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For the Legendre transform, maximize $px-f(x)$ over $x>0$. Its derivative is $p-1-\log x$, and strict concavity makes the stationary point $x=e^{p-1}$ the unique maximizer. Substitution gives the convex conjugate of x log xThe range of $f^{\prime }$ is all of $\mathbb{R}$, so all real dual arguments have this finite value. Every point of the original graph is supported by its tangent of slope $f^{\prime }(x)$; the conjugate supremum therefore recovers $f$ from these supporting lines. Equivalently, the inverse derivative relation for this strictly convex differentiable Legendre transform gives $f^{\ast \ast }=f$ on $x>0$. If an extended-real function on all of $\mathbb{R}$ is wanted, its closed convex extension has value zero at $x=0$ and infinity for $x<0$; biconjugation includes that endpoint closure.