fx)是奇函数，且当x>0时,f(x)=x^2+sinx，则当x属于R时，f(x)为？？
\u82e5f\uff08x\uff09\u4e3a\u5947\u51fd\u6570\uff0c\u4e14x\uff1e0\u65f6\uff0cf\uff08x\uff09=x2-sinx\uff0c\u5219x\uff1c0\u65f6\uff0cf\uff08x\uff09=_____\u8bbex\uff1c0\uff0c\u5219-x\uff1e0\uff0c\u6240\u4ee5f\uff08-x\uff09=\uff08-x\uff092-sin\uff08-x\uff09=x2+sinx\uff0c\u53c8\u56e0\u4e3af\uff08x\uff09\u4e3a\u5947\u51fd\u6570\uff0c\u5219-f\uff08x\uff09=f\uff08-x\uff09=x2+sinx\uff0c\u6240\u4ee5f\uff08x\uff09=-x2-sinx\uff0e\u6545\u5e94\u586b-x2-sinx\uff0e
\u56e0\u4e3a\u51fd\u6570\u662f\u5947\u51fd\u6570\uff0c\u90a3\u4e48f(-x)=-f(x)
\u8bbea0
\u7531\u9898\u610f\u5f97\uff0cf(-a)=(-a)^2+1/(-a)=-f(a)
\u6240\u4ee5f(a)=-[(-a)^2+1/(-a)]
f(-1)=-(1+1)=-2
\u6216f(-1)=-f(1)=-2
-f(x)=x^2-sinx， f(x)=-x^2+sinx，
f(x)=x^2+sinx(x>0) f(x)=-x^2+sinx(x<0)
绛旓細x=0锛宖(0)=0锛泋>0锛宖(x)=e^x+a锛屽崟璋冮掑锛 x鈫0+锛宖(x)鈫1+a锛泋<0锛宖(-x)=e^(-x)+a= -f(x)锛宖(x)=-e^(-x)-a锛屽崟璋冮掑锛寈鈫0-锛宖(x)鈫-1-a.瑕佷娇f(x)鍦≧涓婃槸澧鍑芥暟锛鍙 1+a鈮-1-a锛屼笖1+a鈮0锛屾墍浠鈮-1.
绛旓細鍥犱负鏄鍑芥暟 x锛0.f(x)=e^x 鎵浠锛0鏃讹紝f(-x)=-f(x)e^(-x)=-f(x)f(x)=-1/(e^x)
绛旓細f(0)=0<濂囧嚱鏁鐗规>f(1)=f(-1+2)=f(-1)<鍛ㄦ湡鎬>f(1)=-f(-1)<濂囧嚱鏁扮壒鎬>锛岀敱浜巉(x)鏄疄鏁帮紝鎵浠(1)=0=f(-1).2.璁-1<x1<x2<0锛宖(x1)=-1-2^(x1) f(x2)=-1-2^(x2)f(x1)-f(x2)=2^(x2)-2^(x1),鐢变簬2^x涓洪掑鍑芥暟锛鎵浠褰搙2<x1鏃讹紝2^(x2)-2...
绛旓細褰揳=-2鏃讹紝x>=0,f(x)=-x^2-2x 褰搙<0鏃讹紝-x>0,f(-x)=-(-x)^2-2(-x)=-x^2+2x 鍙鏄鍑芥暟锛鏁呮湁x<0鏃讹紝鏈塮(x)=-f(-x)=-(-x^2+2x)=x^2-2x 鍗宠В鏋愬紡鏄 f(x)=-x^2-2x,(x>=0 =x^2-2x,(x<0)(2)f(x)鏄崟璋冨噺鍑芥暟锛屽垯鏈塧<0 f(m-1)+f(m^2+t...
绛旓細鍥犱负鍛ㄦ湡涓2,f(-9/2)=f(-1/2-2*2)=f(-1/2)=1-2sin(-蟺/2)=3 鎵浠(x)鏄鍑芥暟,鎵浠(9/2)=-f(-9/2)=-3
绛旓細鍗砯锛-x锛=锛-x锛夛紙1-鏍瑰彿x鐨勪笁娆℃柟锛.(*)鍙堢敱f锛坸锛夋槸濂囧嚱鏁,鏁咃紙*锛夊紡鍙樹负 -f锛坸锛=锛-x锛夛紙1-鏍瑰彿x鐨勪笁娆℃柟锛夊嵆f锛坸锛=锛坸锛夛紙1-鏍瑰彿x鐨勪笁娆℃柟锛夋晠x灞炰簬锛堣礋鏃犵┓澶,0锛夋椂 f锛坸锛=锛坸锛夛紙1-鏍瑰彿x鐨勪笁娆℃柟锛2瑙ｇ敱fx鏄伓鍑芥暟,鍏跺畾涔夊煙涓篟,涓鍦紙璐熸棤绌,0锛変笂鏄鍑芥暟,...
绛旓細鏄撶煡杩欎釜鍑芥暟鏄弗鏍煎崟璋冪殑 鑰宖(x+t)>=2f(x)绛変环浜巉(x+t)鈮(鈭2*x)鏁呴棶棰樼瓑浠蜂簬褰搙灞炰簬[t,t+2]鏃 x+t鈮モ垰2*x 鎭掓垚绔 灏唜+t鈮モ垰2*x鍙樺舰涓(鈭2+1)t鈮 鏁呭彧闇(鈭2+1)t鈮+2 瑙ｅ緱t鈮モ垰2
绛旓細x>=0鏃讹紝f(x)=-(x-1)^2+1=x(2-x)鍥犱负鍊煎煙[1/b,1/a]鍦ㄦ鏁板尯闂达紝鎵浠ユ湁0<x<2 鍥犱负f(x)<=1, 鎵浠ユ湁1/a<=1,寰楋細a>=1 鍥犳鍦╗a,b]鍖洪棿锛屽嚱鏁鍗曡皟鍑忥紝鏈塮(a)=1/a, f(b)=1/b 鍗砤,b涓烘柟绋2t-t^2=1/t鐨勪袱涓牴 t^3-2t^2+1=0 t^3-t^2-t^2+1=0 t^2(...
绛旓細绛旓細f(x)鏄瀹氫箟鍦≧涓婄殑濂囧嚱鏁帮紝鍒欐湁锛歠(-x)=-f(x)f(0)=0 x>=0鏃讹紝f(x)=x^2-2x 鍒檟<=0鏃讹紝-x>=0浠ｅ叆涓婂紡寰楋細f(-x)=x^2+2x=-f(x)鎵浠ワ細x<=0鏃讹紝f(x)=-x^2-2x 鎵浠ワ細x>=0锛宖(x)=x^2-2x x<=0锛宖(x)=-x^2-2x ...
绛旓細x0,鍒檉(x)=-f(-x)=-[2^(-x)-1]=1-1/2^x
