Show that for $\;\dfrac12

Question

I have the problem of unique. The existence is follows since using sided LHospital

$$\begin{cases}\lim\limits_{x\to0^+}f(x)=1\\{}\\\lim\limits_{x\to1^-}f(x)=\dfrac12\end{cases}\implies \;\text{the IVT for continuous functions proves}$$ .

For unique I did: evaluate differential

$$f'(x)=-\frac1{x\log^2x}+\frac1{(x-1)^2}=\frac{x\log^2x-(x-1)^2}{x(x-1)^2\log^2x}$$

Now, if I show $\;f'<0\;$ then $\;f\;$ is monotone going down and there is unique solution, but I'm having lots of problem for this: the denominator is always positive, but I can't prove numerator is negative.

After lots of tryings I tried direct proof:

$$f(x)=f(y)\iff \frac{\log\dfrac yx}{\log x\log y}=\frac{x-y}{(x-1)(y-1)}$$

and I won't have way to show $\;x=y\;$ .

Any helps is greatly appreciated.

Answer 1

In order to prove that $$ f(x)=\frac{1}{\log x}-\frac{1}{x-1}$$ is decreasing over $(0,1)$, we just need to prove that: $$ g(t) = \frac{1}{t}-\frac{1}{e^t-1} $$ is decreasing over $\mathbb{R}^-$. It is sufficient to prove that: $$ \forall t<0,\quad g'(t) = \frac{e^t}{(1-e^t)^2}-\frac{1}{t^2}<0,$$ or: $$ \forall t<0,\quad e^t+e^{-t}>2+t^2 $$ that is equivalent to: $$ \forall t<0,\quad \cosh t > 1+\frac{t^2}{2} $$ that is trivial, since for any $t$: $$ \cosh t = \sum_{n\geq 0}\frac{t^{2n}}{(2n)!} = 1+\frac{t^2}{2}+\frac{t^4}{24}+\ldots.$$

Answer 2

We need $x\log^2 x-(x-1)^2<0$ for $0<x< 1$.
It's like $\frac{1}{x}\log^2 \frac{1}{x}-(\frac{1}{x}-1)^2<0$ for $x>1$. $$\begin{aligned}\frac{1}{x}\log^2 \frac{1}{x}-(\frac{1}{x}-1)^2<0 \Leftrightarrow\\ \log^2 \frac{1}{x}-x(\frac{1}{x}-1)^2<0 \Leftrightarrow\\ (-\log x)^2<\frac{(1-x)^2}{x} \Leftarrow \log x<\frac{x-1}{\sqrt{x}}\end{aligned}$$ $t=\sqrt{x},x=t^2, \log x=2\log t, x>1\Rightarrow t>1$ $$2\log t<\frac{t^2-1}{t}=t-\frac{1}{t}\text{ for }t>1$$ LHS=RHS=0 when $t=1$ and $\frac{2}{t}=(2\log t)'<\left(t-\frac{1}{t}\right)'=1+\frac{1}{t^2}$ when $t>1$.

Answer 3

Let us consider $$f(e^x)=f(e^y)\implies\dfrac{1}{x}-\dfrac{1}{e^x-1}=\dfrac{1}{y}-\dfrac{1}{e^y-1}$$ with $x,y\in(-\infty,0).$ $$y(e^y-1)(e^x-x-1)=x(e^x-1)(e^y-y-1)$$ After simplifying this expression you can obtain $$(y-x)(1-e^x)(1-e^y)+xy(e^y-e^x)=0$$ Suppose $y\gt x\gt0.$ Then $(y-x)(1-e^x)(1-e^y)+xy(e^y-e^x)\gt0.$
Therefore this expression is zore if and only if $$\color{Green}{x=y.}$$

Answer 4

$x.(\log x)^2$ has limit 0 near 0 because we can write it as ($\log x)^2/(1/x)$ and use L'Hospital's Rule to replace it by $-2 \log x/(1/x)$ and use the same rule again to get the 0 limit.