Cho hàm số f ( x ) = { 2 x + 3 k h i x ≥ 1 x 3 + 2 x 2 − 7 x + 4 x − 1 k h i x < 1 . Giá trị của f′(1) bằng:

\[\mathop {\lim }\limits_{{\rm{x}} \to {1^ + }} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{1}} \right)}}{{{\rm{x}} - 1}} = \mathop {\lim }\limits_{{\rm{x}} \to {1^ + }} \frac{{2{\rm{x}} + 3 - 5}}{{{\rm{x}} - 1}} = \mathop {\lim }\limits_{{\rm{x}} \to {1^ + }} \frac{{2{\rm{x}} - 2}}{{{\rm{x}} - 1}} = 2\]

\[\mathop {\lim }\limits_{{\rm{x}} \to {1^ - }} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{1}} \right)}}{{{\rm{x}} - 1}} = \mathop {\lim }\limits_{{\rm{x}} \to {1^ - }} \frac{{\frac{{{{\rm{x}}^3} + 2{{\rm{x}}^2} - 7{\rm{x}} + 4}}{{{\rm{x}} - 1}} - 5}}{{{\rm{x}} - 1}}\]

\[ = \mathop {\lim }\limits_{{\rm{x}} \to {1^ - }} \frac{{{{\rm{x}}^3} + 2{{\rm{x}}^2} - 12{\rm{x}} + 9}}{{{{\left( {{\rm{x}} - 1} \right)}^2}}} = \mathop {\lim }\limits_{{\rm{x}} \to {1^ - }} \frac{{\left( {{\rm{x}} - 1} \right)\left( {{{\rm{x}}^2} + 3{\rm{x}} - 9} \right)}}{{{{\left( {{\rm{x}} - 1} \right)}^2}}} = \mathop {\lim }\limits_{{\rm{x}} \to {1^ - }} \frac{{{{\rm{x}}^2} + 3{\rm{x}} - 9}}{{{\rm{x}} - 1}} = + \infty \]

\[ \Rightarrow \mathop {\lim }\limits_{{\rm{x}} \to {1^ + }} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{1}} \right)}}{{{\rm{x}} - 1}} \ne \mathop {\lim }\limits_{{\rm{x}} \to {1^ - }} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{1}} \right)}}{{{\rm{x}} - 1}}\]

Vậy hàm số không tồn tại đạo hàm tại x = 1.

Đáp án cần chọn là: D