函数f(x)=ax2+bx满足:1≤f(-1)≤2,2≤f(1)≤4,求f(-2)的取值范围.
问题描述:
函数f(x)=ax2+bx满足:1≤f(-1)≤2,2≤f(1)≤4,求f(-2)的取值范围.
答
由f (x)=ax2+bx得 f (-1)=a-b,f (1)=a+b,f (-2)=4a-2b∴a= [f (1)+f(-1)],b= [f (1)-f(-1)] 则f(-2)=2[f (1)+f (-1)]-[f (1)-f (-1)]=3f (-1)+f (1)由条件1≤f(-1)≤2,2≤f (1)≤4可得3×1+2≤3f(-1)...