Question

偶函数f(x)在(0,正无穷)尚未增函数,f(2)=0，解不等式 f[log2 (x^2+5x+4]≥0

f(x)在(0,正无穷)上为增函数,

Answer 1

[log2 (x^2+5x+4）]≥2
(x^2+5x+4）≥4
x^2+5x≥0
（负无穷，-5】～【0，正哗局侍无腊饥穷乱吵）

Answer 2

f(x)在(0,正无穷)上为增函数,则有在(-无穷,0)上是减函数.
f[log2 (x^2+5x+4]≥0=f(2)

log2(x^2+5x+4)>=2或log2(x^2+5x+4)<=-2
可得培败(x^2+5x+4)>=4或0<(x^2+5x+4)<=0.25
x^2+5x>=0
x(x+5)>=0
x>=0,x<=-5

0<粗樱x^2+5x+4<=0.25
解出以上就行了.最后岩中丛取并集.