Cho hai hàm số f(x) = {x^3} + a{x^2} + bx + c\) và g(x)= x + 4 / x^2

Ta có: \(f'\left( x \right) = 3{x^2} + 2ax + b\,;\,\,g'\left( x \right) = 1 - \frac{8}{{{x^3}}}\).

Xét \(g'\left( x \right) = 0 \Leftrightarrow 1 - \frac{8}{{{x^3}}} = 0 \Leftrightarrow x = 2\).

Ta có: \(g\left( 1 \right) = 5\,;\,\,g\left( 2 \right) = 3\,;\,\,g\left( 4 \right) = \frac{{17}}{4} \Rightarrow \mathop {\min }\limits_{\left[ {1;4} \right]} g\left( x \right) = 3\) tại \(x = 2.\)

\( \Rightarrow \left\{ {\begin{array}{*{20}{l}}{f'\left( 2 \right) = 0}\\{f\left( 1 \right) = 4}\\{f\left( 2 \right) = 3}\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{12 + 4a + b = 0}\\{a + b + c = 3}\\{8 + 4a + 2b + c = 3}\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{a = - 4}\\{b = 4.}\\{c = 3}\end{array}} \right.} \right.} \right.\)

Khi đó, \(f'\left( x \right) = 3{x^2} - 8x + 4 = 0 \Leftrightarrow x = 2\) vì \(x \in \left( {1;4} \right)\). Có \(f\left( 1 \right) = 4;f\left( 2 \right) = 3;f\left( 4 \right) = 19\).

Vậy \(\mathop {\max }\limits_{\left[ {1;4} \right]} f\left( x \right) = 19.\) Chọn D.