解釋下什么是集中變元法解?
f(x)=2*m*x^2-2*(4-m)*x+1>0或g(x)=mx>0
其中對(duì)于g(x)=mx,x=0,g(0)=0;
1.當(dāng)m>0時(shí),g(x)=mx中x>0有g(shù)(x)>0,x≤0時(shí)有g(shù)(x)≤0,此時(shí)只要保證x≤0時(shí),f(x)>0
f(x)=2*m*x^2-2*(4-m)*x+1>0(x≤0)中a=2m>0,b=-2(4-m),c=1
對(duì)-b/(2a)=(4-m)/2m進(jìn)行討論
當(dāng)00,故可行.
當(dāng)m>4時(shí),二次函數(shù)的對(duì)稱軸在y軸右邊,
只要滿足最小值f(-b/(2a))=-(4-m)^2/(2m)+1>0即可,解(m-4)^2-2m