作业帮高手进来看看已知奇函数f(x)是定义在[-1,1]上的减函数,若f(a^2-a-1)+f(4a-5)>0,求实数a的取值范围?
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![高手进来看看已知奇函数f(x)是定义在[-1,1]上的减函数,若f(a^2-a-1)+f(4a-5)>0,求实数a的取值范围?](/uploads/image/z/9603149-5-9.jpg?t=%E9%AB%98%E6%89%8B%E8%BF%9B%E6%9D%A5%E7%9C%8B%E7%9C%8B%E5%B7%B2%E7%9F%A5%E5%A5%87%E5%87%BD%E6%95%B0f%28x%29%E6%98%AF%E5%AE%9A%E4%B9%89%E5%9C%A8%5B-1%2C1%5D%E4%B8%8A%E7%9A%84%E5%87%8F%E5%87%BD%E6%95%B0%2C%E8%8B%A5f%28a%5E2-a-1%29%2Bf%284a-5%29%3E0%2C%E6%B1%82%E5%AE%9E%E6%95%B0a%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4%3F)
高手进来看看已知奇函数f(x)是定义在[-1,1]上的减函数,若f(a^2-a-1)+f(4a-5)>0,求实数a的取值范围?
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已知奇函数f(x)是定义在[-1,1]上的减函数,若f(a^2-a-1)+f(4a-5)>0,求实数a的取值范围?
高手进来看看已知奇函数f(x)是定义在[-1,1]上的减函数,若f(a^2-a-1)+f(4a-5)>0,求实数a的取值范围?
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
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+0.5√33)
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...
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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
1<=a<(-3+√33)/2
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