已知函数f(x)对一切x,y属于R,都有f(x+y)=f(x)+f(y).(1)求f(-x)+f(x)的值.(2)若f(-3)=a,用a表示f(12)
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![已知函数f(x)对一切x,y属于R,都有f(x+y)=f(x)+f(y).(1)求f(-x)+f(x)的值.(2)若f(-3)=a,用a表示f(12)](/uploads/image/z/5627938-58-8.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%E5%AF%B9%E4%B8%80%E5%88%87x%2Cy%E5%B1%9E%E4%BA%8ER%2C%E9%83%BD%E6%9C%89f%28x%2By%29%3Df%28x%29%2Bf%28y%29.%281%29%E6%B1%82f%28-x%29%2Bf%28x%29%E7%9A%84%E5%80%BC.%EF%BC%882%EF%BC%89%E8%8B%A5f%28-3%29%3Da%2C%E7%94%A8a%E8%A1%A8%E7%A4%BAf%2812%EF%BC%89)
已知函数f(x)对一切x,y属于R,都有f(x+y)=f(x)+f(y).(1)求f(-x)+f(x)的值.(2)若f(-3)=a,用a表示f(12)
已知函数f(x)对一切x,y属于R,都有f(x+y)=f(x)+f(y).(1)求f(-x)+f(x)的值.(2)若f(-3)=a,用a表示f(12)
已知函数f(x)对一切x,y属于R,都有f(x+y)=f(x)+f(y).(1)求f(-x)+f(x)的值.(2)若f(-3)=a,用a表示f(12)
函数f(x)对一切x,y属于R,都有f(x+y)=f(x)+f(y),
令x=y=0,则:f(0+0)=f(0)+f(0),得:f(0)=0
令y=-x,则: f(x-x)=f(x)+f(-x)
f(x)+f(-x)=f(0)=0
由上式得:f(-x)=-f(x)
f(12)=f(6+6)=f(6)+f(6)=2f(6)=4f(3)=-4f(-3)=-4a
令 x = y = 0
所以 f(0+0) = f(0) + f(0)
即 f(0) = 0
令 y = -x
所以 f(x - x) = f(x) + f(-x)
即 0 = f(x) + f(-x)
由上小题可知
f(-x) = -f(x)
所以 f(-3) = -f(3...
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令 x = y = 0
所以 f(0+0) = f(0) + f(0)
即 f(0) = 0
令 y = -x
所以 f(x - x) = f(x) + f(-x)
即 0 = f(x) + f(-x)
由上小题可知
f(-x) = -f(x)
所以 f(-3) = -f(3) = a
即 f(3) = -a
又因为 f(12) = f(6) + f(6)
= 2[f(3) + f(3)]
= 4f(3)
= -4a
收起
(1) y=0时,f(x+0)=f(x)+f(0) 则f(0)=0
y=-x时 f(x-x)=f(x)+f(-x)
所以f(x)+f(-x)=f(0)=0
(2) f(-3)=a
由(1) f(3)+f(-3)=0
所以f(3)=-a
f(12)=f(6+6)=f(6)+f(6)=2f(6)
=2f(3+3)=2[f(3)+f(3)]
=4f(3)
=-4a
由于f(x+y)=f(x)+f(y),所以f(-x)+f(x)=f(-x+x)=f(0)。而f(0)=f(0+0)=2f
(0),所以f(0)=0.故有f(-x)+f(x)=0
由上有f(3)=-a,所以有f(x+y+m+n)=f(x)+f(y)+f(m)+f(n)所以f(12)=4f(3)=-4a