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Hey I have found the derivative to this equation: $$2x\cdot\sin\left(\frac{1}{x}\right)+2x\cdot\cos\left(\frac{1}{x}\right)+\sin\left(\frac{1}{x}\right)-\cos\left(\frac{1}{x}\right)$$

The things is that i can already see that the endpoint of the interval will end up as the biggest value of the all $x$-values. I want and need to prove that here is no other extreme points inside the interval that is bigger. The only problem is that there are multiple and will take a lot of time prove. Is there a simpler way to do this?

Jochen
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SliVetle
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  • What is the exact statement that you need to prove/show? – student91 Feb 14 '23 at 12:35
  • I would try to show that $f(x)$ is increasing from something like $x=0.5$ onwards, and that $|f(x)| \leqslant 1$ for all values less than that. Then it follows that the extreme value is attained at $x=1$. – Luke Collins Feb 14 '23 at 12:39
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    Also the derivative is wrong... unless you meant $f(x)=x^2\big(\sin\big(\tfrac1x\big)+\cos\big(\tfrac1x\big)\big)$? – Luke Collins Feb 14 '23 at 12:45
  • I guess, there is no closed formula for the local extreme points. And for the global maximum, you have $f'(x)=(2x-1)\cos(1/x)+(2x+1)\sin(1/x)\geq 2x+1>0$ for $x\in[\frac 2\pi,1]$ since $x\mapsto\sin(1/x)$ is monotonic decreasing on $[\frac 2\pi,1]$ and $\cos(1/x)>0$ for $x\in[\frac 2\pi,1]$. – Jochen Feb 14 '23 at 13:19

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