Question

Y=X^2+1/X^4+1解值域

Answer 1

Y=（X^2+1）/（X^4+1）
令t=x²+1≥1
则x²=t-1
y = t / {(t-1)²+1}
= t / {t²-2t+2}
= 1 / {t-2+2/t}
= 1 / {t+2/t-2}
= 1 / {(√t-√2/√t)²+2√2-2}
当t=√2时，分母有最小值2√2-2
即：(√t-√2/√t)²+2√2-2 ≥· 2√2-1
0＜ 1 / {(√t-√2/√t)²+2√2-2} ≤ 1/(2√2-2)
即：0＜ 1 / {(√t-√2/√t)²+2√2-2} ≤ (√2+1)/2
即：Y=（X^2+1）/（X^4+1）的值域为：（ 0， (√2+1)/2 】

Answer 2

原题是:Y=(X²+1)/(X^4+1),求值域。
值域C={Y|Y=(X²+1)/(X^4+1)}
={Y|Y=u/(u-1)²+1),u≥1} (设u=X²+1)
={Y|Y=1/(u+(2/u)-2),u≥1}
={Y|Y=1/t,t≥-2+2√2} (设t=u+(2/u)-2)
={Y|0<Y≤1/(-2+2√2)}
={Y|0<Y≤(2+√2)/2}
所以Y=(X²+1)/(X^4+1)的值域是(0,(2+√2)/2]

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