Question

判断并证明函数f(x)=x-1/x+2在(2,正无穷大)上的单调性,并画出其大致图像

Answer 1

判断并证明函数f(x)=x-1/x+2在(2,正无穷大)上的单调性,并画出其大致图像

x在(2,正无穷大)
设2<x1<x2<∞
x1-x2<0

f(x1)=(x1-1)/(x1+2)
f(x2)=(x2-1)/(x2+2)
f(x1)-f(x2)=((x1-1)/(x1+2)-(x2-1)/(x2+2)
=[((x1-1)(x2+2)-(x2-1)/(x1+2)]/(x1+2)(x2+2)
=3(x1-x2)/(x1+2)(x2+2)
<0
f(x1)<f(x2)
f(x)是增函数

图像自己画

Answer 2

f(x)在(2,+∞)上为增函数
设2<x1<x2

f(x1)-f(x2)
=x1-1/x1+2-(x2-1/x2+2)
=x1-x2+1/x2-1/x1
=(x1-x2)+(x1-x2)/(x1x2)
=(x1-x2)[1+1/(x1x2)]
∵2<x1<x2
∴x1-x2<0.1+1/(x1x2)>0
∴(x1-x2)[1+1/(x1x2)]<0
即f(x1)-f(x2)<0,f(x1)<f(x2)
∴f(x)在(2,+∞)上为增函数

Answer 3

单调增，图像就是把y=3/x向左平移两个单位，向上平移一个单位，然后取x大于2的点