Question

2已知,函数 y=2x+3+1/(x+1) ,函数y的取值范围？

Answer 1

y=2x+3+1/(x+1)
lim(x->-1+) [2x+3+1/(x+1)] ->+∞
lim(x->-1-) [2x+3+1/(x+1)] ->-∞
lim(x->+∞) [2x+3+1/(x+1)] ->+∞
lim(x->-∞) [2x+3+1/(x+1)] ->-∞
y'=2-1/(x+1)^2
y'=0
2-1/(x+1)^2 =0
2(x+1)^2 -1 =0
x+1 = √2/2 or -√2/2
x= (-2+√2)/2 or (-2-√2)/2
y''=2/(x+1)^3
y''[(-2+√2)/2] >0 (min)
y''[(-2-√2)/2] <0 (max)
y=2x+3+1/(x+1)

max y
= y((-2-√2)/2)
=(-2-√2) +3 +1/((-2-√2)/2+1)
=1
min y
=y((-2+√2)/2)
=(-2+√2)+3+1/((-2+√2)/2+1)
=1+2√2
y的取值范围 =(-∞, 1] U [ 1+2√2, +∞)