判断并证明y=x/x^+1在(0,+∞)上的单调性
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判断并证明y=x/x^+1在(0,+∞)上的单调性
判断并证明y=x/x^+1在(0,+∞)上的单调性
判断并证明y=x/x^+1在(0,+∞)上的单调性
y = x/x^2+1 = 1/(x+1/x)
分母x+1/x为对勾函数:
x<-1时x+1/x单调增,y = 1/(x+1/x)单调减;
-1<x<0时x+1/x单调减,y = 1/(x+1/x)单调增;
0<x<1时x+1/x单调减,y = 1/(x+1/x)单调增;
x>1时x+1/x单调增,y = 1/(x+1/x)单调减.
x∈R,∴y = 1/(x+1/x)单调减区间(-∞,-1)U(1,+∞),单调增区间(-1,1)
证明:
(一)x∈(-∞,-1),令x1<x2<-1
f(x2)-f(x1) = x2/(x2^2+1)-x1/(x1^2+1)
= (x1^2x2+x2-x1x2^2-x1)/{(x2^2+1)(x1^2+1)}
= (x2-x1)(1-x1x2)/{(x2^2+1)(x1^2+1)}
∵x1<x2<-1
∴x2-x1>0,1-x1x2<0
∴f(x2)-f(x1)=(x2-x1)(1-x1x2)/{(x2^2+1)(x1^2+1)}<0,
∴x∈(-∞,-1)时单调减.
(二)x∈(-1,1),令-1<x1<x2<1
f(x2)-f(x1) = x2/(x2^2+1)-x1/(x1^2+1)
= (x1^2x2+x2-x1x2^2-x1)/{(x2^2+1)(x1^2+1)}
= (x2-x1)(1-x1x2)/{(x2^2+1)(x1^2+1)}
∵-1<x1<x2<1
∴x2-x1>0,1-x1x2>0
∴f(x2)-f(x1)=(x2-x1)(1-x1x2)/{(x2^2+1)(x1^2+1)}>0,
∴x∈(-1,1)时单调增.
(三)x∈(1,∞1),令1<x1<x2
f(x2)-f(x1) = x2/(x2^2+1)-x1/(x1^2+1)
= (x1^2x2+x2-x1x2^2-x1)/{(x2^2+1)(x1^2+1)}
= (x2-x1)(1-x1x2)/{(x2^2+1)(x1^2+1)}
∵1<x1<x2
∴x2-x1>0,1-x1x2<0
∴f(x2)-f(x1)=(x2-x1)(1-x1x2)/{(x2^2+1)(x1^2+1)}<0,
∴x∈(1,+∞)时单调减.