I want to study the variation of $f(x)=(1+\frac{1}{x})^x$ which is defined on $(-\infty,-1) \cup(0,+\infty)$. So,I first computed its derivative $$f'(x)= (\ln(1+\frac{1}{x})- \frac{1}{1+x}) f(x).$$ However the only way I have found to determine the sign of $f'(x)$ is to study the derivative of $\ln(1+\frac{1}{x})- \frac{1}{1+x}):$ $$k(x)= \frac{-1}{x(x+1)^2}.$$ So I would like to know if there is another faster and more efficient way to study the variations of this function without going through the second derivatives?
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You have, if $x>0$,\begin{align}\log\left(1+\frac1x\right)&=\log\left(\frac{x+1}x\right)\\&=\log(x+1)-\log x\\&=\int_x^{x+1}\frac1t\,\mathrm dt\\&>\frac1{x+1}\bigl((x+1)-x\bigr)\\&=\frac1{x+1}.\end{align}Therefore $f'(x)>0$ in $(0,\infty)$.
And, if $x<-1$,\begin{align}\log\left(1+\frac1x\right)&=\log\left(\frac{x+1}x\right)\\&=-\log\left(\frac x{x+1}\right)\\&=-\log\left(\frac{-x}{-x-1}\right)\\&=-\bigl(\log(-x)-\log(-x-1)\bigr)\\&=-\int_{-x-1}^{-x}\frac1t\,\mathrm dt\\&>-\frac1{-x-1}\\&=\frac1{x+1}.\end{align}And then the same argument as before applies.
answered Nov 4, 2021 at 12:13
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$\begingroup$ Thank for the detailed reply. However, I dont understand why $\int_{x}^{x+1}\frac{1}{t}dt>\frac{1}{x+1}((x+1)-x)$ ? $\endgroup$ Commented Nov 4, 2021 at 15:16
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1$\begingroup$ Because you always have$$\int_a^bf(t)\,\mathrm dt>(b-a)\min f,$$when $f$ is continuous and not constant. $\endgroup$ Commented Nov 4, 2021 at 15:17
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$\begingroup$ Could you have any link or can you explaind me in details why this is true ? I would like to better understand this inequality, it's really not obvious for me $\endgroup$ Commented Nov 4, 2021 at 15:24
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$\begingroup$ and why $\frac{1}{1+x}$ is the min int this context ? $\endgroup$ Commented Nov 4, 2021 at 15:25
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$\begingroup$ Let $m=\min f$. Then$$\int_a^bf(t)\,\mathrm dt>\int_a^bm\,\mathrm dt=(b-a)m.$$ $\endgroup$ Commented Nov 4, 2021 at 15:25