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Find the maximum points of $$f(x)=e^{-x}\sin^2(\pi x) \hspace{0.4cm},0<x<10$$

My calculations:I have calculated

$f'(x)=\pi e^{-x}\sin(2\pi x)-e^{-x}\sin^2(\pi x)$

$f''(x)=e^{-x}\sin^2(\pi x)-2\pi e^{-x}\sin(2\pi x)+2\pi^2e^{-x}\cos(2\pi x)$

I found the critical points are all integers and $x=\frac{\arctan(2\pi)}{\pi}$

My problem:Are these critical points?Further what is the behaviour of $f(x)$ at these points and which are maxima for $f(x)$?

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  • $\begingroup$ Look at the sign of the derivative to either side of the critical points, you can get that by plugging in values. Note you can factor out the $e^{-x}$ which is always positive, so just look at the other parts. Your relative maxima occur when you go for + to -. $\endgroup$ – Alan Oct 2 '14 at 6:31
  • $\begingroup$ my calculations gives that all integers are minima. $\endgroup$ – Flip Oct 2 '14 at 6:45
  • $\begingroup$ Why all the computation?Isn't it obvious its $x=1/2$? $\endgroup$ – Troy Woo Oct 2 '14 at 7:54
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If I may suggest, factor things and use trigonometric identities. If $$f(x)=e^{-x}\sin^2(\pi x)$$ $$f'(x)=e^{-x} \sin (\pi x) (2 \pi \cos (\pi x)-\sin (\pi x))$$ $$f''(x)=\frac{1}{2} e^{-x} \left(-4 \pi \sin (2 \pi x)+\left(4 \pi ^2-1\right) \cos (2 \pi x)+1\right)$$ From $f'(x)=0$, you obtain your solutions; $f''(x)=0$ provide you other. As said, for the analysis, forget the exponential term which is always positive.

When $x$ is an integer (first class of solutions for $f'(x)=0$,forgetting the exponential term, $f''(x)={2\pi^2}$ which is positive. So, these points effectively correspond to minima.

For the other solutions of $f'(x)=0$, use $t=\tan(\pi x)$ in the expression of $f''(x)$ and so, still forgetting the exponential term, the second derivative write, after removing the common denominator $(1+t^2)>0$, $$\left(2-4 \pi ^2\right) t^2-8 \pi t+4 \pi ^2$$ and since $t=2\pi$ the result is $-4 \left(\pi ^2+4 \pi ^4\right) <0$ and so they are maxima.

You could even continue using the last expression and show that the inflexion points corresponds to the solution of the previous quadratic and the solutions are given by $$\tan(\pi x)=\pm \frac{\pi \left(\sqrt{2+4 \pi ^2}-2\right)}{2 \pi ^2-1}$$

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as you found out , the critical points are right , so $f(x)=e^{−x}sin^2(πx)$ now $f(x)>=0$ so at integral points the value of function is $0$ which is minimum possible so rest points are maxima's . and as for you solution of critical points $tan(πx)=2π$ , does not mean $x$ has only one solution , arctan reduces then number of solutions , the solution of $x$ is $n+\frac{arctan(2π)}{π}$ where $n$ belongs to whole numbers , notice i have not dealt with $f''(x)$ as between two minima's there is only 1 critical point making it the maxima. hope this clears your doubt

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