What are the local extrema of #f(x)= lnx/e^x#?

1 Answer
Apr 14, 2017

Answer:

#x=1.763#

Explanation:

Take the derivative of #lnx/e^x# using quotient rule:

#f'(x)=((1/x)e^x-ln(x)(e^x))/e^(2x)#

Take out a #e^x# from the top and move it down to the denominator:

#f'(x)=((1/x)-ln(x))/e^x#

Find when #f'(x)=0# This only happens when the numerator is #0#:

#0=(1/x-ln(x))#

You're going to need a graphing calculator for this one.

#x=1.763#

Plugging in a number under #1.763# would give you a positive outcome while plugging a number above #1.763# would give you a negative outcome. So this is a local maximum.