Return to Revisions

In fact $\ln(x^2 + x^\beta)$ is concave for $x > 0$ iff $3-2\sqrt{2} \le \beta \le 3 + 2 \sqrt{2}$. This comes from writing $$ \dfrac{d^2}{dx^2} \ln(x^2 + x^\beta) = \dfrac{x^2}{(x^2 + x^\beta)^2} \left(-\beta x^{2\beta - 4} + (\beta^2 - 5 \beta + 2) x^{\beta - 2} -2\right)$$ and noting that the discriminant of $-\beta t^2 + (\beta^2 - 5 \beta + 2) t - 2$ with respect to $t$ is $(\beta^2 - 6 \beta + 1) (\beta - 2)^2$.