Cho hàm số $f\left( x \right)=\left| 2{{x}^{2}}+\left( a+4...

Cách giải:
Xét $2{{x}^{2}}+\left( a+4 \right)x+b+3$ trên $\left[ -2;3 \right]$ ta có $u'=0$ có $x=-\dfrac{a+4}{4}$
+) TH1: Nếu $-\dfrac{a+4}{4}\notin \left[ -2;3 \right]$
$\Rightarrow \left[ \begin{aligned}
& a<-16 \\
& a>4 \\
\end{aligned} \right.$
Ta có $M=\max \left( \left| u\left( -2 \right) \right|;\left| u\left( 3 \right) \right| \right)=\max \left\{ \left| 2a-b-3 \right|;\left| 3a+b+33 \right| \right\}$
$\ge \dfrac{\left| 2a-b-3 \right|+\left| 3a+b+33 \right|}{2}\ge \dfrac{\left| 2a-b-3+3a+b+33 \right|}{2}=\dfrac{5\left| a+6 \right|}{2}>25$
+) TH2: $-\dfrac{a+4}{4}\in \left[ -2;3 \right]\Rightarrow -16\le a\le 4$
Ta có $M=\max \left\{ \left| u\left( -2 \right) \right|;\left| u\left( 3 \right) \right|;\left| u\left( -\dfrac{a+4}{4} \right) \right| \right\}$
$=\max \left\{ \left| -2a+b+3 \right|;\left| 3a+b+33 \right|;\left| \dfrac{{{a}^{2}}}{8}+a-b-1 \right| \right\}$
$\ge \dfrac{\left| -2a+b+3 \right|+\left| 3a+b+33 \right|+\left| \dfrac{{{a}^{2}}}{8}+a-b-1 \right|}{4}$
$\ge \dfrac{{{\left( a+b \right)}^{2}}}{16}+\dfrac{25}{4}\ge \dfrac{25}{4}$
Vậy $\min M=\dfrac{25}{4}.$
Dấu bằng xảy ra khi $\left| -2a+b+3 \right|=\left| 3a+b+33 \right|=\left| \dfrac{{{a}^{2}}}{3}+a-b-1 \right|\Leftrightarrow \left\{ \begin{aligned}
& a=-6 \\
& b=-\dfrac{35}{4} \\
\end{aligned} \right.\Rightarrow a+4b=-41.$