已知函数y=f(x)对任意实数x,y都有f(x+y)=f(x)+f(y)-1且当x>0时f(x)>1,f(3)=4(1)...已知函数y=f(x)对任意实数x,y都有f(x+y)=f(x)+f(y)-1且当x>0时f(x)>1,f(3)=4(1)证明f(x)为曾函数(2)求函数在区间[1,2]上
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![已知函数y=f(x)对任意实数x,y都有f(x+y)=f(x)+f(y)-1且当x>0时f(x)>1,f(3)=4(1)...已知函数y=f(x)对任意实数x,y都有f(x+y)=f(x)+f(y)-1且当x>0时f(x)>1,f(3)=4(1)证明f(x)为曾函数(2)求函数在区间[1,2]上](/uploads/image/z/13623456-48-6.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0y%3Df%28x%29%E5%AF%B9%E4%BB%BB%E6%84%8F%E5%AE%9E%E6%95%B0x%2Cy%E9%83%BD%E6%9C%89f%28x%2By%29%3Df%28x%29%2Bf%28y%29-1%E4%B8%94%E5%BD%93x%EF%BC%9E0%E6%97%B6f%EF%BC%88x%EF%BC%89%3E1%2Cf%283%29%3D4%281%29...%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0y%3Df%28x%29%E5%AF%B9%E4%BB%BB%E6%84%8F%E5%AE%9E%E6%95%B0x%2Cy%E9%83%BD%E6%9C%89f%28x%2By%29%3Df%28x%29%2Bf%28y%29-1%E4%B8%94%E5%BD%93x%EF%BC%9E0%E6%97%B6f%EF%BC%88x%EF%BC%89%3E1%2Cf%283%29%3D4%281%29%E8%AF%81%E6%98%8Ef%EF%BC%88x%EF%BC%89%E4%B8%BA%E6%9B%BE%E5%87%BD%E6%95%B0%EF%BC%882%EF%BC%89%E6%B1%82%E5%87%BD%E6%95%B0%E5%9C%A8%E5%8C%BA%E9%97%B4%5B1%2C2%5D%E4%B8%8A)
已知函数y=f(x)对任意实数x,y都有f(x+y)=f(x)+f(y)-1且当x>0时f(x)>1,f(3)=4(1)...已知函数y=f(x)对任意实数x,y都有f(x+y)=f(x)+f(y)-1且当x>0时f(x)>1,f(3)=4(1)证明f(x)为曾函数(2)求函数在区间[1,2]上
已知函数y=f(x)对任意实数x,y都有f(x+y)=f(x)+f(y)-1且当x>0时f(x)>1,f(3)=4(1)...
已知函数y=f(x)对任意实数x,y都有f(x+y)=f(x)+f(y)-1且当x>0时f(x)>1,f(3)=4(1)证明f(x)为曾函数(2)求函数在区间[1,2]上的最大值与最小值
已知函数y=f(x)对任意实数x,y都有f(x+y)=f(x)+f(y)-1且当x>0时f(x)>1,f(3)=4(1)...已知函数y=f(x)对任意实数x,y都有f(x+y)=f(x)+f(y)-1且当x>0时f(x)>1,f(3)=4(1)证明f(x)为曾函数(2)求函数在区间[1,2]上
对于任意x2 > x1, 设 x2 = x1 + a , a > 0
f(x2)=f(x1+a)=f(x1)+f(a) -1>f(x1)+1-1=f(x1)
so, f(x2) > f(x1), y = f(x) 是增
f (3) = f (2) + f (1) - 1 = 3f(1) - 2 = 4
f(1) = 2 最小
f(2)= 2f(1) - 1 = 3 最大
1,x1
f(x1)-f(x2)=f(x1)-f[(x2-x1)+x1]=1-f(x2-x1)<0
f(x)增函数。
2,易知;f(0)=1
f(2)=2f(1)-1,f(3)=f(2)+f(1)-1=3f(1)-2=4
f(1)=2,f(2)=3
f(x)增函数,
最大f(2)=3
最小f(1)=2
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1