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Prove that the function \( f(x)=x^{x^{x}}-x^x \) has an inflection point at \(x=1 \).


I may be overlooking something here.. but I cannot prove that directly.

I presume you can show that \(f'(1) = 1\) showing that \(x=1\) is a stationary point. Now observe that for \(0 < x < 1\) we have \(x^x < x^1 = x\). So \(x^{x^x} < x^x\), i.e. \(f(x) < 0 = f(1).\) Also, for \(x > 1\) we have \(x^x > x\) giving \(f(x) > f(1)\). So \(x=1\) is a point of inflection.

Demetres I cannot follow your solution but I have another approach.
A bit tedious however.

Finding the second derivative (easy but a tedious task) we see that \( f''(1)=0 \).
In order to show that \( f \) has an inflection point at \( x=1 \) we're envoking a theorem that states:

Theorem: If \( g \) is differentiable , \(g(a)=0 \) and \(g'(a) \neq 0 \) then \(g \) changes sign at a neigbourhood of \(a \).

To complete the proof we have to prove that \( g'''(1) \neq 0 \) . Of course we are not computing the third derivative. Instead we are expanding the function at a Taylor series around \( x=1 \) and see that the coefficient of the third power is indeed non zero and we are done.