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题目求函数y=x+(x~2+x+1)~(1/2)的值域.解y-x=(x~2+x+1)~(1/2) =((x+1/2)~2+3/4)~(1/2)≥3~(1/2)/2.又(y-x)~2=x~2+x+1(?)x=(y~2-1)/(2y+1),
Find the range of the function y=x+(x~2+x+1)~(1/2). Solution yx=(x~2+x+1)~(1/2)=((x+1/ 2)~2+3/4)~(1/2)≥3~(1/2)/2.(yx)~2=x~2+x+1(?)x=(y~2- 1)/(2y+1),