已知g(x)=-x^2-3,f(x)是二次函數(shù)，當(dāng)x∈[-1,2]時(shí)，f(x)的最小值為1，且g(x)+f(x)是奇函數(shù)，求f(x)的表達(dá)式。請(qǐng)?jiān)敿?xì)說(shuō)明

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設(shè): f(x)=ax^2+bx+c則: g(x)+f(x)=(a-1)x^2+bx+(c-3)由于g(x)+f(x)是奇函數(shù)因此: (a-1)(-x)^2+b(-x)+(c-3) = -[)=(a-1)x^2+bx+(c-3)(a-1)x^2 + (c-3) = 0因此: a=1, c=3f(x) = x^2+bx+3 = (x+b/2)^2 + (3-b^2/4)f(-1) = 4-b, f(-1)取最小值1時(shí): b=3f(2) = 2b+7, f(2)取最小值1時(shí): b=-3f(-b/2) = 3-b^2/4, f(x)取最小值1時(shí): b=2*genhao(2)或-2*genhao(2)顯然, f(x)在x∈[-1,2]上, 于x=-1或2時(shí), 最小值1因此:f(x) = x^2+3x+3, 或: f(x) = x^2-3x+3