Doubt regarding one-one onto function So lets have a function: $f:\mathbb R\rightarrow \mathbb R$, which is one-one and onto and symmetric about origin. Now its given that we have a number $x_1>0$ for which $f(x_1)>0$, then can we generalise $f(x)>0\,\,\forall x\in \mathbb R^+$?. It seems that by drawing graph, we can somehow satisfy ourselves, but I couldn't prove it mathematically, so this is my doubt here.
And also, lets say we have $g(x)=f'(x)$('$'$' refers to differentiation), and similarly, its given that for a number $g(x_1)>0$, then can we generalise for all $x>0$, in similar fashion? i am also a bit unsure that whether $g(x)$, will remain symmetric about origin or not, so this is my doubt too.
Any help is massively appreciated. Note that I am comfortable with basic highschool calculus, so high level maths won't be preferable for me.
 A: Suppose $f$ is continuous. Symmetric around origin means $f(-x)=-f(x)$ for all $x\in\mathbb{R}$. Taking limit as $x\to 0$ we get $f(0)=0$. We have $f(x_1)>0$ for some $x_1>0$. Take some $x_2>0$ and $x_2\neq x_1$. Then $f(x_2)$ cannot be $0$ because $f(0)=0$ and $f$ is one-one. If $f(x_2)<0$ then by continuity, there will be some $x_3$ between $x_1$ and $x_2$ such that $f(x_3)=0$. But that is again impossible due to $f(0)$ is already $0$ and $f$ is one-one. So $f(x_2)$ must be positive. This is true for all $x_2>0$.
If we don't assume $f$ is continuous then we can find counterexamples. For example
$f(x)=x$ for $x$ rational, $f(x)=-x$ for $x$ irrational. This is symmetric around the origin, positive for some positive $x$ but not positive for all positive $x$.
Now let's consider the problem regarding the derivative. The derivative being positive for some $x>0$ does not imply it is positive for all $x$. We can flatten out the curve so that the derivative becomes $0$. For example $f(x)=x+\sin(x)$.
A: It is not true. Take, for instance$$\begin{array}{rccc}f\colon&\Bbb R&\longrightarrow&\Bbb R\\&x&\mapsto&\begin{cases}x&\text{ if }x\ne\pm1\\-x&\text{ if }x=1\text{ or }x=-1.\end{cases}\end{array}$$It is an odd function, and therefore it is symmetric about the origin. But $f(2)>0$ and $f(1)<0$.
