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Let $f(x)=\left(1+\dfrac1x\right)^x$ and $g(x)=\left(1+\dfrac1x\right)^{x+1}$, both $f$ and $g$ being defined for $x\gt0$, then comment about the increasing/decreasing nature of $f(x)$ and $g(x)$.

$$f(x)=e^{x\ln\left(1+\dfrac1x\right)}\\f'(x)=\left(1+\dfrac1x\right)^x\left(\frac{x}{1+\frac1x}\cdot-\frac1{x^2}+\ln\left(1+\dfrac1x\right)\right)\\f'(x)=\left(1+\dfrac1x\right)^x\left(-\frac1{x+1}+\ln\left(1+\dfrac1x\right)\right)$$

Not able to determine the nature of $f(x)$ from this.

Also, $$g(x)=f(x)\left(1+\frac1x\right)\\g'(x)=f'(x)\left(1+\frac1x\right)+f(x)\cdot-\frac1{x^2}$$

Even if I know the nature of $f'(x)$, not sure how to comment about $g'(x)$.

aarbee
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  • To show $f$ is increasing, you might like to see https://math.stackexchange.com/questions/83035/how-to-prove-11-xx-is-increasing-when-x0/4186249#4186249 – zhw. Sep 01 '21 at 19:53

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The sign of $f'(x)$ is the same as the sign of

$$\ln\left(1+\frac1x\right) - \frac{1}{x+1} = \ln\frac{x+1}{x} - \frac{1}{1+x}$$

So all you have to do is examine this sign. Now, you can first play around and plug a couple of values of $x$ into the expression. Try plugging in $x=0.1, 1, 10, 100$. You might notice that in all cases, you get a positive result, which should give you the idea that the expression will likely always be positive.

Now, of course, you still have to prove that the expression is positive, but at least now you have an idea of what to aim for.

To prove the expression is always positive, the way I would go about it is by proving the following:

  1. The function $x\mapsto \ln\left(1+\frac1x\right) - \frac{1}{x+1}$ is decreasing.
  2. The limit of the function as $x\to \infty$ is $0$.

If you prove both points above, you can conclude the function $x\mapsto \ln\left(1+\frac1x\right) - \frac{1}{x+1}$ is always positive.

5xum
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  • Thanks. So, $f'(x), \left(1+\frac1x\right), f(x), \frac1{x^2}$ are positive. From this, can we comment about $g'(x)$? – aarbee Sep 01 '21 at 08:22
  • $\lim_{x\to\infty}g'(x)=0$. Also, $g'(x)=f'(x)(-\frac1{x^2})+f''(x)(1+\frac1x)+f'(x)(-\frac1{x^2})+f(x)(\frac2{x^3})$. First three terms are negative. Last term is positive. So, not able to conclude. Am I doing it right? – aarbee Sep 01 '21 at 08:37
  • @aarbee $g(x)=f(x)\cdot \left(1+\frac{1}{x}\right)$. You know that $f$ is a nonnegative decreasing function, and $x\mapsto 1+\frac1x$ is a nonnegative decreasing function. The product of decreasing functions is... – 5xum Sep 01 '21 at 08:40
  • Got it, thanks. – aarbee Sep 01 '21 at 08:44
  • Hi. In your last comment, you have written $f$ is a nonnegative decreasing function. Do you mean it? Or do you mean it's an increasing function? Haven't we proved that $f$ is an increasing function since $f'\gt0$? – aarbee Sep 04 '21 at 10:32
  • Also, are we getting $g(x)=$ product of an increasing and deceasing function? Can we conclude anything from this? – aarbee Sep 04 '21 at 10:36
  • @aarbee Yeah, it was a typo, $f$ is an increasing function. It was a typo because the derivative is decreasing. My bad. – 5xum Sep 04 '21 at 20:19
  • Thank you for clarifying but sorry I am still not clear about $g(x)$, which is a product of an increasing and a decreasing function. Also, I have written $g'(x)$ above. Can we conclude anything from this? – aarbee Sep 05 '21 at 05:21