求函数y=x^2/(x+1)的单调性
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求函数y=x^2/(x+1)的单调性
求单调性,图像,极值
求单调性,图像,极值
![求函数y=x^2/(x+1)的单调性](/uploads/image/z/20084224-40-4.jpg?t=%E6%B1%82%E5%87%BD%E6%95%B0y%3Dx%5E2%2F%28x%2B1%29%E7%9A%84%E5%8D%95%E8%B0%83%E6%80%A7)
答:
y=x^2/(x+1)
=(x+1-1)^2/(x+1)
=[(x+1)^2-2(x+1)+1]/(x+1)
=(x+1)+1/(x+1)-2
1)x+1>0即x>-1时:
y>=2√[(x+1)*1/(x+1)]-2
=2-2
=0
2)x+1<0即x<-1时:
y<=-2√[(x+1)*1/(x+1)]-2
=-2-2
=-4
所以:y>=0或者y<=-4
单调递增区间(-∞,-2)或者(0,+∞)
单调递减区间(-2,-1)或者(-1,0)
极小值为0,极大值为-4
![](http://img.wesiedu.com/upload/2/39/2391e2244d1e905e6c1b8b86aec1817d.jpg)
y=x^2/(x+1)
=(x+1-1)^2/(x+1)
=[(x+1)^2-2(x+1)+1]/(x+1)
=(x+1)+1/(x+1)-2
1)x+1>0即x>-1时:
y>=2√[(x+1)*1/(x+1)]-2
=2-2
=0
2)x+1<0即x<-1时:
y<=-2√[(x+1)*1/(x+1)]-2
=-2-2
=-4
所以:y>=0或者y<=-4
单调递增区间(-∞,-2)或者(0,+∞)
单调递减区间(-2,-1)或者(-1,0)
极小值为0,极大值为-4
![](http://img.wesiedu.com/upload/2/39/2391e2244d1e905e6c1b8b86aec1817d.jpg)