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Without plotting graph prove that $f: \mathbb{R}\to \mathbb{R}$ $$f(x)=\sin x+x$$ is an Onto function

Let $y_0$ be any number in codomain Then we require atleast one $x_0$ in Domain such that $y_0=\sin x_0 +x_0$

Now since $\sin x_0$ is bounded in $\left[-1 \:\: 1\right]$ we have

$$x_0=y_0-\sin x_0$$

hence there is an $x_0$ in Domain. So $f$ is surjective. But i feel something missing in this proof? Any other way to prove?

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marked as duplicate by Martin R, Nosrati, Jack D'Aurizio algebra-precalculus Nov 9 '17 at 4:54

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    $\begingroup$ How do you know such an $x_0$ exists? That's the part you're missing. Essentially: for any number $a$, does $sin(x)+x-a=0$ have a root? $\endgroup$ – Chris C Nov 9 '17 at 3:59
  • $\begingroup$ For all $n$, $\sin (n\pi)=0$ , so $f(n\pi)=n\pi$. Now you can use intermediate value theorem to conclude (using continuity) $\endgroup$ – user160738 Nov 9 '17 at 4:02
  • $\begingroup$ use the Intermediate value thm. $\endgroup$ – Matematleta Nov 9 '17 at 4:02
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$f$ is an odd function so it suffices to prove surjectivity onto $[0,\infty).$ Note that $f$ is everywhere continuous, that $f(0)=0$ and that $f(x)\to \infty$ as $x\to \infty.$

Now, if $y\in \mathbb R^+$, since$f(x)\to \infty$ as $x\to \infty$ there is an $x\in R^+$ such that $f(x)> y$, so we can apply the Intermediate Value Theorem to get an $0\le c\le x$ such that $f(c)=y$.

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