How does one raise one's self?

Let $f(x) = x^x$ for $x>0$ . Then its first derivative $f'(x) = (\ln x + 1)x^x$ . Note that $f'(x) < 0$ , when $x < \dfrac 1e$ . Therefore, $x^x$ is decreasing between $0 < x < \dfrac 1e$ and increasing when $x > \dfrac 1e$ . $\boxed{\text{No, she is not correct}}$ .

It really suffices to give a concrete counterexample . $\left(\frac 1 4\right)^{1/4} = \frac 1{\sqrt 2};\ \ \ \ \left(\frac 1 2\right)^{1/2} = \frac 1{\sqrt 2}.$ Clearly, the function is not strictly increasing between $x = \tfrac14$ and $x = \tfrac12$ .

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Calculus proof

To simplify work, I consider the (natural) logarithm, $f(x) = \ln x^x = x\ln x.$ Since the logarithm, defined on positive $x$ , is a strictly increasing function, we can consider the behavior of $f$ instead of $x^x$ .

Now $f'(x) = 1 + \ln x,$ so that the function $f$ may have an extreme value if $f'(x) = 0 \ \ \therefore\ \ x = e^{-1} \approx 0.37.$

Taking second derivatives, we find $f''(x) = \frac 1 x > 0,$ so that $f$ is concave on the entire domain. This shows that $f'(x) < 0$ (and therefore $x^x$ is decreasing) for $0 < x < e^{-1}$ .

Victoria's claim is incorrect.

Here's what Chew-Seong Cheong is talking about, see his solution