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Question 1209856: For each of the following functions, determine if the function is increasing, decreasing, even, odd, and/or invertible on its natural domain. Select all the properties that apply.
f(x) = \frac{1}{\sqrt{x^2 + 1}} - \frac{1}{x}
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Let's analyze the function f(x) = 1/√(x² + 1) - 1/x to determine its properties.
**1. Natural Domain:**
* **√(x² + 1):** x² + 1 is always positive for any real number x, so the square root is defined for all real numbers.
* **1/x:** This is undefined when x = 0.
Therefore, the natural domain of f(x) is all real numbers except 0, or (-∞, 0) U (0, ∞).
**2. Even or Odd:**
* **Even:** f(-x) = f(x)
* **Odd:** f(-x) = -f(x)
Let's test f(-x):
f(-x) = 1/√((-x)² + 1) - 1/(-x)
f(-x) = 1/√(x² + 1) + 1/x
Now, let's see if f(-x) = f(x) or f(-x) = -f(x):
* f(x) = 1/√(x² + 1) - 1/x
* -f(x) = -1/√(x² + 1) + 1/x
Comparing:
* f(-x) ≠ f(x) (Not even)
* f(-x) ≠ -f(x) (Not odd)
Therefore, the function is **neither even nor odd**.
**3. Increasing or Decreasing:**
* To determine if the function is increasing or decreasing, we need to analyze its derivative, f'(x).
f(x) = (x² + 1)^(-1/2) - x^(-1)
Now, find the derivative:
f'(x) = (-1/2)(x² + 1)^(-3/2)(2x) + x^(-2)
f'(x) = -x / (x² + 1)^(3/2) + 1/x²
To analyze this, we need to find when f'(x) > 0 (increasing) and when f'(x) < 0 (decreasing).
f'(x) = (-x * x²) + (x² + 1)^(3/2) / (x² * (x² + 1)^(3/2))
f'(x) = (-x³ + (x² + 1)^(3/2)) / (x² * (x² + 1)^(3/2))
* **For x > 0:**
* x² * (x² + 1)^(3/2) is always positive.
* We need to compare x³ and (x² + 1)^(3/2).
* When x is small, (x² + 1)^(3/2) will be larger than x³.
* When x is large, (x² + 1)^(3/2) will be larger than x³.
* Therefore, f'(x) is likely positive for x>0, thus increasing.
* **For x < 0:**
* x² * (x² + 1)^(3/2) is always positive.
* -x³ is positive.
* (x² + 1)^(3/2) is positive.
* Therefore f'(x) is likely positive for x<0, thus increasing.
It appears the function is increasing over its entire domain.
**4. Invertibility:**
* If a function is strictly increasing or strictly decreasing, it is invertible.
* Since the function appears to be strictly increasing over its domain, it is **invertible**.
**Conclusion:**
* **Domain:** (-∞, 0) U (0, ∞)
* **Neither even nor odd**
* **Invertible:** Yes
* **Increasing:** Yes
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