求下列函数的单调区间及极值点。f(x)=(2x-5)3√x^2

2024-12-23 17:43:18
推荐回答(1个)
回答1:

f(x) = (2x-5)*x^(2/3) = 2x^(5/3) -5x^(2/3)
f ′(x) = (10/3)x^(2/3)-(10/3)x^(-1/3)
= (10/3)x^(-1/3) * (x-1)

单调减区间:(-∞,1)
单调增区间:(1,+∞)

当x=1时,极小值f(1)=(2-5)*1^(2/3) = -3