詳解：(1)在 \(x \le 2\) 時， \(f(x) = {x^2} + x - 1\) 為多項式函數，必連續且可微分

(2)在 \(x > 2\) 時， \(f(x) = ax + b\) 為多項式函數，必連續且可微分

(3)唯一可能有問題的位置是 \(x = 2\) ，檢驗左極限是否等於右極限?

\(\lim\limits_{x \to {{\rm{2}}^ - }} f(x) = \lim\limits_{x \to {{\rm{2}}^ - }} ({x^2} + x - 1) = 5\)

\(\lim\limits_{x \to {{\rm{2}}^ + }} f(x) = \lim\limits_{x \to {{\rm{2}}^ + }} (ax + b) = 2a + b\)

\(2a + b = 5\)

檢驗左導數是否等於右導數?

\({f'_ - }(2) = {\left. {(2x + 1)} \right|_{x = 2}} = 5\)

\({f'_ + }(2) = {\left. a \right|_{x = 2}} = a\)

得到

\(\left\{ {\begin{array}{*{20}{c}}{{\rm{2}}a + b = 5 \cdots  \cdots (1)}\\{a = 5 \cdots  \cdots (2)}\end{array}} \right.\)

\(a = 5\)，\(b =  - 5\)