1.已知函数y=f(x),x∈[a,b],那么集合{(x,y)|y=f(x),x∈[a,b]}∩{(x,y)|x=2}中元素的个数为.2.已知定义在R上的奇函数f(x)满足f(x+2)=-f(x),则f(6)=3.奇函数f(x)在(-无穷大,0)上单调递减,f(2)=0,不等式(x-1)f(x-1)>0
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/30 03:23:53
![1.已知函数y=f(x),x∈[a,b],那么集合{(x,y)|y=f(x),x∈[a,b]}∩{(x,y)|x=2}中元素的个数为.2.已知定义在R上的奇函数f(x)满足f(x+2)=-f(x),则f(6)=3.奇函数f(x)在(-无穷大,0)上单调递减,f(2)=0,不等式(x-1)f(x-1)>0](/uploads/image/z/3978479-47-9.jpg?t=1.%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0y%3Df%28x%29%2Cx%E2%88%88%5Ba%2Cb%5D%2C%E9%82%A3%E4%B9%88%E9%9B%86%E5%90%88%EF%BD%9B%28x%2Cy%29%7Cy%3Df%28x%29%2Cx%E2%88%88%5Ba%2Cb%5D%EF%BD%9D%E2%88%A9%7B%28x%2Cy%29%7Cx%3D2%7D%E4%B8%AD%E5%85%83%E7%B4%A0%E7%9A%84%E4%B8%AA%E6%95%B0%E4%B8%BA.2.%E5%B7%B2%E7%9F%A5%E5%AE%9A%E4%B9%89%E5%9C%A8R%E4%B8%8A%E7%9A%84%E5%A5%87%E5%87%BD%E6%95%B0f%28x%29%E6%BB%A1%E8%B6%B3f%28x%2B2%29%3D-f%28x%29%2C%E5%88%99f%286%29%3D3.%E5%A5%87%E5%87%BD%E6%95%B0f%28x%29%E5%9C%A8%28-%E6%97%A0%E7%A9%B7%E5%A4%A7%2C0%29%E4%B8%8A%E5%8D%95%E8%B0%83%E9%80%92%E5%87%8F%2Cf%282%29%3D0%2C%E4%B8%8D%E7%AD%89%E5%BC%8F%28x-1%29f%28x-1%29%EF%BC%9E0)
1.已知函数y=f(x),x∈[a,b],那么集合{(x,y)|y=f(x),x∈[a,b]}∩{(x,y)|x=2}中元素的个数为.2.已知定义在R上的奇函数f(x)满足f(x+2)=-f(x),则f(6)=3.奇函数f(x)在(-无穷大,0)上单调递减,f(2)=0,不等式(x-1)f(x-1)>0
1.已知函数y=f(x),x∈[a,b],那么集合{(x,y)|y=f(x),x∈[a,b]}∩{(x,y)|x=2}中元素的个数为.
2.已知定义在R上的奇函数f(x)满足f(x+2)=-f(x),则f(6)=
3.奇函数f(x)在(-无穷大,0)上单调递减,f(2)=0,不等式(x-1)f(x-1)>0的解集是.
4.已知函数y=f(x)是定义在R上的偶函数,当x<0时,f(x)是单调递增的,则不等式f(x+1)>f(1-2x)的解集是.
5.若f(x)为R上的奇函数,且在(0,+无穷大)内是增函数,又f(-3)=0则(x-1)f(x)<0的解集为.
6.设f(x)是(负无穷大,正无穷大)上的奇函数,f(x+2)=-f(x),0≤x≤1时,f(x)=x,则f(7.5)=.
7.定义在(0,正无穷大)上的函数y=f(x)满足f(xy)=f(x)+f(y),f(1/3)=1,且x>1时,f(x)<0
(1)证明y=f(x)为(0,正无穷大)上的单调减函数
(2)如果f(x)+f(2/3-x)≤2,求x的值
1.已知函数y=f(x),x∈[a,b],那么集合{(x,y)|y=f(x),x∈[a,b]}∩{(x,y)|x=2}中元素的个数为.2.已知定义在R上的奇函数f(x)满足f(x+2)=-f(x),则f(6)=3.奇函数f(x)在(-无穷大,0)上单调递减,f(2)=0,不等式(x-1)f(x-1)>0
1.如果2不在区间[a,b]之间,那么个数是0,如果2在区间[a,b]之间,那么个数是1.
2.由于是奇函数,就有f(0)=0,f(6)=-f(4)=f(2)=-f(0)=0.
3.f(x)是奇函数,关于原点对称,所以f(x)=-f(-x),所以f(-2)=-f(2)=0.f(x)在(0,+无穷大)上单调递增.在区间(-无穷大,-2)、(0,2)上f(x)>0,在(-2,0)、(2,+无穷大)上f(x)0时,f(x)单调递减.由于是偶函数,原不等式转换成f(x+1)>f(2x-1),
所以解集是x∈(-无穷,-1)∪(2,+无穷).
5.解集是x∈(-3,0)∪(1,3).
6.f(7.5)=-f(5.5)=f(3.5)=-f(1.5)=f(-0.5)=-f(0.5)=0.5.
7.(1)令x,y∈(0,+无穷),y1,f(x)=f(y*x/y)=f(y)+f(x/y),而x>1时,f(x)<0,所以f(x)-f(y)=f(x/y)<0.也就 是y=f(x)为(0,正无穷大)上的单调减函数.
(2)由f(xy)=f(x)+f(y),f(1/3)=1得到f(1/9)=2,f(x)+f(2/3-x)=f(x(2/3-x))≤2=f(1/9).
所以x(2/3-x)≥1/9.所以x=1/3.
不知道这样回答,你满不满意.
额