We need
#(u/v)'=(u'v-uv')/(v^2)#
Calculate the first and second derivatives
Let #f(x)=e^x/(5+e^x)#
#u(x)=e^x#, #=>#, #u'(x)=e^x#
#v(x)=5+e^x#, #=>#, #v'(x)=e^x#
Therefore,
#f'(x)=(e^x(5+e^x)-e^x*e^x)/(5+e^x)^2=(5e^x)/(5+e^x)^2#
#AA x in RR, |, f'(x)>0#
No critical points.
Therefore,
The sign chart is
#color(white)(aaaa)##Interval##color(white)(aaaa)##(-oo,+oo)#
#color(white)(aaaa)##sign f'(x)##color(white)(aaaaaaaa)##+#
#color(white)(aaaaaaa)## f(x)##color(white)(aaaaaaaaaa)##↗#
Now, calculate the second derivative
#u(x)=5e^x#, #=>#, #u'(x)=5e^x#
#v(x)=(5+e^x)^2#, #=>#, #v'(x)=2e^x(5+e^x)#
#f''(x)=(5e^x(5+e^x)^2-5e^x(2e^x(5+e^x)))/(5+e^x)^4#
#=(25e^x+5e^(2x)-10e^(2x))/((5+e^x)^3)#
#=(25e^x-5e^(2x))/(((5+e^x)^3))#
#=(5e^x(5-e^(x)))/(((5+e^x)^3))#
The point of inflection is when #f''(x)=0#
#=>#, #5-e^x=0#, #x=ln5#
We can make the chart
#color(white)(aaaa)##Interval##color(white)(aaaa)##(-oo,ln5)##color(white)(aaaa)##(ln5,+oo)#
#color(white)(aaaa)##sign f''(x)##color(white)(aaaaaa)##+##color(white)(aaaaaaaaaaa)##-#
#color(white)(aaaa)## f(x)##color(white)(aaaaaaaaaaaa)##uu##color(white)(aaaaaaaaaaa)##nn#
See the graph of the function
graph{e^x/(5+e^x) [-8.89, 8.89, -4.444, 4.445]}
