The function is:
$$ f(x) = \frac{2x}{mx^2 + 1} $$
The answer sheet says the answer is: $ m \ge 0 $ but I don't get it. My answer was $ m \in \mathbb{R} \setminus \{-1, 1\} $, because then, when $x=1$ or $x=-1$ the denominator may come out as $0$. Would anyone care to explain?
For $m\geq 0,$ $mx^2+1\geq 1,$ so the denominator is never $0$, and the formula makes sense on all of $\mathbb{R}$.
For $m<0$, $mx^2+1=0$ for $x=\sqrt{-1/m},$ which is now a real number, since $m$ is negative. Hence, the above formula cannot be defined for all $x\in \mathbb{R}$ for $m<0$.
