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Mathematics Question on Exponential and Logarithmic Functions

The function f(x)=log⁡(1+x)−2x2+xf(x) = \log(1+x) - \frac{2x}{2+x} is increasing on

A

(−∞,∞)(- \infty, \infty)

B

(−1,∞)(- 1, \infty)

C

(∞,−1)( \infty, -1)

D

(−∞,0)(- \infty, 0)

Answer

(−1,∞)(- 1, \infty)

Explanation

Solution

f(x)=log⁡(1+x)−2x2+xf(x) = \log(1+x) - \frac{2x}{2+x}
To simplify the expression, we can rewrite it as:
f(x)=log⁡(1+x)−2x2+xf(x) = \log(1+x) - \frac{2x}{2+x}
To find the derivative, we apply the logarithmic differentiation and the quotient rule:
f′(x)=11+x−2(12+x)−2x(−1(2+x)2)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)=11+x−22+x+2x(2+x)2f'(x) = \frac{1}{1+x} - \frac{2}{2+x} + \frac{2x}{(2+x)^2}
Now, we need to determine where f′(x)>0.f'(x) > 0.
To analyze the sign of f'(x), we can find the common denominator for the fractions:
f′(x)=(2+x)(2+x)−2(1+x)+2x(1+x)(2+x)2f'(x) = \frac{(2+x)(2+x) - 2(1+x) + 2x}{(1+x)(2+x)^2}

Expanding the numerator:
f′(x)=4+4x+x2−2−2x+2x(1+x)(2+x)2f'(x) = \frac{4+4x+x^2 - 2 - 2x + 2x}{(1+x)(2+x)^2}

f′(x)=x2(1+x)(2+x)2f'(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, x2x^2 is always non-negative.
For the denominator, (1+x)(1+x) and (2+x)2(2+x)^2 are also always non-negative, except when x = -1.
Therefore, f'(x) is positive (greater than zero) when x≠−1.x ≠ -1.
Hence, the function f(x)=log⁡(1+x)−2x2+xf(x) = \log(1+x) - \frac{2x}{2+x} is increasing on the interval (−1,∞)(-1, \infty), which corresponds to option (B) (−1,∞)(-1, \infty).