(1)
y= x/(1+x^2)
y' =[(1+x^2)- 2x^2]/(1+x^2)^2
=(1-x^2)/(1+x^2)^2
y'=0
x=1 or -1
y'|x=1+ <0
y'|x=1- >0
x=1 (max)
y'|x=-1+ >0
y'|x=-1- <0
x=-1 (min)
max f(x)= f(1) =1/2
min f(x)= f(-1) =-1/2
(2)
y= sinx +cosx
= √2 sin(x+ π/4)
min y= -√2
max y= √2
(3)
y=ln(1+x^2)
y' = 2x/(x^2+1)
y'=0
x=0
y'|x=0+ >0
y'|x=0- < 0
=> x=0 (min)
min y
=y(0)
=0
y(-1)= ln2
y(2)=ln5
最大值= ln5
最小值 = 0
追问第一个x属于0到1不包括负一呀
追答(1)
y= x/(1+x^2)
y' =[(1+x^2)- 2x^2]/(1+x^2)^2
=(1-x^2)/(1+x^2)^2
y'=0
x=1
y'|x=1+ 0
x=1 (max)
max f(x)= f(1) =1/2
f(0) = 0
=> min f(x)= f(0) =0