Question

求导数学题？

Answer 1

y= (x+1)^(2/3).(x-5)^2
y'= (2/3)(x+1)^(-1/3).(x-5)^2 + 2(x+1)^(2/3).(x-5)
y'=0
(2/3)(x+1)^(-1/3).(x-5)^2 + 2(x+1)^(2/3).(x-5) =0
(x-5)^2 + 3(x+1).(x-5) =0
(x-5) ( x-5 +3x+3) =0
(x-5) ( 2x-1) =0
x=5 or 1/2
y'|x=5+ >0 , y'|x=5- <0
x=5 (min)
y'|x=1/2+ <0, y'|x=1/2- >0
x=1/2 (max)
min f(x) = f(5) =0
max f(x)
=f(1/2)
= (3/2)^(2/3).(-9/2)^2
= (27/2) .(3/2)^(1/3)

追问

令y'=0时求x，化简的那步没看懂

这一步怎么来的？

追答

y'=0
(2/3)(x+1)^(-1/3).(x-5)^2 + 2(x+1)^(2/3).(x-5) =0
(x+1)^(-1/3).(x-5)^2 + 3(x+1)^(2/3).(x-5) =0
(x+1)^(1/3) .[(x+1)^(-1/3).(x-5)^2 + 3(x+1)^(2/3).(x-5)] =0
(x-5)^2 + 3(x+1).(x-5) =0
(x-5) ( x-5 +3x+3) =0
(x-5) ( 2x-1) =0
x=5 or 1/2

追问

还有x=–1处取极小值