Is #f(x)=e^x/(x-e^x)# concave or convex at #x=0#?

Question

1204 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-15T18:08:43

The curve is convex at #x=0#

Let's calculate the derivative

#(u/v)'=(u'v-uv')/v^2#

Here, #u=e^x# ; #=>##u'=e^x#

#v=x-e^x# #=># ; #v'=1-e^x#

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

#=(xe^x-e^(2x)-e^x+e^(2x))/(x-e^x)^2=(xe^x-e^x)/(x-e^x)^2#

#=(e^x(x-1))/(x-e^x)^2#

#f'(x)=0# when #x=1#

Let's do a sign chart

#color(white)(aaaa)##x##color(white)(aaaaa)##-oo##color(white)(aaaaa)##0##color(white)(aaaaa)##1##color(white)(aaaaa)##+oo#

#color(white)(aaaa)##f'(x)##color(white)(aaaaaaa)##-##color(white)(aaaaa)##-##color(white)(aaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaaa)##darr##color(white)(aaaaa)##darr##color(white)(aaa)##uarr#

As the curve is decreasing from #-oo# to #1#, the curve is convex at #x=0#

graph{e^x/(x-e^x) [-5.546, 5.55, -2.773, 2.774]}