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.


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 $f(0)=0$.

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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.