18. Given the quadratic function $f(x)=4 x^{2}-4 a x+\left(a^{2}-2 a\right.$ +2 ) has a minimum value of 2 on $0 \leqslant x \leqslant 1$. Find the value of $a$.
$$
\text { Three, 18. } f(x)=4\left(x-\frac{a}{2}\right)^{2}-2 a+2 \text {. }
$$

When $0 \leqslant \frac{a}{2} \leqslant 1$ i.e., $0 \leqslant a \leqslant 2$, $f_{\text {min}}(x)=-2 a+2=2$. This gives $a=0$. When $\frac{a}{2}>1$, i.e., $a>2$,
$$
f_{12}(x)=f(1)=4-4 a+a^{2}-2 a+2=2 \text {. }
$$

We can find $a=3-\sqrt{5}$ (discard) or $a=3+\sqrt{5}$.