Problem source

The functions $\f$ and $\g$ are defined, for $x>0$, by
\[ \f(x) = x^x\,, \ \ \ \ \ \g(x) = x^{\f(x)}\,. \]
\begin{questionparts}
\item By taking logarithms, or otherwise, show that $\f(x) > x$ for $0 < x < 1\,$. Show further that $x < \g(x) < \f(x)$ for $0 < x < 1\,$. 
Write down the corresponding results for $x > 1 \,$. 
\item Find the value of $x$ for which $\f'(x)=0\,$. 
\item Use the result $x\ln x \to 0$ as $x\to 0$ to find  $\lim\limits_{x\to0}\f(x)$, and write down  $\lim\limits_{x\to0}\g(x)\,$. 
\item Show that $  x^{-1}+\, \ln x \ge 1\,$ for $x>0$. 
Using this result, or otherwise, show that $\g'(x) > 0\,$.
\end{questionparts}
Sketch the graphs, for $x > 0$,  of $y=x$, $y=\f(x)$ and $y=\g(x)$ on the same axes.

Solution source

\begin{questionparts}
\item $\,$
\begin{align*}
&& \ln f(x) &= x \ln x \\
&&&> \ln x  \quad (\text{if } 0 < x < 1)\\
\Rightarrow && f(x) &> x\quad\quad (\text{if } 0 < x < 1)\\
\Rightarrow && x^{f(x)} &< x^x \\
&& g(x) &< f(x) \\
&& 1&>f(x) \\
\Rightarrow && x &< x^{f(x)} = g(x)
\end{align*}

\item $\,$ \begin{align*}
&& f(x) &= e^{x \ln x} \\
\Rightarrow && f'(x) &= (\ln x + 1)e^{x \ln x} \\
\Rightarrow && f'(x) = 0 &\Leftrightarrow x = \frac1e
\end{align*}

\item $\,$ \begin{align*}
&& \lim_{x \to 0} f(x) &= \lim_{x \to 0} \exp \left ( x \ln x \right ) \\
&&&= \exp \left ( \lim_{x \to 0} \left ( x \ln x \right )\right) \\
&&&= \exp \left ( 0 \right)  = 1\\
\\
&& \lim_{x \to 0} g(x) &= \lim_{x \to 0} \exp \left ( f(x) \ln x\right) \\
&&&= \exp \left (\lim_{x \to 0}  \ln x f(x)\right) \\
&&&= \exp \left (\lim_{x \to 0}  \ln x \lim_{x \to 0}f(x)\right) \\
&&&= \exp \left (\lim_{x \to 0}  \ln x\right) \\
&&&= 0
\end{align*}

\item $y = x^{-1} + \ln x \Rightarrow y' = -x^{-2} + x^{-1}$ which has roots at $x =1$, therefore the minimum value is $1$. (We can see it's a minimum by considering $x \to 0, x \to \infty$.

So 
\begin{align*}
&& g'(x) &= x^{f(x)} \cdot (f'(x) \ln x + f(x) x^{-1})\\
&&&= x^{f(x)} \cdot f(x) \cdot ((1+\ln x) \ln x + x^{-1}) \\
&&&= x^{f(x)} \cdot f(x) \cdot (\ln x + x^{-1} + (\ln x)^2) \\
&&&\geq x^{f(x)} \cdot f(x) > 0
\end{align*}

\end{questionparts}


\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){((#1)+.5)*((#1)-1)*((#1)-2.1)};
    \def\xl{-1};
    \def\xu{3};
    \def\yl{-1};
    \def\yu{3};
    
    % Calculate scaling factors to make the plot square
    \pgfmathsetmacro{\xrange}{\xu-\xl}
    \pgfmathsetmacro{\yrange}{\yu-\yl}
    \pgfmathsetmacro{\xscale}{10/\xrange}
    \pgfmathsetmacro{\yscale}{10/\yrange}
    
    % Define the styles for the axes and grid
    \tikzset{
        axis/.style={very thick, ->},
        grid/.style={thin, gray!30},
        x=\xscale cm,
        y=\yscale cm
    }
    
    % Define the bounding region with clip
    \begin{scope}
        % You can modify these values to change your plotting region
        \clip (\xl,\yl) rectangle (\xu,\yu);
        
        % Draw a grid (optional)
        % \draw[grid] (-5,-3) grid (5,3);

        \filldraw ({1/e}, {exp(-1/e)}) circle (1.5pt) node[above]{$(\frac1e, e^{-1/e})$};
        \filldraw (1, 1) circle (1.5pt) node[below right]{$(1,1)$};
        % \filldraw (2.1, 0) circle (1.5pt) node[below]{$r$};
        
        \draw[thick, blue, smooth, domain=0:\xu, samples=100] 
            plot (\x, {\x});
        \draw[thick, green, smooth, domain=0.00001:2, samples=100] 
            plot (\x, {exp(ln(\x)*\x)});
        \draw[thick, red, smooth, domain=0.00001:2, samples=100] 
            plot (\x, {exp(ln(\x)*exp(ln(\x)*\x))});

        \node[blue, above, rotate=45] at (2,2) {\tiny $y = x$}; 
        \node[green, below, rotate=70] at (1.6,{exp(ln(1.6)*1.6)}) {\tiny $y = x^x$}; 
        \node[red, above, rotate=60] at (1.3,{exp(ln(1.3)*exp(ln(1.3)*1.3))}) {\tiny $y = x^{x^x}$}; 
    \end{scope}
    
    % Set up axes
    \draw[axis] (\xl,0) -- (\xu,0) node[right] {$x$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above] {$y$};
    
    \end{tikzpicture}
\end{center}