Question

The function $y = \frac{x}{\left( 1 + x^{2} \right)}$decreases in the interval

• A. (-1,1)
• B. $\lbrack 1,\infty)$
• C. $( - \infty,1\rbrack$
• D. $( - \infty,\infty)$

Answer 1

Answer: (B)

$\lbrack 1,\infty)$

Answer 2

$y' = \frac{\left( 1 + x^{2} \right) - 2x.x}{\left( 1 + x^{2} \right)^{2}} = \frac{1 - x^{2}}{\left( 1 + x^{2} \right)^{2}}$

Since $\left( 1 + x^{2} \right)^{2} > 0$ for all $x$, we have $y' < 0$

That is $1 - x^{2} < 0$, whence $x > 1$ or $x < - 1$ .

Therefore, $x \in ( - \infty, - 1)$ or $x \in (1,\infty)$

Thus $\frac{x}{\left( 1 + x^{2} \right)}$ decreases in $( - \infty, - 1\rbrack$or $\lbrack 1,\infty)$