What are the absolute extrema of $f (x) = \frac{sin x}{x e^{x}} \in [ln 5, ln 30]$ $f(x)=(sinx) / (xe^x) in[ln5,ln30]$ ?

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Answer 1 · 2015-12-18T23:51:14

$x = ln (5)$ $x = ln(5)$ and $x = ln (30)$ $x = ln(30)$

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

You need $f'$ : $f'(x) = (xcos(x)e^x - sin(x)(e^x + xe^x))/(xe^x)^2$

$f'(x) = (xcos(x) - sin(x)(1 + x))/(x^2e^x)$

$AAx in [ln(5),ln(30)], x^2e^x > 0$ so we need $sign(xcos(x) - sin(x)(1 + x))$ in order to have the variations of $f$ .

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

Its max is $f(ln(5)) = sin(ln(5))/(ln(25))$ and its min is $f(ln(30)) = sin(ln(30))/(30ln(30))$

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