Question

已知f(x)=px2+2/3x+q是奇函数,且f(2)=5/3.

Answer 1

f(x)=(px^2+2)/(3x+q)

f(-x)=(px^2+2)/(-3x+q)=-f(x)=(px^2+2)/(-3x-q)

所以有q=0

f(x)=(px^2+2)/(3x)

代入f(2)=(4p+2)/6=5/3,得：p=2

f(x)=2(x^2+1)/(3x)

令t=2x-3,当x∈[-0.5,1.25]时,t∈[-4,-0.5]

g(x)=f(t)=2(t^2+1)/(3t)=2/3*(t+1/t)

由于 t