求函数y=(x4+2x2+1)/(x2+2)的最小值
求函数y=(x4+2x2+1)/(x2+2)的最小值
解: 原式变形后,可用均值不等式: y=(x^4+2x^2+1)/(x^2+2) =[x^2(x^2+2)+1]/(x^2+2) =x^2+[1/(x^2+2)] =(x^2+2)+[1/(x^2+2)]-2 =(x^2+2)/4+(x^2+2)/4+(x^2+2)/4+(x^2+2)/4+[1/(x^2+2)]-2 =3/4*(x^2+2)+(x^2+2)/4+[1/(x^2+2)]-2 >=3/4*2+2根[(x^2+2)/4*1/(x^2+2)]-2 =1/2 故上式取等号,得y|min=1/2, 此时有且只有(x^2+2)/4=1/(x^2+2),即x=0.
记u=x^2+2,u≥2,所以 y=(x^4+2x^2+1)/(x^2+2)=[(x^2+1)^2]/(x^2+2) =(u-1)^2/u=u-2+1/u , 根据对勾函数性质,在u≥2时,y=u-2+1/u 单调增加, u=2,即x=0时,y有最小值1/2.
答:解: y=(x-x^3)/(1+2x^2+x^4) --->y=x(1-x^2)/(1+x^2)^2 设x=tan(t/2),则 y=1/2*sintcost ...详情>>
答:详情>>