Question

若函数y=(mx^2+4√3x+n)/(x^2+1)的最大值为7，最小值为-1，求此函数式

导数没学过

Answer 1

y=(mx^2+4√3x+n)/(x^2+1)
(m-y)x^2+4√3x+n-y=0
上方程未知数为x的判别式△≥0
即(4√3)^2-4(m-y)*(n-y)≥0
y^2-(m+n)y+mn-12≤0
[m+n-√(m^2+n^2-2mn+48)]/2≤y≤[m+n+√(m^2+n^2-2mn+48)]/2
已知y=(mx^2+4√3x+n)/(x^2+1)(m<n)的值域为[-1，7],可知
[m+n-√(m^2+n^2-2mn+48)]/2=-1......(1)
[m+n+√(m^2+n^2-2mn+48)]/2=7......(2)
(1)+(2)
m+n=6......(3)
(2)-(1)
√(m^2+n^2-2mn+48)=8
m^2+n^2-2mn+48=64
(n-m)^2=16
m<n
-m+n=4......(4)
(3)+(4)
n=5
(3)-(4)
m=1
答:m=1,n=5