函数y=2x+4(根号下1-x)的值域为?
答案是将根号下1-x设为t,最后得出-2(t-1)^2+4,又t≥0.所以为什么得出了t=1,y=4,∴y≤4,最后的结果都是怎么求出来的?
人气:456 ℃ 时间:2019-09-30 19:30:57
解答
设√(1-x)=t,则x=1-t²
==>>y=2(1-t²)+4t
=-2t²+4t+2
=-2(t-1)²+4
∵-2(t-1)²≤0
∴y=-2(t-1)²+4≤4,且当t=1时,y=4
即y的值域为(-∝,4]
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