已知f(x)是奇函数,当x≥0时,f(x)=x²-2x,求当x<0时,f(x)的解析式,并画出图像 已知f(x)是奇函数,当x<0时,f(x)=-x2-2x 当...
1\u3001\u5df2\u77e5f(x)\u662f\u5947\u51fd\u6570,\u4e14\u5f53x>0\u65f6,f(x)=x²-2x,\u6c42\u5f53x<0\u65f6f(x)\u7684\u8868\u8fbe\u5f0f1.\u5947\u51fd\u6570\u5229\u7528f(-x)=-f(x)\uff0c\u5f53x0\uff0c\u4ee3\u5165\u8868\u8fbe\u5f0ff(-x)=\uff08-x\uff09²-2\uff08-x\uff09=x²+2x,\u6240\u4ee5f(x)=-x²-2x\u3002
2.\u5076\u51fd\u6570\u5229\u7528f(-x)=f(x)\uff0c\u5219\uff08k-2\uff09\uff08-x\uff09²-\uff08k-3\uff09x+3=\uff08k-2\uff09x²+\uff08k-3\uff09x+3\uff0c\u4e0a\u5f0f\u5bf9\u4e8e\u4efb\u4f55x\u90fd\u6210\u7acb\uff0c\u6545k=3\uff0c\u6240\u4ee5f\uff08x\uff09=-x²+3\uff0c\u6c42\u5bfc\u5373\u5f97\u5728\uff08\u8d1f\u65e0\u7a77\uff0c0)\u5355\u8c03\u9012\u589e\uff0c\u5728[0,\u6b63\u65e0\u7a77)\u5355\u8c03\u9012\u51cf\u3002
3.\u5947\u51fd\u6570\u5229\u7528f(-x)=-f(x)\uff0c\u5373f\uff080\uff09=0\uff0c\u5f53x0\uff0c\u4ee3\u5165\u8868\u8fbe\u5f0ff(-x)=\uff08-x\uff09²+2x+3=x²+2x+3\uff0c\u6545f\uff08x\uff09=-x²-2x-3\u3002\u7efc\u5408\u5f97\u5206\u6bb5\u51fd\u6570\uff1ax>0\u65f6\uff0cf\uff08x\uff09=x²-2x+3\uff1bf\uff080\uff09=0\uff1bx<0\u65f6,f\uff08x\uff09=-x²-2x-3\u3002
\u2235x>0 \u2234-x<0\uff0c\u5219f(-x)=-(-x)²-2(-x)=-x²+2x
\u2235f(x)\u662f\u5947\u51fd\u6570 \u2234f(x)=-f(-x)=-(-x²+2x)=x²-2x
f(x)是奇函数→f(-x)=-f(x)
f(x)=x²-2x x≥0
f(-x)=-[(-x)²-2(-x)]=-x²-2x x<0
绛旓細锛1锛夆埖褰搙鈮0鏃讹紝f锛坸锛=x锛坸-2锛夆埓褰搙锛0鏃讹紝f锛-x锛=-x锛-x-2锛=x锛坸+2锛夊張鈭f锛坸锛夋槸瀹氫箟鍦≧涓濂囧嚱鏁帮紝鈭村綋x锛0鏃讹紝f锛坸锛=-f锛-x锛=-x锛坸+2锛夊洜姝わ紝鍑芥暟f锛坸锛夌殑瑙ｆ瀽寮忎负f锛坸锛=x(x?2) (x鈮0)?x(x+2) (x锛0)锛 锛2锛夆埖褰搙鈮0鏃讹紝f锛坸锛=x...
绛旓細绠鍗曞垎鏋愪竴涓嬶紝璇︽儏濡傚浘鎵绀
绛旓細褰揦0,鍒欐湁f(-x)=(-x)^2-2x 鍙堝嚱鏁鏄鍑芥暟,鍒欐湁f(x)=-f(-x)鏁呭綋X<0鏃舵湁f(x)=-f(-x)=-(x^2-2x)=-x^2+2x
绛旓細瑙ｏ細鍥犱负锛f(x)鏄瀹氫箟鍦≧涓婄殑濂囧嚱鏁 鎵浠ワ細f(x)鐨勫浘鍍忔槸鍏充簬鍘熺偣鎴愪腑蹇冨绉扮殑鍥惧儚銆傝岋細褰搙鈮o鏃,f(x)=x²-x 鎵浠ワ細褰搙<0鏃,f(x)鐨勮〃杈惧紡涓篺(x)=-(x²-x)锛屽嵆锛歠(x)=-x²+x (x<0)
绛旓細f锛坸锛=锕n锛堬梗x+1锛
绛旓細璁X0.鍗虫弧瓒矲锛-X锛=锛-X锛夛紙-X+1锛夊張鍥犱负F锛圶锛変负R涓濂囧嚱鏁,鍒橣锛-X锛=-F锛圶锛夊彲寰桭锛-X锛=-F锛圶锛=锛-X锛夛紙-X+1锛夊彲寰桭锛圶锛=X锛-X+1锛夌患涓,褰揦>=0鏃,F(X)=X(X+1锛夊綋X
绛旓細鍥犱负f锛坸锛夋槸濂囧嚱鏁帮紝鎵浠(-x)=-f(x)锛屾墍浠(x)=-f(-x)锛屼笖鏈塮(0)=0锛屽張鍥犱负褰揦锛0鏃秄锛坸锛夛紳X²锛屾墍浠ュ綋X<0鏃讹紝f锛坸锛=-f(-x)=-X²鎵浠鈮0鏃讹紝f锛坸锛夌瓑浜-X²銆
绛旓細璁x0 f(-x)=-x[1+(-x)^(-3)]=-x+x^(-2)鍙堝洜涓哄嚱鏁皔=f(x)鏄鍑芥暟 鎵浠(-x)=-f(x)=-x+x^(-2)鎵浠(x)=x-x^(-2)鍥犳,褰搙
绛旓細瑙ｏ細鐢遍鍙煡锛f(x)涓哄鍑芥暟锛鍒檉(-x)= - f(x)锛屽嵆锛歠(x)= - f(-x)锛涗负浜嗕究浜庡尯鍒嗭紝浠= - t锛涗唬鍏ュ緱锛歠(-t)=(-t)^2 -2(-t)=t^2+2t锛涙墍浠 f(t)= - f(-t)= -t^2 - 2t锛涙崲鍏冿紝寰楋細f(x)= - x^2-2x (x<0)f(x)鍦≧涓婄殑瑙ｆ瀽寮忓彲浠ョ敤鍒嗘鍑芥暟琛ㄧず锛-x...
绛旓細瑙ｈx锛0 鏁-x锛0 鍙堢敱x锛0鏃讹紝f(x)=(1/3)x 鍒檉(-x)=(1/3)(-x)鍙堢敱f(x)鏄鍑芥暟 鍒-f(x)=-1/3x 鍒檟锛0鏃讹紝f(x)=1/3 鍙堢敱鍑芥暟f(x)鏄畾涔夊湪r涓婄殑濂囧嚱鏁 鍒檉(0)=0 鏁呭嚱鏁癴(x)鐨勫嚱鏁拌В鏋愬紡f(x)=x/3.
