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1、1 抽象函数问题1、()()()f xyfxf y，类比函数()f xkx类比正比例函数()f xkx的性质（1）、过原点()f xkx的图像过原点对应性质：已知函数()fx对任意,x yR恒有()()()f xyf xf y.求证(0)0f证明:令0 xy得(0)(0)(0)fff,所以(0)0f(2)、奇偶性()f xkx是奇函数对应性质：已知函数()fx对任意,x yR恒有()()()f xyf xf y.求证()fx是奇函数。证明：在()()()f xyf xf y中，令yx得(0)()()0ff xfx，所以()()fxf x，即()f x是奇函数（3）()()()f xyf xf
2、y()()()f xyf xfy证明略。（4）、单调性()f xkx，当0k时，()f x单调递增，当0k时，()f x单调递减对应性质已知函数()f x对任意,x yR恒有()()()f xyf xfy，且当0 x时()0f x求证()f x在,单调递增证明一任取12,x x并且120 xx则210 xx21()f xx0又由（2）知()fx是奇函数所以21()f xx=21()()f xfx=21()()f xf x0.所以()f x在,单调递增。证明二直接用（3）2、()()()f xyfx f y，类比函数()f x=xa类比函数()f x=xa的性质（1）()f x=xa过(0,1)
3、对应性质：已知函数()f x对任意,x yR恒有()()()fxyf x fy，并且()f x不恒为 0，求(0)f解：因 为()f x不 恒 为0，不 妨 设()f a0，在()()()fxyfxfy中 令,0 xa y得名师资料总结-精品资料欢迎下载-名师精心整理-第 1 页，共 4 页 -2()()(0)f af a f所以(0)1f.(2)()f x=xa满足()()1f x fx对应性质已知函数()fx对任意,x yR恒有()()()fxyf x f y.并且()f x不恒为 0，求证()()1f x fx证明 在()()()f xyf x fy中，令yx得，()()(0)1f x
4、fxf（3）()()()f xyf x f y()()()f xf xyfy证明：在()()()f xyfx fy中，以y代y得：()()()f xyf x fy由(2)得1()()fyfy所以()()()f xf xyfy，反之亦然。（4）单调性()fx=xa当1a时是增函数对应性质已知函数()f x对任意,x yR恒有()()()f xyf x fy；对任意xR，()0f x并且当0 x时，()1fx，求证()f x在,单调递增。证明一：任取12,x x并且12xx则21210,()1xxf xx所以2121121()()()()f xf xxxf xf xx1()f x.所以()f x在
5、,单调递增。证明二：直接用（3）任取12,x x并且12xx则21210,()1xxf xx所以2211()()1()f xf xxf x又()0f x，所以21()()f xf x.所以()f x在,单调递增。3、()()()f xyf xfy，类比函数()f x=logax类比函数()f x=logax的性质（1）()f x=logax过点（1，0）对应性质已知函数()f x对任意,x yR恒有()()()f xyf xfy，求证(1)0f名师资料总结-精品资料欢迎下载-名师精心整理-第 2 页，共 4 页 -3 证明 在()()()f xyfxf y中，令1xy得(1)0f（2）对于()
6、f x=logax，1()()0f xfx对应性质已知 函数 f(x)对任意,x yR恒有()()()f xyf xfy，求证1()()0f xfx证明 在()()()f xyfxf y中，令1yx得1(1)()()ff xfx，由（1）知(1)0f所以1()()0f xfx（3）()()()f xyf xf y()()()xff xf yy证明：在()()()f xyf xf y中，以1y代y得：()xfy()fx1()fy由（2）知1()()ff yy所以()xfy()f x()f y（4）当1a时，()f x=logax是增函数对应性质函数()f x对任意,0 x y恒有()()()fx
7、yf xfy，当1x时()0fx证明一：任取12,x x并且120 xx，211xx21()0 xfx，2221111()()()()xxf xf xf xfxx1()f x所以()fx在0,上单调递增证明二：任取12,x x并且120 xx，211xx，21()0 xfx，又由（3）21()xfx21()()f xf x所以21()()f xf x，所以()f x在0,上单调递增4、()()()f xyf x f y，类比性质()fx=x类比函数()f x=x的性质（1）()f x=x满足过点(1,1)对应性质函数()f x对任意,0 x y恒有()()()fxyf x fy，并且()f x
8、不恒为 0，求证(1)1f名师资料总结-精品资料欢迎下载-名师精心整理-第 3 页，共 4 页 -4 证 明因 为()f x不 恒 为0，不 妨 设()f a0，在()()()fx yfxfy中 令,1xa y得()()(1)f af a f所以(1)1f.(2)对于函数()f x=x，11()()1f x fxxx对应性质函数()f x对任意,0 x y恒有()()()fxyf x fy，求证1()()f x fx1证明 由（1）知(1)1f在()()()f xyf x fy中，令1yx得11()()(1)1f x ffxfxx。（3）()()()f xyf x f y()()()xf xf
9、yfy证明：在()()()f xyf x fy中，以1y代y得：1()()()xff x fyy，又由（2）知：11()()fyfy，所以()xfy()()f xfy（4）函数()f x=x，当0时，()f x=x在第一象限是增函数对应性质函数()f x对任意,0 x y恒有()()()fxyf x fy，对任意xR，()0f x并且当1x时()f x1，求证 当0 x时，()f x是增函数证明一：任取12,x x并且120 xx，211xx，21()1xfx，2221111()()()()xxf xf xf xfxx1()f x所以()fx在0,上单调递增证明二：任取12,x x并且120 xx，211xx，21()1xfx，又由(3)2211)()()(xf xfxf x即21)()1(f xf x又()0f x所以21()()f xf x所以()fx在0,上单调递增名师资料总结-精品资料欢迎下载-名师精心整理-第 4 页，共 4 页 -
2022年抽象函数问题山东省莱芜市第一中学刘
