Question

The real valued function \( f(x) = \frac{x^2}{2} - \log(x^2+x+1) \) is

• A. Strictly decreasing in (1, \(\infty\))
• B. Strictly increasing in (1, \(\infty\))
• C. Strictly increasing in (-\(\infty\), 0)
• D. Strictly decreasing in (0, \(\infty\))

Hint:

To find intervals of increase/decrease for a function \(f(x)\), find the derivative \(f'(x)\) and determine its sign. The function is increasing where \(f'(x)>0\) and decreasing where \(f'(x)<0\). The critical points where the sign might change are where \(f'(x)=0\) or \(f'(x)\) is undefined.

Answer 1

The Correct Option is B

To determine where the function is increasing or decreasing, we need to analyze the sign of its first derivative, \(f'(x)\).
First, find the derivative of \(f(x)\).
\( f'(x) = \frac{d}{dx}\left(\frac{x^2}{2}\right) - \frac{d}{dx}\left(\ln(x^2+x+1)\right) \).
\( f'(x) = x - \frac{2x+1}{x^2+x+1} \).
Combine the terms into a single fraction:
\( f'(x) = \frac{x(x^2+x+1) - (2x+1)}{x^2+x+1} = \frac{x^3+x^2+x - 2x - 1}{x^2+x+1} = \frac{x^3+x^2-x-1}{x^2+x+1} \).
Let's analyze the sign of the numerator and denominator.
The denominator, \(x^2+x+1\), has a discriminant \(D = 1^2 - 4(1)(1) = -3<0\). Since the leading coefficient is positive, the denominator is always positive for all real x.
So, the sign of \(f'(x)\) is determined by the sign of the numerator, \(N(x) = x^3+x^2-x-1\).
Let's factor the numerator: \( N(x) = x^2(x+1) - 1(x+1) = (x^2-1)(x+1) = (x-1)(x+1)(x+1) = (x-1)(x+1)^2 \).
So, \( f'(x) = \frac{(x-1)(x+1)^2}{x^2+x+1} \).
The term \((x+1)^2\) is always non-negative, and the denominator is always positive.
Therefore, the sign of \(f'(x)\) is determined by the sign of the term \((x-1)\).
If \(x>1\), then \(x-1>0\), so \(f'(x)>0\). This means \(f(x)\) is strictly increasing in \((1, \infty)\).
If \(x<1\) (and \(x \neq -1\)), then \(x-1<0\), so \(f'(x)<0\). This means \(f(x)\) is strictly decreasing in \((-\infty, -1)\) and \((-1, 1)\).
Checking the options, option (B) is correct.