Is #f(x)=x/e^x # increasing or decreasing at #x=1 #?

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Answer 1 · 2017-03-23T14:36:55

For #x=1# the function #f(x)# has a local maximum.

Calculate the first derivative of the function:

#f(x) = x/e^x = xe^(-x)#

#(df)/dx = e^(-x)-xe^(-x) = e^(-x)(1-x) =(1-x)/e^x#

We can therefore see that #x=1# is a stationary point for #f(x)#, so the function is neither increasing nor decreasing but has a local extremum.

Looking at #(df)/dx# around #x=1# we can see that:

#(df)/dx > 0 # for #x < 1#

#(df)/dx < 0 # for #x > 1#

which means that #f(x)# is increasing from #-oo# to #x=1# and decreasing from #x=1# to #+oo#, so for #x=1# it has a local maximum.

graph{x/e^x [-2.23, 7.77, -3.42, 1.58]}