一道高一数学题目,请教一下
1个回答
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1.
f(x^2-5)=loga [x^2/(10-x^2)]
取t=x^2-5
f(t)=loga [(t+5)/(5-t)]
所以f(x)=loga [(x+5)/(5-x)]
(x+5)/(5-x)>0且x-5≠0时,f(x)有意义
10/(5-x)<1
所以x>(-5),
x≠5
2.
f(0)=0
f(-x)=loga[(5-x)/(5+x)]
=-loga [(5+x)/(5-x)]
=-f(x)
奇函数
3.
f(x)=log a [(5+x)/(5-x)]
f(x)≥0
loga[(5+x)/(5-x)]≥0
a>1时
a^[f(x)]≥1
(5+x)/(5-x)≥1
-1+10/(5-x)≥1
5/(5-x)≥1
(5-x)<5
x≥0且x ≠5
f(x^2-5)=loga [x^2/(10-x^2)]
取t=x^2-5
f(t)=loga [(t+5)/(5-t)]
所以f(x)=loga [(x+5)/(5-x)]
(x+5)/(5-x)>0且x-5≠0时,f(x)有意义
10/(5-x)<1
所以x>(-5),
x≠5
2.
f(0)=0
f(-x)=loga[(5-x)/(5+x)]
=-loga [(5+x)/(5-x)]
=-f(x)
奇函数
3.
f(x)=log a [(5+x)/(5-x)]
f(x)≥0
loga[(5+x)/(5-x)]≥0
a>1时
a^[f(x)]≥1
(5+x)/(5-x)≥1
-1+10/(5-x)≥1
5/(5-x)≥1
(5-x)<5
x≥0且x ≠5
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