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

Given the function:

\[ f(x) = \log(1+x) - \frac{2x}{2+x} \]

We need to find the derivative of this function. To do so, we apply logarithmic differentiation for the first term and the quotient rule for the second term:

\[ f'(x) = \frac{1}{1+x} - 2\left(\frac{1}{2+x}\right) - 2x \left(-\frac{1}{(2+x)^2}\right) \]

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 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:

\[ f'(x) = \frac{4 + 4x + x^2 - 2 - 2x + 2x}{(1+x)(2+x)^2} \]

Simplifying the numerator:

\[ f'(x) = \frac{x^2}{(1+x)(2+x)^2} \]

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 \(x \neq -1\).

Hence, the function \(f(x) = \log(1+x) - \frac{2x}{2+x}\) is increasing on the interval \((-1, \infty)\), which corresponds to option (B) \((-1, \infty)\).

Answer 2

We are given: \[ f(x) = \log(1 + x) - \frac{2x}{2 + x} \] 

Step 1: Domain
For \( \log(1 + x) \) to be defined: \[ 1 + x > 0 \Rightarrow x > -1 \] For \( \frac{2x}{2 + x} \) to be defined: \[ 2 + x \ne 0 \Rightarrow x \ne -2 \] Thus, domain is \( x \in (-1, \infty) \) 

Step 2: First derivative
\[ f'(x) = \frac{d}{dx} \left[\log(1 + x)\right] - \frac{d}{dx} \left[\frac{2x}{2 + x}\right] \] Differentiate: \[ f'(x) = \frac{1}{1 + x} - \frac{(2)(2 + x) - 2x(1)}{(2 + x)^2} = \frac{1}{1 + x} - \frac{4 + 2x - 2x}{(2 + x)^2} = \frac{1}{1 + x} - \frac{4}{(2 + x)^2} \] 

Step 3: Sign of derivative
Let: \[ f'(x) = \frac{1}{1 + x} - \frac{4}{(2 + x)^2} \] Find common denominator: \[ f'(x) = \frac{(2 + x)^2 - 4(1 + x)}{(1 + x)(2 + x)^2} \] Numerator: \[ (2 + x)^2 = x^2 + 4x + 4 \quad ; \quad 4(1 + x) = 4 + 4x \] \[ \Rightarrow x^2 + 4x + 4 - (4 + 4x) = x^2 \] So: \[ f'(x) = \frac{x^2}{(1 + x)(2 + x)^2} \] 

Step 4: Analyze sign
Since the numerator is \(x^2 \ge 0\), and the denominator is positive for \(x > -1\), we conclude: \[ f'(x) \ge 0 \text{ for all } x > -1 \] 

Therefore, the function is increasing on: \( \mathbf{(-1, \infty)} \)

Answer 3

We are given the function \(f(x) = \log(1+x) - \frac{2x}{2+x}\). To determine where the function is increasing, we need to find the intervals where its derivative, \(f'(x)\), is positive.

First, find the derivative \(f'(x)\):

\(f'(x) = \frac{1}{1+x} - \frac{d}{dx}\left(\frac{2x}{2+x}\right)\)

Use the quotient rule to differentiate \(\frac{2x}{2+x}\): \(\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}\)

\(\frac{d}{dx}\left(\frac{2x}{2+x}\right) = \frac{(2+x)(2) - (2x)(1)}{(2+x)^2} = \frac{4 + 2x - 2x}{(2+x)^2} = \frac{4}{(2+x)^2}\)

So, \(f'(x) = \frac{1}{1+x} - \frac{4}{(2+x)^2}\)

Find a common denominator:

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

Now, we need to find where \(f'(x) > 0\). First, note that for log(1+x) to be defined 1+x>0, so x>-1.

The numerator \(x^2\) is always non-negative (and 0 only at x=0).

The denominator is \((1+x)(2+x)^2\). For \(f'(x) > 0\), we need \((1+x)(2+x)^2 > 0\).

• x cannot be -2 as this will be undefined. Also x > -1. If -2-1 (1+x) >0.
• If \(x = 0\) then this equals to zero. So it cant be increasing when x=0.
• And \((2+x)^2 > 0 \) for \(x \neq -2\)

So, we require \(1+x > 0\), which means \(x > -1\). And, since [x] cannot be equal to -2, The domain of \(x \in (-1, \infty)\). But x=0 should be excluded.

So at (-1Therefore, f(x) is increasing on \((-1, \infty)\)

Answer: \((- 1, \infty)\)