MT【136】一道三次函数的最佳逼近问题

已知函数\(f(x)=-x^3-3x^2+(1+a)x+b(a<0,b\in R)\),
若\(|f(x)|\)在\([-2,0]\)上的最大值为\(M(a,b)\),求\(M(a,b)\)的最小值

\[\begin{align*} \textbf{解答:记}M&=M(a,b)\textbf{则}\\ 3M&\ge|f(-2)|+\dfrac{1}{2}|f(-\frac{1}{2})|+\dfrac{3}{2}|f(-\frac{3}{2})| \\ &=|-6-2a+b|+\dfrac{1}{2}|-\dfrac{9}{8}-\dfrac{1}{2}a+b|+\dfrac{3}{2}|-\dfrac{39}{8}-\dfrac{3}{2}a+b|\\ &\ge|-6-2a+b-\dfrac{9}{16}-\dfrac{1}{4}a+\dfrac{1}{2}b+\dfrac{117}{16}+\dfrac{9}{4}a-\dfrac{3}{2}b| \\ &=\dfrac{3}{4}\\ \therefore M&=\dfrac{1}{4}\textbf{当}a=-\dfrac{13}{4},b=-\dfrac{1}{4}\textbf{时等号取到}\\ \\ \textbf{或者}\\ \\ 3M&\ge|f(0)|+\dfrac{3}{2}|f(-\frac{1}{2})|+\dfrac{1}{2}|f(-\frac{3}{2})| \\ &=|b|+\dfrac{3}{2}|-\dfrac{9}{8}-\dfrac{1}{2}a+b|+\dfrac{1}{2}|-\dfrac{39}{8}-\dfrac{3}{2}a+b|\\ &\ge|b+\dfrac{27}{16}+\dfrac{3}{4}a-\dfrac{3}{2}b-\dfrac{39}{16}-\dfrac{3}{4}a+\dfrac{1}{2}b| \\ &=\dfrac{3}{4}\\ \therefore M&=\dfrac{1}{4}\textbf{当}a=-\dfrac{13}{4},b=-\dfrac{1}{4}\textbf{时等号取到} \end{align*}\]

评：如果是选择题,画图更快捷点,二次的考察较多.