Cho hàm số \(f\left( x \right) = \left\{ \begin{array}{l}2x + 3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,khi\,\,x \ge 1\\\dfrac{{{x^3} + 2{x^2} - 7x + 4}}{{x - 1}}\,\,khi\,\,x < 1\end{array} \right.\). Giá trị của \(f'\left( 1 \right)\) bằng:

\( - \dfrac{3}{{{{\left( {x + 2} \right)}^2}}}\)             

\(\dfrac{3}{{x + 2}}\) 

\(\dfrac{3}{{{{\left( {x + 2} \right)}^2}}}\) 

\(\dfrac{2}{{{{\left( {x + 2} \right)}^2}}}\)

\(\dfrac{a}{c}\) 

\(\dfrac{{ad - bc}}{{{{\left( {cx + d} \right)}^2}}}\) 

\(\dfrac{{ad + bc}}{{{{\left( {cx + d} \right)}^2}}}\)                         

\(\dfrac{{ad - bc}}{{cx + d}}\) 

\(y' = \dfrac{{{x^2} - 2x}}{{{{\left( {x - 1} \right)}^2}}}\) 

\(y' = \dfrac{{{x^2} + 2x}}{{{{\left( {x - 1} \right)}^2}}}\) 

\(y' = \dfrac{{{x^2} + 2x}}{{{{\left( {x + 1} \right)}^2}}}\) 

\(y' = \dfrac{{ - 2x - 2}}{{{{\left( {x - 1} \right)}^2}}}\)

Bước 1

Bước 2

Bước 3

Tính toán đúng

\(\dfrac{1}{6}\) 

\(\dfrac{1}{{12}}\) 

\( - \dfrac{1}{6}\) 

\( - \dfrac{1}{{12}}\)

\(\,\dfrac{{\Delta y}}{{\Delta x}} = \dfrac{{2{x^3} - 2{{\left( {\Delta x} \right)}^3}}}{{\Delta x}}.\)

\(\dfrac{{\Delta y}}{{\Delta x}} = 2{\left( {\Delta x} \right)^2}.\)

\(\dfrac{{\Delta y}}{{\Delta x}} = 6{x^2} + 6x\Delta x + 2{\left( {\Delta x} \right)^2}.\)

\(\dfrac{{\Delta y}}{{\Delta x}} = 3{x^2} + 3x\Delta x + {\left( {\Delta x} \right)^2}.\)