已知函数(f(x)=-x^3-3x^2+(1+a)x+b(a<0,bin R)),
若(|f(x)|)在([-2,0])上的最大值为(M(a,b)),求(M(a,b))的最小值
[egin{align*}
extbf{解答:记}M&=M(a,b) extbf{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}\
herefore M&=dfrac{1}{4} extbf{当}a=-dfrac{13}{4},b=-dfrac{1}{4} extbf{时等号取到}\ \
extbf{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}\
herefore M&=dfrac{1}{4} extbf{当}a=-dfrac{13}{4},b=-dfrac{1}{4} extbf{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.