The interval for which the function \( f(x)=\displaystyle\frac{x}{x^2+x+1} \) is an increasing function, is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( (-1,1) \)
By the quotient rule,
\( \displaystyle f'(x)=\frac{(x^2+x+1)\cdot 1-x(2x+1)}{(x^2+x+1)^2} \)
\( \displaystyle =\frac{1-x^2}{(x^2+x+1)^2} \)
The denominator is always positive (\( x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\gt 0 \)), so the sign of f'(x) is the sign of \( 1-x^2 \).
\( \displaystyle f'(x)\gt 0\ \Leftrightarrow\ 1-x^2\gt 0 \)
\( \displaystyle \Leftrightarrow\ -1\lt x\lt 1 \)
So f is increasing on (−1, 1).
Why the other options are incorrect
Option (1) \( (0,\infty) \) — Incorrect. For x > 1, f'(x) < 0, so f is decreasing on part of \( (0,\infty) \).
Option (2) \( (-\infty,-1)\cup(1,\infty) \) — Incorrect. This is where \( 1-x^2\lt 0 \), i.e. where f is decreasing.
Option (4) \( (-\infty,0) \) — Incorrect. For x < −1, f'(x) < 0, so f is decreasing on part of \( (-\infty,0) \).