Use differential to find an approximate value of √(2.99)^2 + √(4.03)^2.

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.

Learn how to find an approximate value for √(2.99)^2 + √(4.03)^2 using differentials and linear approximation. This step-by-step guide covers differential calculus, square root approximations, and key formulas, making it suitable for students in Grades 10-12.

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