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

1 Answer
Dec 18, 2015

Answer:

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

Explanation:

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))#