x = 3^2 = 9$, we have: $\sqrt{2.99^2} = f(2.99^2) \approx f(9) + f'(9) \cdot \Delta x$ where $\Delta x = 2.99^2 - 9 \approx -0.06$.

Since $f(9) = \sqrt{9} = 3$, and $f'(9) = \frac{1}{2\sqrt{9}} = \frac{1}{6}$

Using the differential: $\sqrt{2.99^2} \approx 3 + \frac{1}{6} \cdot (-0.06) \approx 3 - 0.01 = 2.99$

Step 4: Approximate $\sqrt{4.03^2}$

At $x = 4^2 = 16$, we have: $\sqrt{4.03^2} = f(4.03^2) \approx f(16) + f'(16) \cdot \Delta x$ where $\Delta x = 4.03^2 - 16 \approx 0.06$.

Since $f(16) = \sqrt{16} = 4$, and $f'(16) = \frac{1}{2\sqrt{16}} = \frac{1}{8}$

Using the differential: $\sqrt{4.03^2} \approx 4 + \frac{1}{8} \cdot 0.06 \approx 4 + 0.0075 = 4.0075$

Step 5: Add the results

Now, approximate the sum: $\sqrt{(2.99)^2} + \sqrt{(4.03)^2} \approx 2.99 + 4.0075 = 6.9975$

Thus, the approximate value is: $\boxed{6.9975}$

Would you like more details on any part of this solution?

Relative Questions:

1. How do you compute differentials for more complex functions?
2. Why is the linear approximation method valid for small changes in $x$?
3. Can you explain how higher-order approximations work?
4. What is the significance of differentials in real-world applications?
5. How would this approach change for approximating cube roots?

Tip:

Using differentials for approximation works best when the change in $x$ is small, as it provides a quick and efficient estimate without needing full computations.