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Let $a \in \mathbb{R}$ and \begin{align*} g: \mathbb{R} &\rightarrow \mathbb{R},&\ g(x) = x+\cos x - a. \end{align*}

I want to prove that $g$ is strictly increasing. But the derivative is not everywhere zero. I tried the intermediate theorem, and also tried proving that $x<y\implies g(x)<g(y),$ but no luck.

ryang
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Dimi
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    @mathcounterexamples.net: I think the asker's concern is that the derivative isn't positive everywhere, but becomes zero at isolated points. – Hans Lundmark Mar 04 '23 at 20:45
  • Also related: https://math.stackexchange.com/questions/3489495/non-negative-derivative-not-being-zero-function-on-all-subintervals-implies-st, https://math.stackexchange.com/questions/4143209/prove-that-f-is-always-strictly-increasing, https://math.stackexchange.com/questions/1845927/monotone-functions-and-non-vanishing-of-derivative. – Hans Lundmark Mar 04 '23 at 20:51
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    Consider the function $f(x) = x^3$, which is strictly increasing everywhere, despite the presence of the isolated point $~x=0. ~$ At this isolated point, $~f'(x) = 0.$ – user2661923 Mar 05 '23 at 00:54

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