On what interval is the curve y = x ln(x - 1) concave upward
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On what interval is the curve y = x ln(x - 1) concave upward?
题目意思是求自变量x的单调递增区间,因为y = x ln(x - 1)函数有意义,x>1,解决函数单调性的通用方法是求导(微分),一次求导y'=ln(x-1)+1/(x-1),此时判断y'的符号仍然有困难,二次求导
y''=1/(x-1)-1/(x-1)^2=(x-2)/(x-1)^2,此时容易得x>2时y">0,所以y'在x=2时有最小值,带进去y'最小值为0,也就是y'大于等于0,所以函数在自变量有意义(x大于1的前提下)都是单调递增的,因此单调递增区间应该是(1,正无穷)
y''=1/(x-1)-1/(x-1)^2=(x-2)/(x-1)^2,此时容易得x>2时y">0,所以y'在x=2时有最小值,带进去y'最小值为0,也就是y'大于等于0,所以函数在自变量有意义(x大于1的前提下)都是单调递增的,因此单调递增区间应该是(1,正无穷)
On what interval is the curve y = x ln(x - 1) concave upward
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