# Prove without using graphing calculators that $f: \mathbb R\to \mathbb R,\,f(x)=x+\sin x$ is both one-to-one, onto (bijective) function.

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.

• 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. – j0equ1nn Dec 15 '15 at 12:10

## 2 Answers

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 $f'(x)=0$ )

• 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 – Vinod Kumar Punia Dec 15 '15 at 11:09
• $C^{1}$ means derivable and its derivative is continuous. – Stravog Dec 15 '15 at 11:32
• @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 – Tryss Dec 15 '15 at 11:41

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.