If # e^x / (5+e^x)#, what are the points of inflection, concavity and critical points?

1 Answer
Oct 27, 2017

The point of inflection is #(ln5,1/2)# and no critical points. The concavities are shown below

Explanation:

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]}