Find the maximum of the function

Question

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!

Answer 1

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

Answer 2

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.