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For integers $n \geq3$, I want to prove that $f(x)=n^x-x^n$ is increasing on the interval $x\geq n$.

Also I want to prove that for integers $x\geq 3$, $f(x)=x^{x+1}-(x+1)^x$ is increasing.

I differentiatied the function but $x$ being in both powers and bases, I couldn't find critical points etc.

Can anyone give me hints?

zxcvber
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1 Answers1

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$f(x) = n^x - x^n = e^{x \ln n} - e^{n \ln x}$

Clearly $f(n) =0 $

We are given that $x \ge n\ge 3$ with $n\in \mathbb Z$ and note that $x >\ln x \ge \ln n > 1$

Consider $ g_1(x) = x \ln n$ and $g_2(x) = n \ln x$

$g_1'(x) = \ln n $ and $g_2'(x) = \frac nx $

For $x>n, g_1'(x)>1$ and $g_2'(x)<1 \implies g_1(x)-g_2(x)$ increasing and $\therefore f(x)$ increasing.

\begin{align}f(x) &=x^{x+1}-(x+1)^x \\[3ex] &= x^x\left(x - \left(\frac{x+1}{x}\right)^x\right) \\[3ex] &= x^x\left(x - \left(1+\frac{1}{x}\right)^x\right) \\[3ex] \end{align}

Clearly $x^x$ is increasing and also since $2.37 < \left(1+\frac{1}{x}\right)^x <e$ for $x \ge 3$, the second term is increasing also, giving the result.

Joffan
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  • You seem to argue that if $a(x) - b(x)$ is increasing then $\log a(x) - \log b(x)$ is increasing? This is not true. – Winther Jun 19 '16 at 04:19
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    @Winther at minimum I think that it is true, in fact, at the point where $a(x) =b(x)$; I guess it might not be true everywhere though, I'll have to think about that. – Joffan Jun 19 '16 at 04:59
  • Yes it is true where $a(x) = b(x)$ (if $a=b>0$), but a simple counter-example is $a(x) = \frac{1}{x^2}$ and $b(x) = \frac{1}{x}$ for $x\geq 2$. There $a(x) - b(x)$ is increasing but $\log a(x) - \log b(x)$ is decreasing. – Winther Jun 19 '16 at 05:04
  • @Joffan Can you help me out with showing why $x^{x+1}-(x+1)^x$ increases please? – zxcvber Jun 23 '16 at 13:15