回復 1# son249 的帖子
(1)
\(\begin{align}
& \left| f\left( -1 \right) \right|=\left| a-b+c \right|\le 5 \\
& \left| f\left( 0 \right) \right|=\left| c \right|\le 5 \\
& \left| f\left( 1 \right) \right|=\left| a+b+c \right|\le 5 \\
& \\
& \left| g\left( 1 \right) \right|=\left| a+b \right|=\left| a+b+c-c \right|\le \left| a+b+c \right|+\left| c \right|\le10 \\
& \left| g\left( -1 \right) \right|=\left| -a+b \right|=\left| a-b \right|=\left| a-b+c-c \right|\le \left| a-b+c \right|+\left| c \right|\le10 \\
& -1\le x\le 1,\left| g\left( x \right) \right|\le 10 \\
\end{align}\)
(2)先考慮\(a>0\)
\(g\left( x \right)\)在\(\left[ -1,1 \right]\)的最大值是\(g\left( 1 \right)=a+b=10\)
\(\begin{align}
& -5\le c=f\left( 0 \right)=f\left( 1 \right)-\left( a+b \right)\le 5-10=-5 \\
& c=-5 \\
\end{align}\)
由於\(-1\le x\le 1,f\left( x \right)\ge -5=f\left( 0 \right)\)，\(y\)軸是拋物線的對稱軸
\(\begin{align}
& b=0,a=10 \\
& f\left( x \right)=10{{x}^{2}}-5 \\
\end{align}\)
\(a<0\)時，就不寫了，答案是\(f\left( x \right)=-10{{x}^{2}}+5\)
[ 本帖最後由 thepiano 於 2016-3-1 11:19 AM 編輯 ]
