设函数f(x)对任意x,y都有f(x+y)=f(x)+f(y) 且x大于零时f(X)小于零,f(1)=-2 ①求证f(x)是奇函数.②判断f(x)的单调性 ③在{-3,3}之间是否有最值
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![设函数f(x)对任意x,y都有f(x+y)=f(x)+f(y) 且x大于零时f(X)小于零,f(1)=-2 ①求证f(x)是奇函数.②判断f(x)的单调性 ③在{-3,3}之间是否有最值](/uploads/image/z/62799-15-9.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%E5%AF%B9%E4%BB%BB%E6%84%8Fx%2Cy%E9%83%BD%E6%9C%89f%28x%2By%29%3Df%28x%29%2Bf%28y%29+%E4%B8%94x%E5%A4%A7%E4%BA%8E%E9%9B%B6%E6%97%B6f%28X%29%E5%B0%8F%E4%BA%8E%E9%9B%B6%2Cf%281%29%3D-2+%E2%91%A0%E6%B1%82%E8%AF%81f%28x%29%E6%98%AF%E5%A5%87%E5%87%BD%E6%95%B0.%E2%91%A1%E5%88%A4%E6%96%ADf%28x%EF%BC%89%E7%9A%84%E5%8D%95%E8%B0%83%E6%80%A7+%E2%91%A2%E5%9C%A8%7B-3%2C3%7D%E4%B9%8B%E9%97%B4%E6%98%AF%E5%90%A6%E6%9C%89%E6%9C%80%E5%80%BC)
设函数f(x)对任意x,y都有f(x+y)=f(x)+f(y) 且x大于零时f(X)小于零,f(1)=-2 ①求证f(x)是奇函数.②判断f(x)的单调性 ③在{-3,3}之间是否有最值
设函数f(x)对任意x,y都有f(x+y)=f(x)+f(y) 且x大于零时f(X)小于零,f(1)=-2 ①求证f(x)是奇函数.②判断f(x)的单调性 ③在{-3,3}之间是否有最值
设函数f(x)对任意x,y都有f(x+y)=f(x)+f(y) 且x大于零时f(X)小于零,f(1)=-2 ①求证f(x)是奇函数.②判断f(x)的单调性 ③在{-3,3}之间是否有最值
(1)、设x=y=0,则f(0)=f(0)+f(0),得f(0)=0;再设y=-x,代入得:f(0)=f(x)+f(-x),所以f(-x)=-f(x),即函数为奇函数.
(2)、设x2>x1,则有f(x2)-f(x1)=f(x2)+f(-x1)=f(x2-x1),又x2-x1>0,则f(x2-x1)<0,即f(x2)-f(x1)<0,函数为单调递减函数.
(3)、因为函数为奇函数且单调递减,则函数有最值,f(3)=f(2)+f(1)=3f(1)=-6,由单调性得f(-3)=6,为最大值,f(3)=-6为最小值.
1.
f(0) = f(0+0) = f(0) + f(0) = 2f(0) → f(0) = 0
f(-x) = f(x-x-x) = f(x) + f(-x) + f(-x) → f(x) + f(-x) = 0
故f(x)是奇函数
2.
设0≤x≤y≤3
f(y) = f(x+t) = f(x) + f(t) ≤ f(x),再结合奇...
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1.
f(0) = f(0+0) = f(0) + f(0) = 2f(0) → f(0) = 0
f(-x) = f(x-x-x) = f(x) + f(-x) + f(-x) → f(x) + f(-x) = 0
故f(x)是奇函数
2.
设0≤x≤y≤3
f(y) = f(x+t) = f(x) + f(t) ≤ f(x),再结合奇偶性可知f(x)在-3≤x≤3上是单调递减函数,所以最大值为f(-3),最小值为f(3)。
f(3) = f(1+1+1) = 3f(1) = -6
f(-3) = -f(3) = 6
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