The maximum value of $[x(x-1)+1]^{1/3}$ for $0\le x\le 1$ is

1. Let $g(x)=x(x-1)+1=x^2-x+1$; maximise its cube root. 2. $g'(x)=2x-1=0 \Rightarrow x=\tfrac{1}{2}$ (this gives the minimum of $g$). 3. Check endpoints: $g(0)=1$, $g(1)=1$, $g(\tfrac{1}{2})=\tfrac{3}{4}$. 4. Maximum of $g$ on $[0,1]$ is $1$, so $[g]^{1/3}=1^{1/3}=1$. _Source: syllabus Mathematics Ch 6 "Application of Derivatives", p.30_