Solution verification : infimum of $\{x(\alpha-x)~|~x\in(0,1) \}$

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$$B=\{x(\alpha-x)~|~x\in(0,1)\},$$ where $~\alpha \in \mathbb Q $ is a fixed parameter

My attempt :

  • $\alpha=0$

$$x(\alpha-x)=-x^2 $$ $$ \inf(B)=-1$$

  • $\alpha<0$ $$x(\alpha-x)\le0 $$ $$ \inf(B)=\alpha -1$$
  • $\alpha\ge1$ $$x(\alpha-x)\ge x $$ $$\inf(B)=0$$
  • $\alpha \in(0,1) $ $$x(\alpha-x),~~\alpha-1<0 $$ $$ \inf(B)=\alpha-1$$