$x(a^{1/x}-1)$ is decreasing

Question

Prove that $f(x)=x(a^{1/x}-1)$ is decreasing on the positive $x$ axis for $a\geq 0$.

My Try:

I wanted to prove the first derivative is negative.

$\displaystyle f'(x)=-\frac{1}{x}a^{1/x}\ln a+a^{1/x}-1$. But it was very difficult to show this is negative. Any suggestion please.

Around zero. With $a^x = e^{\ln(a) \cdot x} = 1 + \ln(a) \cdot x + ...$. — aventurin, Jul 08 '16 at 22:29
But I don't see how its useful. Here I have $a^{1/x}$, not $a^x$ — Extremal, Jul 08 '16 at 22:31
@EpsilonDelta You can prove that $f(u)$ where $u=1/x$ is increasing instead, it's essentially the same. — Burrrrb, Jul 08 '16 at 22:35

score 1 · Accepted Answer · answered Jul 08 '16 at 22:38

1

If $f$ is convex on $\mathbb R,$ then $(f(y) - f(0))/y$ increases as $y$ increases from $0.$ The function $y \to a^y$ is convex on $\mathbb R$ for any $a\ge 0.$ The result falls right out.

answered Jul 08 '16 at 22:38

zhw.

105,693

Just wondering: Why the down vote? – zhw. Jul 09 '16 at 03:15
Welcome to the club of victims of unjustified downvotes ! Cheers. – Claude Leibovici Jul 09 '16 at 05:54
Rough crowd.$,,$ – zhw. Jul 09 '16 at 16:20

score 0 · Answer 2 · answered Jul 08 '16 at 22:39

Assume that $a\ne 0,1$, since $f$ is clearly decreasing for $x>0$ in these two cases.

Let $g(t)=a^t(1-t\ln a)$; then $g(0)=1$ and $g^{\prime}(t)=-a^t t(\ln a)^2<0$ for $t>0$, so $g(t)<1$ for $t>0$.

Letting $t=\frac{1}{x},\;\;$ $f^{\prime}(x)=a^{1/x}\left(1-\frac{\ln a}{x}\right)-1=a^t(1-t\ln a)-1=g(t)-1<0$ for $x>0$

score 0 · Answer 3 · edited Apr 13 '17 at 12:19

0

Using simple algebra it is possible to prove that $g(x) = f(1/x) = \dfrac{a^{x} - 1}{x}$ is strictly increasing for $x > 0, a > 0, a \neq 1$ and $x$ being rational. The extension to irrational values of $x$ is easily done by considering sequences of rationals converging to $x$.

edited Apr 13 '17 at 12:19

Community

1

answered Jul 09 '16 at 10:25

Paramanand Singh

87,309

$x(a^{1/x}-1)$ is decreasing

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