设函数f（x）在（-∞，+∞）内有定义，下列函数（1）y=-|f（x）|；（2）y=xf（x 2 ）；（3）y=-f（-x）； 设函数y=f（x）在（-∞，+∞）内有定义，对于给定的实数k...
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| y=-|f（x）|中不论x取任何值，|f（x）|所对的函数值均不变，故（1）为偶函数； y=xf（x 2 ）可以看成为两个函数的乘积，其中，y=x是奇函数，y=f（x 2） 是偶函数，故（2）是奇函数． y=-f（-x）奇偶性没办法确定．故（3）不是奇函数． 令F（x）=y=f（x）-f（-x）因为F（-x）=f（-x）-f（x）=-（f（x）-f（-x））=-F（x），故（4）是奇函数 故答案为：（2）（4） |
绛旓細绠鍗曞垎鏋愪竴涓嬶紝璇︽儏濡傚浘鎵绀
绛旓細绠鍗曞垎鏋愪竴涓嬶紝璇︽儏濡傚浘鎵绀
绛旓細銆愮瓟妗堛戯細鐢变簬f(x)鍦(-鈭烇紝+鈭)鍐呮湁瀹氫箟锛屼笖 f(x+y)=f(x)+f(y)浠=y=0锛屽彲寰 f(0)=f(0)+f(0)=2f(0)鐢辨鍙緱鐭 f(0)=0 瀵逛换鎰忕粰瀹氱殑x锛岃冨療 f(x+鈻硏)=f(x)+f(鈻硏)锛宖(x+鈻硏)-f(x)=f(鈻硏)浠庤 姝(x)=k锛屽張f(0)=0锛屽彲鐭 f(x)=0鍙互璇佹槑f(x)鍦(...
绛旓細銆愮瓟妗堛戯細B 鎻愮ず 宸茬煡f(x)鍦(-鈭锛+鈭)涓婃槸鍋鍑芥暟锛屽嚱鏁板浘鍍忓叧浜巠杞村绉帮紝宸茬煡鍑芥暟鍦(0锛+鈭)锛宖'(x)锛0锛 f''(x)锛0锛岃〃鏄庡湪(0锛+鈭)涓婂嚱鏁板浘鍍忎负鍗曞涓斿嚬鍚戯紝鐢卞绉版у彲鐭ワ紝f(x)鍦(-鈭烇紝0)鍗曞噺涓斿嚬鍚戯紝鎵浠'(x)锛0锛 f''(x)锛0銆
绛旓細绠鍗曞垎鏋愪竴涓嬶紝璇︽儏濡傚浘鎵绀
绛旓細銆愮瓟妗堛戯細D 銆愯冩儏鐐规嫧銆戞湰棰樹富瑕佽冩煡鐨勭煡璇嗙偣涓哄嚱鏁扮殑濂囧伓鎬э紝銆愬簲璇曟寚瀵笺戣冩煡鍑芥暟鐨勫鍋舵э紝鍙渶灏嗭紝(z)涓殑x鎹㈡垚-x锛岃绠楀嚭锛宖(-x)锛岀劧鍚庣敤濂鍑芥暟锛鍋跺嚱鏁板畾涔変笅缁撹锛庡浜嶢銆丅銆丆椤规棤娉曞垽鏂叾濂囧伓鎬э紝鑰岄夐」D鏈墆=f(x)+f(-x)锛屽皢f(x)涓殑x鎹㈠啓鎴-x鏈塮(-x)+f[-(-x)]=f...
绛旓細璁惧湪鏁翠釜鏁拌酱涓奻(x)鈮爔锛屽垯渚f(x)鍦鏁翠釜鏁拌酱涓婄殑杩炵画鎬у彲鐭ワ紝f(x)-x鍦ㄦ暣涓暟杞翠笂蹇呭畾鍚屽彿锛屼笉濡璁緁(x)-x锛0锛屽嵆f(x)锛瀤銆傚洜姝[f(x)]锛瀎(x)锛瀤锛岃繖涓巉[f(x)]=x鐩哥煕鐩撅紝鏁呯煡鍦ㄦ暟杞翠笂瀛樺湪x0锛屼娇f(x0)=x0銆傝嫢灏嗛棶棰樿浆鍖栦负f(x0)-x0=0锛屽垯灞炰簬杩炵画鍑芥暟f(x)-x鍦-鈭锛渪...
绛旓細瑙ｇ瓟锛氬洜涓鍑芥暟F锛圶锛夊湪锛堬紞鈭烇紝锛鈭烇級涓婃湁瀹氫箟锛屼笖瀵逛换浣昘,Y鏈塅锛圶*Y锛夛紳F锛圶锛*F锛圷锛-X锛峐 鍋囪x=0锛寉=0鏃讹紝F锛圶脳Y锛夛紳F锛圶锛壝桭锛圷锛-X锛峐 鏈夛細F锛0锛夛紳F锛0锛壝桭锛0锛-0锛0 鎵浠ワ細F锛0锛=0鎴朏(0)=1 锛1锛変护x=1锛寉=1锛屽垯鏈夛細F锛1锛夛紳F锛1锛壝桭锛1...
绛旓細绠鍗曞垎鏋愪竴涓嬶紝绛旀濡傚浘鎵绀
绛旓細绠鍗曞垎鏋愪竴涓嬶紝璇︽儏濡傚浘鎵绀
