Is $f(x)=|x|(x-\sin x)$ bijective If $f:\mathbb{R}\rightarrow\mathbb{R}$ be defined as $f(x)=|x|\left(x-\sin x\right)$. Is $f$ bijective.
My Attempt:
To prove whether $f$ is bijective we need to prove that $f$ is both one-one(injective) and onto(surjective).
Since $f$ is odd and continuous it will clearly extend from $-\infty$ to $\infty$ so $f$ can be regarded as onto. But I am not able to get a rigorous proof for this or is my argument sufficient.
Also to prove that $f$ is one-one let us assume that $f$ is many-one.
So, there would exist $x_1,x_2\in\mathbb{R^+}$ such that $x_1\neq x_2$ but $f(x_1)=f(x_2)$
$|x_1|\left(x_1-\sin x_1\right)=|x_2|\left(x_2-\sin x_2\right)$
$x_1^2-x_2^2=x_1\sin x_1-x_2\sin x_2$
$(x_1+x_2)(x_1-x_2)=(x_1-x_2)\sin x_1+x_2(\sin x_1-\sin x_2)$
$x_1+x_2=\sin x_1+x_2\left(\frac{\sin x_1-\sin x_2}{x_1-x_2}\right)$
$x_1+x_2=\sin x_1+x_2\cos\left(\frac{x_1+x_2}{2}\right)\frac{\sin(\frac{x_1-x_2}{2})}{\frac{x_1-x_2}{2}}<x_1+x_2$
which is a contradiction. So $f$ is one-one. Is this method correct.
I was wondering that if I prove $f'(x)>0 \forall x\in\mathbb{R}$ then it could be easily proved that since $f$ is strictly increasing so $f$ is one-one.
But I am not able to justify $f'(x)>0 \forall x\in\mathbb{R}$
 A: You are right that your function is odd. Therefore since it is positive for $x>0$ we cn establish it is one-to-one by showing it is increasing for positive $x.$ For these $x$ you function is $g(x)=x^2-x \sin x.$ The derivative of $g(x)$ can be written as
$$ x(2-\cos x) - \sin x.$$
Now for positive $x$ we have $(2-\cos x) \ge 1$ so that the first term above is $\ge x.$ Now use that for positive $x$ we have $x > \sin x$ to conclude that $g'(x)>0$ which shows your function is indeed strictly increasing for positive $x$ as claimed, therefore one-to one. Together with your original function being odd, it is thus a bijection. [Your approach to showing surjective only needs the added fact that it takes on arbitrarily large values.]
Added later: User @RamanujanXXV has noted a shorter proof, Namely each of the functions $u(x)=x,\ v(x)=x-\sin x$ is strictly increasing on the positive reals, so their product is also.
A: The function $\arctan(x)$ is odd and continuous but doesn't extend from $-\infty$ to $\infty$. You need to do more to prove that the function is surjective.
