Question

If $u\left(x\right)$ is larger of $x-\left(x^3/6\right)$ and ${\tan }^{-1}x$ for $0<x\le 1$ then $48u\left(1/2\right)$ is equal to

Answer 1

$for0<x\le 1f\left(x\right)=x-\frac{x^3}{6}f\left(0\right)=0f^{\prime }\left(x\right)=1-\frac{x^2}{2}>0f\left(1\right)=\frac{5}{6}g\left(x\right)={\tan }^{-1}xg\left(0\right)=0g\left(1\right)=\frac{\pi }{4}andg\left(x\right)isalsoanincreasingfunctionin0<x\le 1but\frac{\pi }{4}<\frac{5}{6}sofor0<x\le 1f\left(x\right)islargerfunctionf\left(\frac{1}{2}\right)=\frac{23}{48}48f\left(\frac{1}{2}\right)=23$