Question

函数f(x)=x²/x+1的单调区间及其极值

Answer 1

x^2
=x(x+1)-x
=x(x+1)-(x+1) +1
f(x)
= x^2/(x+1) ; x≠-1
=x -1 + 1/(x+1)
f'(x) = 1 - 1/(x+1)^2
f'(x) =0
(x+1)^2 -1 =0
x(x+2)=0
x=0 or -2
f''(x) = 2/(x+1)^3

f''(0) = 2 >0 (min)
f''(-2) = -2 < 0 (max)
min f(x) = f(0) = 0/(0+1) =0
max f(x) = f(-2) = 4/(-2+1 ) = -4
f(x) = x^2/(x+1)

lim(x->-1+) x^2/(x+1) ->+∞
lim(x->-1-) x^2/(x+1) ->-∞
单调区间
增加 = (-∞, -2] U [ 0, +∞ )
减小 =[ -2, -1) U (-1, 0]