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I have this equation:

$$9x + \cos x = 0$$ but I need to write out and prove why it has one real root. Could someone maybe give me a few pointers or what do I do exactly?

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Let $f(x)=9x+\cos x$ then $f$ is differentiable and $f'(x)=9-\sin(x)>0$.
So $f$ is strictly increasing, moreover $\displaystyle\lim_{-\infty}f=-\infty$ and $\displaystyle\lim_{+\infty}f=+\infty$ so $f(x)=0$ has exactly one solution (there is a solution by the IVT and the solution is unique by the monotony).

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  • $\begingroup$ Dear JB, functions having a derivative are called differentiable in English, not derivable. The adjective derivable exists, but means "which can be deduced, inferred". But this is nitpicking (du pinaillage!) : +1 for your perfect answer . $\endgroup$ – Georges Elencwajg Jul 2 '12 at 7:52
  • $\begingroup$ Thank you for your comment. I corrected my answer. PS : I didn't know the word "nitpicking", thank you. $\endgroup$ – JBC Jul 2 '12 at 12:43

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