$\min_{x\in(0,1]}f(x)=f\left(\dfrac{1}{2}\right)=-\dfrac{1}{4}$. $\max_{x\in(0,1]}f(x)=f(1)=0$. Let be $k\in\mathbb{Z}$. For $x\in(k,k+1], f(x)=2^k(x-k)(x-k-1)$. Results: $f(x)\ge-\dfrac{1}{4}>-\dfrac{8}{9}, \forall x\le1$. For $x\in(1,2]$ results $f(x)\ge-\dfrac{1}{2}>-\dfrac{8}{9}$. For $x\in(2,3]$ results $f(x)\in[-1,0]$, hence $m\in(2,3)$. $f(x)$ is strictly decreasing for $2<x\le\dfrac{5}{2}$. $f(x)=-\dfrac{8}{9}\Longrightarrow 4(x-2)(x-3)=-\dfrac{8}{9}$, with the solutions $x_1=\dfrac{7}{3}; x_2=\dfrac{8}{3}$. Hence $f(x)\ge-\dfrac{8}{9}$ for $2<x\le\dfrac{7}{3}$ and $f(x)<-\dfrac{8}{9}$ for $\dfrac{7}{3}<x\le\dfrac{5}{2}$. Using the previous results, we can conclude: $f(x)\ge-\dfrac{8}{9}, \forall x\le\dfrac{7}{3}$. $f(x)<-\dfrac{8}{9}, \forall x\in\left(\dfrac{7}{3},\dfrac{5}{2}\right]$. Results: $m=\dfrac{7}{3}$.