若函数f(x)=(ax+b)/(x²+1)的最大值是4,最小值是-1,求实数a,b的值
人气:328 ℃ 时间:2019-10-19 19:43:23
解答
用根的判别式法:
ax+b=yx²+y
yx²-ax+(y-b)=0
因为关于x的方程有解,所以判别式
Δx=a²-4y(y-b)≥0
4y²-4by-a²≤0
因为
y(MAX)=4,
y(min)=-1
所以,y=4,y= - 1是方程:
4y²-4by-a²=0的根;
由韦达定理:
4+(-1)=b=3
4*(-1)=-a²
a=±2
b=3
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