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Find the maximum of the function:

$$f(x)=\sin x+\sin\left(\frac{1}{x}\right) \quad x>0$$

My Try :

$$f'(x)=\cos x-\dfrac{\cos(\frac{1}{x})}{x^2}=0 \\\cos x= \dfrac{\cos(\frac{1}{x})}{x^2} \ \ \ \\x^2\cos x=\cos\left(\frac{1}{x}\right)$$

Now what do I do ? Please help me!

Community
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Fricul38
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  • One solution to $x^2\cos x = \cos(1/x)$ is $x=1.$ That's not enough to prove there's a maximum there, nor does it mean there are no other solutions. – Michael Hardy Jan 31 '18 at 17:33
  • i think a global Maximum doesn't exist – Dr. Sonnhard Graubner Jan 31 '18 at 17:38
  • @Dr.SonnhardGraubner ??? Of course it exists and it is at $x=1$. – almagest Jan 31 '18 at 17:39
  • @almagest $x=1$ could be a minimum. – cansomeonehelpmeout Jan 31 '18 at 17:41
  • aha , can you prove this? take $$x=10312,3$$ – Dr. Sonnhard Graubner Jan 31 '18 at 17:41
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    $\sin \frac {1}{x}$ is a decreasing function when $x>\frac {2}{\pi}$ And $\sin x$ is increasing all the way until it hits its max at $x = \frac {\pi}{2}$ this suggests that the max of the function is in $[\frac {2}{\pi},\frac {\pi}{2}]$ – Doug M Jan 31 '18 at 17:43
  • @DougM how local maximum? – Fricul38 Jan 31 '18 at 17:48
  • Why are we sure that the max exist? yes it is bounded does that implies the existence of the maximum? you could first ask for the sup – Guy Fsone Jan 31 '18 at 18:14
  • @GuyFsone see this https://www.desmos.com/calculator – Fricul38 Jan 31 '18 at 18:15
  • @GuyFsone $\sin x$ is periodical and $\sin\frac1x$ is strictly decreasing for $x>2\pi$. So, $\sup_{x\in (0,30\pi]}\sin x+\sin\frac1x> \sup_{x\in (30\pi,\infty)}\sin x+\sin\frac1x$. By making the substitution $y=1/x$, $$\sup_{x\in(0,\frac1{30\pi})}\sin x+\sin\frac1x=\sup_{y\in(30\pi,\infty)}\sin y+\sin\frac1y$$ So we can restrict the inspection to the closed interval $[\frac1{30\pi},30\pi]$... –  Jan 31 '18 at 18:19

2 Answers2

0

For $x \in [0;2/\pi]$, $f(x) \le 1+\sin(x)\le 1+\sin(2/\pi)\le 1.6 \le f(1)$

For $x \ge 2$, $f(x) \le 1+\sin(1/x)\le 1+\sin(0.5)\le 1.6 \le f(1)$

Now you only have to study $f$ on $[0.6;2]$ and you will find that $f(1)$ is the maximum

stity
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For $x>\frac{\pi}{2}\pi$ we obtain: $$f(x)<1+\frac{2}{\pi}<2\sin1.$$ Also, for $0<x<\frac{\pi}{6}$ we have $$f(x)<1+\frac{\pi}{6}<2\sin1.$$

Now, $f''(x)<0$ for all $x\in\left[\frac{\pi}{6},\frac{\pi}{2}\right]$ and $f'(1)=0,$ which says that $2\sin1$ it's the answer.

Michael Rozenberg
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