Cho hàm số f ( x ) { x 3 − 4 x 2 + 3 x x 2 − 3 x + 2 k h i x ≠ 1 0 k h i x = 1 . Giá trị của f′(1) bằng:

Ta có:\[\mathop {\lim }\limits_{{\rm{x}} \to {1^ + }} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{1}} \right)}}{{{\rm{x}} - {\rm{1}}}} = \mathop {\lim }\limits_{{\rm{x}} \to {1^ + }} \frac{{\frac{{{{\rm{x}}^{\rm{3}}} - {\rm{4}}{{\rm{x}}^{\rm{2}}}{\rm{ + 3x}}}}{{{{\rm{x}}^{\rm{2}}} - {\rm{3x + 2}}}} - {\rm{0}}}}{{{\rm{x}} - {\rm{1}}}} = \mathop {\lim }\limits_{{\rm{x}} \to {1^ + }} \frac{{{\rm{x}}\left( {{\rm{x}} - {\rm{3}}} \right)\left( {{\rm{x}} - {\rm{1}}} \right)}}{{{{\left( {{\rm{x}} - {\rm{1}}} \right)}^{\rm{2}}}\left( {{\rm{x}} - {\rm{2}}} \right)}}\]

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

\[\mathop {\lim }\limits_{{\rm{x}} \to {1^ - }} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{1}} \right)}}{{{\rm{x}} - 1}} = \mathop {\lim }\limits_{{\rm{x}} \to {1^ - }} \frac{{\frac{{{{\rm{x}}^{\rm{3}}} - {\rm{4}}{{\rm{x}}^{\rm{2}}}{\rm{ + 3x}}}}{{{{\rm{x}}^{\rm{2}}} - {\rm{3x + 2}}}} - {\rm{0}}}}{{{\rm{x}} - {\rm{1}}}} = \mathop {\lim }\limits_{{\rm{x}} \to {1^ - }} \frac{{{\rm{x}}\left( {{\rm{x}} - {\rm{3}}} \right)\left( {{\rm{x}} - {\rm{1}}} \right)}}{{{{\left( {{\rm{x}} - {\rm{1}}} \right)}^{\rm{2}}}\left( {{\rm{x}} - {\rm{2}}} \right)}}\]

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

Do đó không tồn tại giới hạn \[\mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( 1 \right)}}{{{\rm{x}} - 1}}\]

Vậy hàm số không có đạo hàm tại x = 1

Đáp án cần chọn là: D