4
$\begingroup$

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?

1 Answers 1

9

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).

  • 0
    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 .2012-07-02
  • 0
    Thank you for your comment. I corrected my answer. PS : I didn't know the word "nitpicking", thank you.2012-07-02