Compute the derivative of the function $f(x)=\left(1-\frac1x\right)^x$, $x > 1$, and conclude that $f(x)$ is monotone increasing

So for this question the derivative for this function is $$f'(x)= \left(1-\frac1x\right)^x\left[\log\left(1-\frac1x\right)+\frac{x^2}{(x-1)}\right]$$

but I am not sure how to use the derivative to conclude that $f(x)$ is monotone increasing

Definition for montone increasing is $f(x_1) \leq f(x_2)$

• To prove that a function $f$ is monotonically increasing on $(a,b)$, it is only necessary to show that $f'(x) \ge 0$ on $(a,b)$ – eepperly16 Jun 18 '15 at 1:42
• aaaa your right i missed forgot one piece – ChrisV Jun 18 '15 at 2:03
• The derivative is wrong. – egreg Jun 18 '15 at 6:57

hint:

it is trivial $1-\dfrac{1}{x}>0$, so you only need to prove:

$\log\left(1-\dfrac1x\right)+\dfrac{x^2}{(x-1)}>0$

note: $\log\left(1-\dfrac1x\right)> \dfrac{1}{1-x}$

if you can prove it, rest is easy for you.

\begin{align} \frac{f'(x)}{f(x)} &= \frac{d}{dx}\log{f(x)} = \left( x \log(1 - \frac{1}{x})\right)' = \log(1 - \frac{1}{x}) + x \cdot (\frac{1}{x-1} - \frac{1}{x}) \\ &= \log(1 - \frac{1}{x}) + \frac{1}{x-1} \end{align} So $f'(x) = f(x) \cdot \left( \log(1 - \frac{1}{x}) + \frac{1}{x-1} \right)$

Since $f(x)$ is positive. It suffices to prove that $\log(1 - \frac{1}{x}) + \frac{1}{x-1} \ge 0$. Set $u = \frac{1}{x}$. For $x \in (1,+\infty)$, $u \in (0,1)$. We aim to prove that $$\forall u \in (0,1), \, \, g(u) = \log(1 - u) + \frac{u}{1-u} = \log(1 - u) + \frac{1}{1-u} - 1 \ge 0$$ The derivative of $g$ is $$g'(u) = -\frac{1}{1-u} + \frac{1}{(1-u)^2} = \frac{1-(1-u)}{(1-u)^2} = \frac{u}{(1-u)^2} > 0, \, \, \forall u \in (0,1).$$ Since $g(0) = 0$, $\forall u \in (0,1)$, $g(u) > 0$. That finishes the proof.

Claim: Suppose $f:\mathbb{R}\to\mathbb{R}$ is differentiable. Then $f$ is monotonic if and only if $f'(x)\geq 0$ for all $x\in\mathbb{R}$.

Suppose $f$ is monotonic. Then

$$\frac{f(x+h)-f(x)}{h} \geq 0$$ So $f'(x)\geq 0$ by applying limits to both sides.

Suppose $f'(x)\geq 0$. We need to case here. Suppose $f'(x)>0$. Then for small enough $h$, $$\frac{f(x+h)-f(x)}{h} \geq 0$$ for small enough $h$. But $h$ is positive so $f(x+h)-f(x)\geq 0$ for small $h$. Repeat for every $x$.

Now suppose $f'(x)=0$. Then $f$ cannot be a maximum or a minimum. Otherwise a point nearby has $f'(x+h)<0$. So for $h$ small, $f(x+h)-f(x-h)>0$.

A brute force solution could be to show that the minimum value of $$A(x)=\log\left(1-\frac1x\right)+\frac{x^2}{(x-1)}$$ is positive. Differentiating, we have $$A'(x)=\frac{(x-1)^2 x-1}{(x-1)^2 x}$$ the denominator of which being positive. So, $A'(x)=0$ would impliy $(x-1)^2 x-1=0$; this is a cubic polynomial and Cardano method will show that there is only one positive root $$x=\frac{1}{3} \left(2+\sqrt[3]{\frac{1}{2} \left(25-3 \sqrt{69}\right)}+\sqrt[3]{\frac{1}{2} \left(25+3 \sqrt{69}\right)}\right)\approx 1.75488\approx \frac74$$ $$A(\frac74)=\frac{49}{12}-\log \left(\frac{7}{3}\right)\approx 3.23604$$ while for the exact solution given above $A(x)\approx 3.23600$.