Let $f:\Bbb R-\{0\} \to \Bbb R$ be defined by $f(x)=x+\frac{1}{x^3}$. On which intervals is $f$ one-one?
$a)(-\infty,-1)$
$b)(0,1)$.
$c)(0,2)$.
$d)(0,\infty)$.
I know that this is very basic question but I am not getting how to check domain for this function? Only I can say that f is odd function and hence it is symmetric with respect to origin.By checking it's kernel and to get nullity zero ,$x^2=-1$ which not possible for real numbers.Then how to check function's domain for f to be 1-1?
Observe that
$$f'(x)=1-\frac3{x^4}=\left(1-\frac{\sqrt3}{x^2}\right)\left(1+\frac{\sqrt3}{x^2}\right)\implies f'(x)\ge0\iff1-\frac{\sqrt3}{x^2}\ge0\iff$$
$$\left(1-\frac{\sqrt[4]3}x\right)\left(1+\frac{\sqrt[4]3}x\right)\ge0\stackrel{\cdot x^2}\iff\left(x-\sqrt[4]3\right)\left(x+\sqrt[4]3\right)\ge0\iff$$
$$x\le-\sqrt[4]3\;\;\text{or}\;\;x\ge\sqrt[4]3$$
and now you know exactly where your function is monotonic ascending (descending) and thus $\;1-1\;$ . Try to take it from here