Question

Show that `x/2<``f(x)``<``xf'(x)` , `AAx>0` ?

`f:RR->RR` differentiable in `RR` with

• `f(x)-e^(-f(x))=x-1` , `x` `in` `RR`

Show that

`x/2<` `f(x)` `<` `xf'(x)` , `x>0`

Answer 1

Check below for answer

Explanation:

For `x=0` we have

`f(0)-e^(-f(0))=-1`

We consider a new function `g(x)=x-e^(-x)+1` , `x` `in` `RR`

`g(0)=0` ,

`g'(x)=1+e^(-x)>0` , `x` `in` `RR`

As a result `g` is increasing in `RR`. Thus because it's strictly increasing `g` is "`1-1`" (one to one)

So, `f(0)-e^(-f(0))+1=0` `<=>` `g(f(0))=g(0)` `<=>` `f(0)=0`

We need to show that `x/2<` `f(x)<` `xf'(x)` `<=>^(x>0)`

`1/2<` `f(x)/x<` `f'(x)` `<=>`

`1/2<` `(f(x)-f(0))/(x-0)<` `f'(x)`

• `f` is continuous at `[0,x]`
• `f` is differentiable in `(0,x)`

According to the mean value theorem there is `x_0` `in` `(0,x)`
for which `f'(x_0)=(f(x)-f(0))/(x-0)`

`f(x)-e^(-f(x))=x-1` , `x` `in` `RR` so

by differentiating both parts we get

`f'(x)-e^(-f(x))(-f(x))'=1` `<=>` `f'(x)+f'(x)e^(-f(x))=1` `<=>`

`f'(x)(1+e^(-f(x)))=1` `<=>^(1+e^(-f(x))>0)`

`f'(x)=1/(1+e^(-f(x)))`

The function `1/(1+e^(-f(x)))` is differentiable. As a result `f'` is differentiable and `f` is 2 times differentiable with

`f''(x)=-((1+e^(-f(x)))')/(1+e^(-f(x)))^2` `=`

`(f'(x)e^(-f(x)))/((1+e^(-f(x)))^2` `>0` , `x` `in` `RR`

-> `f'` is strictly increasing in `RR` which means

`x_0` `in` `(0,x)` `<=>` `0<` `x_0<` `x` `<=>`

`f'(0)<` `f'(x_0)<` `f'(x)` `<=>`

`1/(1+e^(-f(0)))` `<` `f(x)/x<` `f'(x)` `<=>`

`1/2<` `f(x)/x<` `f'(x)` `<=>^(x>0)`

`x/2<` `f(x)<` `xf'(x)`