Cho hàm số $y = \left\{ \begin{array}{l}\dfrac{2}{{x - 1}}{\rm{  }},{\rm{  }}x \in \left( { - \infty ;0} \right)\\\sqrt {x + 1} {\rm{ }},{\rm{  }}x \in \left[ {0;2} \right]\\{x^2} - 1{\rm{ }},{\rm{  }}x \in \left( {2;5} \right]\end{array} \right.$. Tính \(f\left( 4 \right)\), ta được kết quả:

\( - 1 \le m \le 1\)

\(\dfrac{3}{2} \le m \le 2\)

\(0 \le m \le 2\)

\(0 \le m < \dfrac{3}{2}\)

\(\overrightarrow v  = \left( { - 1;2} \right)\)    

\(\overrightarrow v  = \left( {1; - 1} \right)\)

\(\overrightarrow v  = \left( {1;1} \right)\)      

\(\overrightarrow v  = \left( {1;0} \right)\)

\({M_1}\left( {2;1} \right)\).

\({M_2}\left( {1;1} \right).\)

\({M_3}\left( {2;0} \right).\)

\({M_4}\left( {0; - 2} \right).\)

\(A\left( {2;0} \right).\)

\(B\left( {3;\dfrac{1}{3}} \right).\)

\(C\left( {1; - 1} \right).\)

\(D\left( { - 1; - 3} \right).\)

\(f\left( { - 1} \right) = 5.\)

\(f\left( 2 \right) = 10.\)

\(f\left( { - 2} \right) = 10.\)

\(f\left( {\dfrac{1}{5}} \right) =  - 1.\)

\({\rm{D}} = \mathbb{R}.\)

\({\rm{D}} = \left( {1; + \infty } \right).\)

\({\rm{D}} = \mathbb{R}\backslash \left\{ 1 \right\}.\)

\({\rm{D}} = \left[ {1; + \infty } \right).\)