 Solution

For $f=x\ln x$, the derivatives are $f^{\prime }=1+\ln x$ and $f^{\prime \prime }=1/x>0$. Its unique stationary point is the global minimum $(e^{-1},-e^{-1})$. For $g=\ln x/x$, $g^{\prime }=(1-\ln x)/x^2$ changes from positive to negative at $x=e$, giving the global maximum $(e,e^{-1})$. Their endpoint limits areAn intersection requires $(x^2-1)\ln x=0$, and on $x>0$ both factors vanish only at $x=1$. There $f=g=0$ and $f^{\prime }=g^{\prime }=1$, proving a common tangent line $y=x-1$. In fact,because the two numerator factors always have the same sign. Thus the curves touch at $(1,0)$ without exchanging their order.