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

## $f : \mathbb{R} \to \mathbb{R}$ differentiable in $\mathbb{R}$ with $f \left(x\right) - {e}^{- f \left(x\right)} = x - 1$ , $x$$\in$$\mathbb{R}$ Show that $\frac{x}{2} <$$f \left(x\right)$$<$$x f ' \left(x\right)$ , $x > 0$

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#### Explanation

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#### Explanation:

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Jim S Share
Mar 8, 2018

#### Explanation:

For $x = 0$ we have

$f \left(0\right) - {e}^{- f \left(0\right)} = - 1$

We consider a new function $g \left(x\right) = x - {e}^{- x} + 1$ , $x$$\in$$\mathbb{R}$

$g \left(0\right) = 0$ ,

$g ' \left(x\right) = 1 + {e}^{- x} > 0$ , $x$$\in$$\mathbb{R}$

As a result $g$ is increasing in $\mathbb{R}$. Thus because it's strictly increasing $g$ is "$1 - 1$" (one to one)

So, $f \left(0\right) - {e}^{- f \left(0\right)} + 1 = 0$ $\iff$ $g \left(f \left(0\right)\right) = g \left(0\right)$ $\iff$ $f \left(0\right) = 0$

We need to show that $\frac{x}{2} <$$f \left(x\right) <$$x f ' \left(x\right)$ ${\iff}^{x > 0}$

$\frac{1}{2} <$$f \frac{x}{x} <$$f ' \left(x\right)$ $\iff$

$\frac{1}{2} <$$\frac{f \left(x\right) - f \left(0\right)}{x - 0} <$$f ' \left(x\right)$

• $f$ is continuous at $\left[0 , x\right]$
• $f$ is differentiable in $\left(0 , x\right)$

According to the mean value theorem there is ${x}_{0}$$\in$$\left(0 , x\right)$
for which $f ' \left({x}_{0}\right) = \frac{f \left(x\right) - f \left(0\right)}{x - 0}$

$f \left(x\right) - {e}^{- f \left(x\right)} = x - 1$ , $x$$\in$$\mathbb{R}$ so

by differentiating both parts we get

$f ' \left(x\right) - {e}^{- f \left(x\right)} \left(- f \left(x\right)\right) ' = 1$ $\iff$ $f ' \left(x\right) + f ' \left(x\right) {e}^{- f \left(x\right)} = 1$ $\iff$

$f ' \left(x\right) \left(1 + {e}^{- f \left(x\right)}\right) = 1$ ${\iff}^{1 + {e}^{- f \left(x\right)} > 0}$

$f ' \left(x\right) = \frac{1}{1 + {e}^{- f \left(x\right)}}$

The function $\frac{1}{1 + {e}^{- f \left(x\right)}}$ is differentiable. As a result $f '$ is differentiable and $f$ is 2 times differentiable with

$f ' ' \left(x\right) = - \frac{\left(1 + {e}^{- f \left(x\right)}\right) '}{1 + {e}^{- f \left(x\right)}} ^ 2$ $=$

(f'(x)e^(-f(x)))/((1+e^(-f(x)))^2 $> 0$ , $x$$\in$$\mathbb{R}$

-> $f '$ is strictly increasing in $\mathbb{R}$ which means

${x}_{0}$$\in$$\left(0 , x\right)$ $\iff$ $0 <$${x}_{0} <$$x$ $\iff$

$f ' \left(0\right) <$$f ' \left({x}_{0}\right) <$$f ' \left(x\right)$ $\iff$

$\frac{1}{1 + {e}^{- f \left(0\right)}}$$<$$f \frac{x}{x} <$$f ' \left(x\right)$ $\iff$

$\frac{1}{2} <$$f \frac{x}{x} <$$f ' \left(x\right)$ ${\iff}^{x > 0}$

$\frac{x}{2} <$$f \left(x\right) <$$x f ' \left(x\right)$

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