Step 1: Understanding the Concept:
To determine whether the function is increasing or decreasing, we need to find its derivative f'(x) and analyze its sign within the function's domain.
The domain is determined by the condition:
(x - 1)/(x + 1) > 0
which implies:
x ∈ (-∞, -1) ∪ (1, ∞)
Step 2: Key Formula or Approach:
1. Derivative of log u is (1/u) · u'.
2. Quotient Rule: (u/v)' = (v u' - u v') / v².
3. Monotonicity condition:
f'(x) > 0 ⇒ increasing,
f'(x) < 0 ⇒ decreasing.
Step 3: Detailed Explanation:
Given:
f(x) = x + log((x - 1)/(x + 1))
Differentiate with respect to x:
f'(x) = 1 + [1 / ((x - 1)/(x + 1))] · d/dx((x - 1)/(x + 1))
Using quotient rule:
d/dx((x - 1)/(x + 1)) = [(1)(x + 1) - (1)(x - 1)] / (x + 1)²
= (x + 1 - x + 1) / (x + 1)²
= 2 / (x + 1)²
Substitute back:
f'(x) = 1 + [(x + 1)/(x - 1)] · [2/(x + 1)²]
= 1 + 2 / [(x - 1)(x + 1)]
= 1 + 2/(x² - 1)
Combine terms:
f'(x) = (x² - 1 + 2) / (x² - 1)
= (x² + 1) / (x² - 1)
Now analyze the sign of f'(x):
• Numerator x² + 1 is always positive.
• Denominator x² - 1 is positive in the domain
(-∞, -1) ∪ (1, ∞).
Thus,
f'(x) > 0 for all values in the domain.
Therefore, the function is strictly increasing.
Step 4: Final Answer:
The function is a monotonically increasing function.