Question 1209855
Let's analyze the function f(x) = √(10 - x) + √(x + 10).

**1. Natural Domain:**

* **√(10 - x):** 10 - x ≥ 0 => x ≤ 10
* **√(x + 10):** x + 10 ≥ 0 => x ≥ -10

Therefore, the domain is the intersection of these two conditions: -10 ≤ x ≤ 10, or [-10, 10].

**2. Even or Odd:**

* **Even:** f(-x) = f(x)
* **Odd:** f(-x) = -f(x)

Let's test f(-x):

f(-x) = √(10 - (-x)) + √((-x) + 10)
f(-x) = √(10 + x) + √(10 - x)

Comparing:

* f(-x) = √(10 + x) + √(10 - x)
* f(x) = √(10 - x) + √(10 + x)

Therefore, f(-x) = f(x), and the function is **even**.

Since the function is even, it cannot be odd.

**3. Increasing or Decreasing:**

* To determine if the function is increasing or decreasing, we need to analyze its derivative, f'(x).

f(x) = (10 - x)^(1/2) + (x + 10)^(1/2)

Now, find the derivative:

f'(x) = (1/2)(10 - x)^(-1/2)(-1) + (1/2)(x + 10)^(-1/2)(1)
f'(x) = -1 / (2√(10 - x)) + 1 / (2√(x + 10))

To find when f'(x) > 0 (increasing) and f'(x) < 0 (decreasing), we need to set f'(x) = 0:

-1 / (2√(10 - x)) + 1 / (2√(x + 10)) = 0
1 / (2√(x + 10)) = 1 / (2√(10 - x))
√(x + 10) = √(10 - x)
x + 10 = 10 - x
2x = 0
x = 0

Now, let's test intervals:

* **For -10 ≤ x < 0:**
    * Let's test x = -9:
    * f'(-9) = -1 / (2√(10 - (-9))) + 1 / (2√(-9 + 10))
    * f'(-9) = -1 / (2√19) + 1 / (2√1)
    * f'(-9) ≈ -0.114 + 0.5
    * f'(-9) > 0 (increasing)

* **For 0 < x ≤ 10:**
    * Let's test x = 9:
    * f'(9) = -1 / (2√(10 - 9)) + 1 / (2√(9 + 10))
    * f'(9) = -1 / (2√1) + 1 / (2√19)
    * f'(9) ≈ -0.5 + 0.114
    * f'(9) < 0 (decreasing)

Therefore, the function is increasing on [-10, 0] and decreasing on [0, 10].

**4. Invertibility:**

* Since the function is not strictly increasing or decreasing over its entire domain, it is **not invertible**.

**Conclusion:**

* **Domain:** [-10, 10]
* **Even**
* **Invertible:** No
* **Increasing:** [-10, 0]
* **Decreasing:** [0, 10]