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当x>1时,求f(x)=x²-3x+1/x+1的值域
人气:180 ℃ 时间:2020-05-01 23:00:03
解答
f(x)=(x^2-3x+1)/(x+1) = (x^2+x-4x-4+5)/(x+1) = x - 4 + 5/(x+1)f'(x) = 1 - 5/(x+1)^2 = { (x+1)^2 - 5 ] / (x+1)^2x>1x∈(1,-1+√5)时,f'(x)<0,f(x)单调减x∈(-1+√5,+∞)时,f'(x)>0,f(x)单调增x=-1+√5时,最小值f(x)min= -1+√5 - 4 + 5/(-1+√5+1) = -5+2√5值域【-5+2√5,+∞)
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