6
$\begingroup$

Prove that the function $f:\mathbb R\to \mathbb R$ defined by $f(x)=x+\sin x$ for $x\in \mathbb R$ is a bijective function.


The codomain of the $f(x)=x+\sin x$ is $\mathbb R$ and the range is also $\mathbb R$. So this function is an onto function.
But I am confused in proving this function is one-to-one.

I know about its graph and I know that if a function passes the horizontal line test (i.e horizontal lines should not cut the function at more than one point), then it is a one-to-one function. The graph of this function looks like the graph of $y=x$ with sinusoids going along the $y=x$ line.

If I use a graphing calculator at hand, then I can tell that it is a one-to-one function and $f(x)=\frac{x}{2}+\sin x$ or $\frac{x}{3}+\sin x$ functions are not, but in the examination I need to prove this function is one-to-one theoritically, without graphing calculators.

I tried the method which we generally use to prove a function is one-to-one but no success.
Let $f(x_1)=f(x_2)$ and we have to prove that $x_1=x_2$ in order fot the function to be one-to-one.
Let $x_1+\sin x_1=x_2+\sin x_2$
But I am stuck here and could not proceed further.

$\endgroup$
  • 2
    $\begingroup$ I just want to say that technically you can't prove anything with a graphing calculator, unless you may a list of the entire algorithm the calculator used to present the displayed phenomenon, which is unrealistic. $\endgroup$ – j0equ1nn Dec 15 '15 at 12:10
9
$\begingroup$

You can prove that this function is strictly increasing :

It's a $C^1$ function and $f'(x) = 1+\cos(x) \geq 0$, so the function is increasing.

$\{x \mid f'(x) = 0 \} = \pi \Bbb Z$ is a discrete set, so $f$ is strictly increasing (if $f$ was locally constant somewhere, there would be an intervall $]a,b[$ where $f'(x)=0$ )

$\endgroup$
  • $\begingroup$ What does it mean to be $C^1$ function and i did not understand how you proved it strictly increasing inspite of $f'(x)\geq 0$@Tryss $\endgroup$ – Vinod Kumar Punia Dec 15 '15 at 11:09
  • 1
    $\begingroup$ $C^{1}$ means derivable and its derivative is continuous. $\endgroup$ – Stravog Dec 15 '15 at 11:32
  • 1
    $\begingroup$ @VinodKumarPunia : if your function is not strictly increasing, by continuity of $f$ there exists an intervall $[a,b]$ where $f$ is constant. This imply that $f'(x)=0$ on $]a,b[$. And you conclude by contraposition $\endgroup$ – Tryss Dec 15 '15 at 11:41
0
$\begingroup$

Assume $f$ is Many-One.

So there do exist some $x_{1}$ and $x_{2}$ where $x_{1}\neq x_{2}$ such that $$f(x_{1})=f(x_{2})$$ $$x_{1}+\sin x_{1}=x_{2}+\sin x_{2}$$$$x_{1}-x_{2}=\sin x_{2}-\sin x_{1}=-2\cos\left(\frac{x_{1}+x_{2}}{2}\right)\sin\left(\frac{x_{1}-x_{2}}{2}\right)$$

$$\left|x_{1}-x_{2}\right|=2\left|\cos\left(\frac{x_{1}+x_{2}}{2}\right)\sin\left(\frac{x_{1}-x_{2}}{2}\right)\right|\leq2\left|\sin\left(\frac{x_{1}-x_{2}}{2}\right)\right|$$

$$\implies\left|\frac{\sin\left(\frac{x_{1}-x_{2}}{2}\right)}{\frac{x_{1}-x_{2}}{2}}\right|\geq 1$$

which is obviously a contradiction.

So our assumption that $f(x)$ is Many-One is false.

$f(x)$ being onto is trivial since on the RHS $x$ can take all real values.

$\endgroup$

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.