Does $f(x)=x(2^{\frac{1}{x}}-1)$ decrease with $x\in[0,\infty)$?
I calculated the derivative of $f$, which is $f^{'}(x)=2^{\frac{1}{x}}(1-\frac{1}{x}\ln2)-1$. But how to prove that $f^{'}(x)<0$ when $x\in[0,\infty)$?
Does anyone give me a hint for this? Thanks a lot!
One way to see it is to switch variables to $u=1/x$, now $\frac{2^u-1}{u}$ is an increasing function of $u$ because of convexity of $2^x$ (because it is the slope of the secant line of $2^x$ between $x=0$ and $x=u$). This gives you that $x(2^{1/x}-1)$ is a decreasing function of $x$ at least for $x>0$ where the change of variable made sense.
Dealing with $x=0$ requires you to tell us what $f(0)$ is; if it is understood as $\lim_{x \to 0^+} f(x)$, then that's $+\infty$, which achieves what you want.