Question

求函数y=-[log以2为底的(x^2+2)]^2-log以1/4为底的(x^2+2)+5的值域

Answer 1

y=-[log₂(x^2+2)]²-log(1/4)(x^2+2)+5
=-[log₂(x²+2)]²+1/2log₂(x^2+2)+5
设t=log₂(x²+2)
∵x²≥0,x²+2≥2
∴t=log₂(x²+2)≥log₂2=1
∴y=-t²+1/2t+5
对称轴为t=1/4,函数在[1,+∞)内为减函数
∴t=1时，函数取得最大值9/2
∴函数值域为(-∞,9/2]

Answer 2

y=-[log2 (x^2+2)]^2-log1/4 (x^2+2)+5
=-[log2 (x^2+2)]^2+log4 (x^2+2)+5
=-[log2 (x^2+2)]^2-1/2*log2 (x^2+2)+5
=-{[log2 (x^2+2)]^2-log2 (x^2+2) /2 +1/16}+5+1/16
=-[log2 (x^2+2)-1/4]^2+81/16
∵x^2>=0
x^2+2>=2

log2 (x^2+2)>=log2 2=1
∴log2 (x^2+2)不可能=1/4
log2 (x^2+2)-1/4>=3/4
[log2 (x^2+2)+1/4]^2>=9/16
-[log2 (x^2+2)+1/4]^2<=-9/16
-[log2 (x^2+2)+1/4]^2+81/16<=81/16-9/16=72/16=4.5
所以值域是(-无穷大，4.5]