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当x>1时,求函数f(x)=(x*x-3x+1)/(x+1)的值域
人气:177 ℃ 时间:2020-05-20 06:14:06
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)^2
x>1
x∈(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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