What are the absolute extrema of f(x)=(sinx) / (xe^x) in[ln5,ln30]?

Dec 18, 2015

$x = \ln \left(5\right)$ and $x = \ln \left(30\right)$

Explanation:

I guess the absolute extrema is the "biggest" one (smallest min or biggest max).

You need $f '$ : $f ' \left(x\right) = \frac{x \cos \left(x\right) {e}^{x} - \sin \left(x\right) \left({e}^{x} + x {e}^{x}\right)}{x {e}^{x}} ^ 2$

$f ' \left(x\right) = \frac{x \cos \left(x\right) - \sin \left(x\right) \left(1 + x\right)}{{x}^{2} {e}^{x}}$

$\forall x \in \left[\ln \left(5\right) , \ln \left(30\right)\right] , {x}^{2} {e}^{x} > 0$ so we need $s i g n \left(x \cos \left(x\right) - \sin \left(x\right) \left(1 + x\right)\right)$ in order to have the variations of $f$.

$\forall x \in \left[\ln \left(5\right) , \ln \left(30\right)\right] , f ' \left(x\right) < 0$ so $f$ is constantly decreasing on $\left[\ln \left(5\right) , \ln \left(30\right)\right]$. It means that its extremas are at $\ln \left(5\right)$ & $\ln \left(30\right)$.

Its max is $f \left(\ln \left(5\right)\right) = \sin \frac{\ln \left(5\right)}{\ln \left(25\right)}$ and its min is $f \left(\ln \left(30\right)\right) = \sin \frac{\ln \left(30\right)}{30 \ln \left(30\right)}$