2. Let $0 \leqslant x \leqslant 2$, the minimum value of the function $y=4^{x-\frac{1}{2}}-3 \cdot 2^{x}+5$ is

保留源文本的换行和格式，翻译结果如下：

2. Let $0 \leqslant x \leqslant 2$, the function $y=4^{x-\frac{1}{2}}-3 \cdot 2^{x}+5$'s minimum value is
2. $\frac{1}{2}$.
$y=\frac{1}{2} \cdot 2^{2 x}-3 ; 2^{x}+5=\frac{1}{2}\left(2^{x}-3\right)^{2}+\frac{1}{2}$, since $2^{x} \in[1,4]$, when $2^{x}=3$, $y_{\min }=$ $\frac{1}{2}$.