函數(shù)F{x}=1/3ax三方-1/4x二方+cx+d {a,c,d屬于R}滿足F{0}=0 F{1}的導(dǎo)數(shù)=0且F{x}的導(dǎo)數(shù)在R上恒大于等于0
1.求a c d 的值
2.若H{x}=3/4x二方-bx+b/2-1/4 ,解關(guān)于x的不等式 F{x}的導(dǎo)數(shù)+H{x}小于0
3.是否存在M,使函數(shù)G{X}=F{X}的導(dǎo)數(shù)-MX在區(qū)間[M,M+2]上有最小值-5 求M
F{0}=0,d=0；F‘{x}=ax方-1/ 2 x+ c,帶入x=1,a+c=1/2,
利用均值不等式ac≤1/16,F{x}的導(dǎo)數(shù)在R上恒大于等于0,方程ax方-1/ 2 x+ c=0判別式≤0,ac≥1/16
ac=1/16,a=c=1/4
2.F‘{x}=1/4x方-1/ 2 x+ 1/4
F{x}的導(dǎo)數(shù)+H{x}=x^2-(0.5+b)x+0.5b<0.(x-0.5)(x-b)<0
當(dāng)b=0.5,不等式無解
當(dāng)b>0.5,不等式0.5
當(dāng)b<0.5,不等式b
3.G(X)=1/4x方-1/ 2 x+ 1/4-Mx
G‘(X)=1//2[x-(1+2m)]
令G‘(X)=0,x=1+2m,若x∈[M,M+2],即-1
當(dāng)-1
當(dāng)m≤-1,函數(shù)G{X}在[M,M+2]單調(diào)遞增,x=m時(shí)函數(shù)G{X}取最小值
當(dāng)m≥1,函數(shù)G{X}在[M,M+2]單調(diào)遞減,x=m+2時(shí)函數(shù)G{X}取最小值