求方程均值f(x)=2x^2-8x+9,0

Question

求方程均值
f(x)=2x^2-8x+9,0

Answer 1

设f(X)在区间[0,3]的平均值为：E
E = (1/3)∫(3,0)(2X^2-8X+9)dX
= (1/3){2X^3/3-4X^2+9X}|(3,0)
= (1/3){18-36+27}
= 3 即：f(x)在[0,3]上的平均值为：3
2）f(c)=3,即：2c^2-8c+9=3 2c^2-8c+6=0 c=[8±√(64-48)]/4
c = (8±4)/4=2±1
c1=3 c2=1
表明在[0,3]上有两个点c1和c2处的函数值f（c1）和f(c2)都等于平均值3.