1) f'(x)=2x+2+a/x=(2x^2+2x+a)/x =(2x^2+2x-4)/x=0 (x>0)
x^2+x-2=0
(x-1)(x+2)=0
所以
有唯一駐點(diǎn)x=1,左邊導(dǎo)數(shù)小于0,右邊導(dǎo)數(shù)大于0,即取極小值f(1)=3.
2) f(2t-1)>=2f(t)-3
2t^2-4t+2+aln(2t-1)=2lnt>=0
2(t-1)^2>+alm(2t-1)-2lnt>=0
設(shè)x=t-1,x>=0,上面不等式等價(jià)于
2x^2+aln(2x+1)-2aln(x+1)>=0
ln(2x+1)=0
ln[(2x+1)/(x^2+2x+1)]=ln[1-x^2/(x^2+2x+1)]>=-x^2/(x^2+2x+1)].
所以如果2x^2-ax^2/(x^2+2x+1)]>=0,即2(x+1)^2-a>=0,那么原不等式自然成立.2(x+1)^2-a>=0恒成立對(duì)x>=0,那么a2,因?yàn)楫?dāng)x--->0+時(shí),極限x^2/ln[1-x^2/(x^2+2x+1)]=-1,因此對(duì)充分小的正數(shù)x,2x^2+aln[1-x^2/(x^2+2x+1)]=ax^2*[2/a+ln[1-x^2/(x^2+2x+1)]/x^2]