Ta có: \(f(x) = {x^2} + 2 + \sqrt {2 - x};\)
\(g(x) = - 2{x^3} - 3x + 5\)
\(u(x) = \left\{ \begin{array}{l}\sqrt {3 - x} ,x < 2\\\sqrt {{x^2} - 4} ,x \ge 2\end{array} \right.\); \(v(x) = \left\{ \begin{array}{l}\sqrt {6 - x} ,x \le 0\\{x^2} + 1,x > 0\end{array} \right.\)
Suy ra
\(f( - 2) - f(1) = {( - 2)^2} + 2 + \sqrt {2 + 2} \)
\(- ({1^2} + 2 + \sqrt {2 - 1} ) = 8 - 4 = 4\)
\(g(3) = - {2.3^3} - 3.3 + 5 = - 58\);
\(f( - 7) - g( - 7) = {( - 7)^2} + 2 + \sqrt {2 + 7}\)
\( - {\rm{[}} - 2.{( - 7)^3} - 3.( - 7) + 5] = - 658\);
\(f( - 1) - u( - 1) = {\left( { - 1} \right)^2} + 2 +\)
\( \sqrt {2 - \left( { - 1} \right)} - \sqrt {3 - \left( { - 1} \right)} \)
\(= 3 + \sqrt 3 - 2 = 1 + \sqrt 3 \)
(do \( - 1 < 2 \Rightarrow u\left( { - 1} \right) = \sqrt {3 - \left( { - 1} \right)} \)
\(u(3) - v(3) = \sqrt {{3^2} - 4} - ({3^2} + 1) \)
\(= \sqrt 5 - 10\); (do \(3 > 2 > 0\))
\(v(0) - g(0) = \sqrt {6 - 0} \)
\(-\left( { - 2.0 - 3.0 + 5} \right) \) \(= \sqrt 6 - 5\);
\(\dfrac{{f(2) - f( - 2)}}{{v(2) - v( - 3)}} = \dfrac{{6 - 8}}{{5 - 3}} = - 1\)
