摘要：求函数f(x)=4x2-4ax+a2-2a+2在［0.2］上的最值. 解析:f(x)=4(x-)2-2a+2. (1)当≤0时.即a≤0,f(x)在［0.2］上递增. ∴f(x)max=f(2)=a2-10a+18. f(x)min=f(0)=a2-2a+2. (2)当≥2时.即a≥4时,f(x)在［0.2］上递减. ∴f(x)max=f(0)=a2-2a+2. f(x)min=f(2)=a2-10a+18. (3)当0≤≤2时.即0≤a≤4时. f(x)min=f()=-2a+2. ①当0≤≤1时.即0≤a≤2时, f(x)max=f(2)=a2-10a+18, ②当1≤≤2时.即2≤a≤4时. f(x)max=a2-2a+2.

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