作业帮已知函数f(x)=ax²+bx+4在[-1,+∞)上是单调递增函数,在(-∞,-1]上是单调递减函数,且当x1＜x2,x1+x2=1-a时,有f(x1)＜f(x2),则实数a的取值范围是?重点解释且当x1＜x2,x1+x2=1-a时,有f(x1)＜f(x2)
来源：学生作业帮助网 编辑：作业帮 时间：2024/04/30 08:25:23
![已知函数f(x)=ax²+bx+4在[-1,+∞)上是单调递增函数,在(-∞,-1]上是单调递减函数,且当x1＜x2,x1+x2=1-a时,有f(x1)＜f(x2),则实数a的取值范围是?重点解释且当x1＜x2,x1+x2=1-a时,有f(x1)＜f(x2)](/uploads/image/z/10216968-24-8.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dax%26%23178%3B%2Bbx%2B4%E5%9C%A8%5B-1%2C%2B%E2%88%9E%29%E4%B8%8A%E6%98%AF%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%E5%87%BD%E6%95%B0%2C%E5%9C%A8%28-%E2%88%9E%2C-1%5D%E4%B8%8A%E6%98%AF%E5%8D%95%E8%B0%83%E9%80%92%E5%87%8F%E5%87%BD%E6%95%B0%2C%E4%B8%94%E5%BD%93x1%EF%BC%9Cx2%2Cx1%2Bx2%3D1-a%E6%97%B6%2C%E6%9C%89f%28x1%29%EF%BC%9Cf%28x2%29%2C%E5%88%99%E5%AE%9E%E6%95%B0a%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4%E6%98%AF%3F%E9%87%8D%E7%82%B9%E8%A7%A3%E9%87%8A%E4%B8%94%E5%BD%93x1%EF%BC%9Cx2%2Cx1%2Bx2%3D1-a%E6%97%B6%2C%E6%9C%89f%28x1%29%EF%BC%9Cf%28x2%29)
已知函数f(x)=ax²+bx+4在[-1,+∞)上是单调递增函数,在(-∞,-1]上是单调递减函数,且当x1＜x2,x1+x2=1-a时,有f(x1)＜f(x2),则实数a的取值范围是?重点解释且当x1＜x2,x1+x2=1-a时,有f(x1)＜f(x2)
已知函数f(x)=ax²+bx+4在[-1,+∞)上是单调递增函数,在(-∞,-1]上是单调递减函数,且当x1＜x2,x1+x2=1-a时,有f(x1)＜f(x2),则实数a的取值范围是?
重点解释且当x1＜x2,x1+x2=1-a时,有f(x1)＜f(x2)
已知函数f(x)=ax²+bx+4在[-1,+∞)上是单调递增函数,在(-∞,-1]上是单调递减函数,且当x1＜x2,x1+x2=1-a时,有f(x1)＜f(x2),则实数a的取值范围是?重点解释且当x1＜x2,x1+x2=1-a时,有f(x1)＜f(x2)