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Let the function $f:\mathbb{R} \to \mathbb{R}$ be defined as $f(x)=2x+\sin x$ then prove that $f$ is bijective.

My try: Given $f(x)=2x+\sin x$ Differentiate both sides w.r.t. $x$ $$f^{(1)}(x) =2+ \cos x$$ As $f^{(1)}(x)$ is always positive, $f$ continuously increases so $f$ is one-one. But I am unable to prove that $f$ is onto.

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Given $y>0$, we know that there is some $c$ such that $f(c)>y$, then by Intermediate Value Theorem we have some $\eta\in(0,c)$ such that $f(\eta)=y$. Similar reasoning applies to $y<0$. Note that $f(0)=0$.

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  • $\begingroup$ Note that $\lim_{x\rightarrow\infty}f(x)=\infty$. $\endgroup$ – user284331 Nov 30 '17 at 4:34
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Correct me if wrong.

1) Onto:

Let $y \in \mathbb{R}.$

Choose $a,b \in \mathbb {R}$ with

$f(a)<y$ and $f(b) >y.$

Possible, since:

$\lim_{ x \rightarrow -\infty } f(x) = -\infty,$ and

$\lim_{ x \rightarrow \infty} f(x) = \infty.$

MVT : $f$ continuos on $[a,b]$ and

$f(a) < y < f(b)$, there exists a

$p\in [a,b]$ with $f(p) =y.$

2) Injective:

Assume $y\not = x,$ say, $y<x$, with:

$f(x)=f(y)$, i.e.

$2(x -y) +\sin x- \sin y =0.$

$\dfrac {\sin x -\sin y}{x -y} = -2.$

MVT : $\dfrac{\sin x -\sin y}{x -y} = \cos(t) , $

$ y< t < x$.

A contradiction (why?), hence injective.

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