求函数的单调区间:

来源:百度知道 编辑:UC知道 时间:2024/05/12 10:33:49
1 求y=(3-2x-x^2)^1/2的增区间

2 求y=1/[(3+2x-x^2)^1/2]的减区间
谢谢ynlalbert,答案很棒~~~但是~~~我是新高一,我只知道导数是求拐点的~~~~请问有没有其它的办法?

1。Y=3-2X-X^2 x的对称轴为x=-1 开口向下
(-无穷,-1)为增区间
2。求y=2/-x^2+2x+3减区间 求-x^2+2x+3的减区间
x的对称轴为1 所以(1,正无穷)

解:1)3-2x-x^2>0, 由导数可知,当y'=0时,x=-1, 综上,函数增区间为(-3,-1]; 2)同理,3 2x-x^2>0, y'=0时解得x=1, 综上,减区间为[1,3);

都先求导数。第一个令导数大于0,第二个令导数小于0。再考虑到导数中的分母都大于0。通过解不等式组,得到两个函数的所求区间都是(-3,1)。
Solve their derivatives,respectively. Let the first derivative is greater than zero and the second smaller than zero. The denominators in the derivatives are greater than zero. By solving the inqualities systems, we can obtain that the intervals are all (-3,0).