高中数学知奇函数求范围？高手进来看看

因为是奇函数，所以-f(x)=f(-x)
f(a^2-a-1)+f(4a-5)>0等价于： f(a^2-a-1)>f(5-4a)
由题意： -1〈=a^2-a-1〈=1
-1<=5-4a<=1
a^2-a-1<5-4a
解得a的范围：[-1，0]U[1,(-3+√33)/2,1]
很高兴为你解决问题！

定义域
-1<=a^2-a-1<=1
-1<=a^2-a-1,a^2-a>=0,a<=0,a>=1
a^2-a-1<=1,a^2-a-2<=0,-1<=a<=2
所以-1<=a<=0,1<=a<=2

-1<=4a-5<=1
4<=4a<=6
1<=a<=3/2

所以1<=a<=3/2

f(a^2-a-1)+f(4a-5)>0
f(a^2-a-1)>-f(4a-5)
奇函数
-f(4a-5)=f[-(4a-5)]=f(5-4a)
f(a^2-a-1)>f(5-4a)
减函数
a^2-a-1<5-4a
a^2+3a-6<0
(-3-√33)/2<a<(-3+√33)/2

综上
1<=a<(-3+√33)/2

f(a^2-a-1)+f(4a-5)>0
f(a^2-a-1)-f(-4a+5)>0
f(a^2-a-1)>f(-4a+5)
a^2-a-1<-4a+5
a^2+3a-6<0
a∈(-1.5-0.5√33,-1.5+0.5√33)
∵a^2-a-1∈[-1,1]
4a-5∈[-1,1]
∴a∈[1,1.5]
综上
a∈[1,-1.5+0.