Lời giải:

• \( D = [-1, 3] \)
• \( y = \sqrt{4 - (x-1)^2} + (x-1)^2 - 3 \)
• Đặt: \( t = (x-1)^2 \)

\( -1 \leq x \leq 3 \Leftrightarrow -2 \leq x-1 \leq 2 \Leftrightarrow 0 \leq t \leq 4 \)

• \( y = \sqrt{4 - t} + t - 3 = g(t) \)
• \( g'(t) = 1 - \frac{1}{2\sqrt{4-t}} = 0 \Leftrightarrow 2\sqrt{4-t} = 1 \) \( \Leftrightarrow 4 - t = \frac{1}{4} \Leftrightarrow t = \frac{15}{4} \)
• \( g(0) = -1, g(4) = 1, g\left(\frac{15}{4}\right) = \frac{5}{4} \)
• \( Min  y = \min\limits_{[0,4]} g(t) = -1\) khi \(t = 0 \Leftrightarrow x = 1 \)
• \( Max  y = \max\limits_{[0,4]} g(t) = \frac{5}{4} \) khi \(x = \frac{2 \pm \sqrt{15}}{2} \)