Question

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

Answer 1

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]}