Question

The function \(f(x) = \log(1+x) - \frac{2x}{2+x}\) is increasing on

• A. \((- \infty, \infty)\)
• B. \((- 1, \infty)\)
• C. \(( \infty, -1)\)
• D. \((- \infty, 0)\)

Answer 1

The Correct Option is B

\(f(x) = \log(1+x) - \frac{2x}{2+x}\)
To simplify the expression, we can rewrite it as:
\(f(x) = \log(1+x) - \frac{2x}{2+x}\)
To find the derivative, we apply the logarithmic differentiation and the quotient rule:
\(f'(x) = \frac{1}{1+x} - 2\left(\frac{1}{2+x}\right) - 2x \left(-\frac{1}{(2+x)^2}\right)\)
Simplifying further:
\(f'(x) = \frac{1}{1+x} - \frac{2}{2+x} + \frac{2x}{(2+x)^2}\)
Now, we need to determine where \(f'(x) > 0.\)
To analyze the sign of f'(x), we can find the common denominator for the fractions:
\(f'(x) = \frac{(2+x)(2+x) - 2(1+x) + 2x}{(1+x)(2+x)^2}\)

Expanding the numerator:
\(f'(x) = \frac{4+4x+x^2 - 2 - 2x + 2x}{(1+x)(2+x)^2}\)

\(f'(x) = \frac{x^2}{(1+x)(2+x)^2}\)
Now, let's analyze the sign of f'(x) based on the numerator and denominator.
For the numerator, \(x^2\) is always non-negative.
For the denominator, \((1+x)\) and \((2+x)^2\) are also always non-negative, except when x = -1.
Therefore, f'(x) is positive (greater than zero) when \(x ≠ -1.\)
Hence, the function \(f(x) = \log(1+x) - \frac{2x}{2+x}\) is increasing on the interval \((-1, \infty)\), which corresponds to option (B) \((-1, \infty)\).