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一九八三年各省、市、自治区联合数学竞赛中,有这样一道试题: 已知函数f(x)=ax~2-c,满足 -4≤f(1)≤-l,-1≤f(2)≤5。求f(3)的范围。用待定系数法能解答本题: 解∵f(1)=a-c (1) f(2)=4a-c (2) 解(1)与(2)联立之方程组得到 a=1/3(f(2)-f(1)), c=1/3(f(2)-4f(1))、∴ f(3)=9a-a -1/3(8f(2)-3f(1)) ∵-4≤f(1)≤-1, -1≤f(2)≤5。∴5/3≤-5/3f(1)≤20/3, -8/3≤8/3f(2)≤40/3。
In the joint mathematics competition of provinces, municipalities, and autonomous regions in 1983, there was a question: Known function f(x)=ax~2-c, satisfies -4≤f(1)≤-l, -1≤f (2) ≤ 5. Find the range of f(3). The undetermined coefficient method can be used to solve this problem: Solution ∵ f(1)=ac (1) f(2)=4a-c (2) Solve equations of (1) and (2) to obtain a=1/3 ( f(2)-f(1)), c=1/3(f(2)-4f(1)), ∴f(3)=9a-a-1/3(8f(2)-3f(1) )) ∵-4≤f(1)≤-1, -1≤f(2)≤5. ∴5/3 ≤ -5/3f(1) ≤ 20/3, -8/3 ≤ 8/3f(2) ≤ 40/3.