(2016天津压轴题)设函数$f(x)=(x-1)^3-ax-b,x\in R$, 其中$a,b\in R$
(1)求$f(x)$的单调区间.
(2)若$f(x)$存在极值点$x_0$,且$f(x_1)=f(x_0),$其中$x_1\ne x_0$,求证:$x_1+2x_0=3$;
(3)设$a>0$,函数$g(x)=|f(x)|,$求证:$g(x)$在区间$[0,2]$上的最大值不小于$\dfrac{1}{4}$
分析:
(1)
当$a\le0,f(x)$在$(-\infty,+\infty)$单调递增.
当$a>0,f(x)$在$\left(-\infty,-\dfrac{\sqrt{3a}}{3}+1\right)\nearrow,\left(-\dfrac{\sqrt{3a}}{3}+1,\dfrac{\sqrt{3a}{3}}+1\right)\searrow,\left(\dfrac{\sqrt{3a}}{3}+1,+\infty\right)\nearrow$
(2)由于$x_0,$是$f(x)$的极值点,故由(1)知$a>0$,且$a=3(x_0-1)^2$,由题意$f(x)=f(x_0)$有且仅有两根$x_0,x_1$,容易验证$f(3-2x_0)-f(x_0)=0$
故$3-2x_0=x_0(\textbf{舍去，此时}a=0) $或$3-2x_0=x_1$即$2x_0+x_1=3$
(3)记$M(a,b)=\max\limits_{a>0,b\in R}|f(x)|$则
$$\begin{cases}
M(a,b)&\ge|f(0)|=|-1-b|\\
M(a,b)&\ge|f(\dfrac{1}{2})|=|-\dfrac{1}{8}-\dfrac{1}{2}a-b|\\
M(a,b)&\ge|f(\dfrac{3}{2})|=|\dfrac{1}{8}-\dfrac{3}{2}a-b|\\
M(a,b)&\ge|f(2)|=|1-2a-b|\\
\end{cases}$$
则\begin{align*}
6M(a,b)&\ge|f(0)|+2|f(\dfrac{1}{2})|+2|f(\dfrac{3}{2})|+|f(2)|\\
&=|-1-b|+2|-\dfrac{1}{8}-\dfrac{1}{2}a-b|+2|\dfrac{1}{8}-\dfrac{3}{2}a-b|+|1-2a-b|\\
&=|-1-b-2(-\dfrac{1}{8}-\dfrac{1}{2}a-b)+2(\dfrac{1}{8}-\dfrac{3}{2}a-b)-(1-2a-b)|\\
&=\dfrac{3}{2}
\end{align*}
故$M(a,b)\ge \dfrac{1}{4}$,当$f(x)=(x-1)^3-\dfrac{3}{4}(x-1)$时取到等号.
注：通过画图，两条直线“夹紧”曲线，得到0，1/2，3/2或者1/2，3/2，2都可以。
$3M\ge |f(0)|+\dfrac{3}{2}|f(\dfrac{1}{2})|+\dfrac{1}{2}|f(\dfrac{3}{2})|$
或者$3M\ge \dfrac{1}{2}|f(\dfrac{1}{2})|+\dfrac{3}{2}|f(\dfrac{3}{2})|+|f(2)|$
两者并起来写就是$6M(a,b)\ge | f(0)|+2|f(\dfrac{1}{2})|+2|f(\dfrac{3}{2})|+|f(2)|$