fx-1是偶函数说明什么
说明x=1。既然y=f(x+1)为偶函数,那么就是说y=f(x+1)关于y轴对称,而y=f(x)相当于是由y=f(x+1)向右平移1个单位得来的,所以对称轴也向右移1个单位,即x=1。
绛旓細褰揦鈭圼2,3]鏃讹紝2-x鈭圼-1,0]婊¤冻F(x)鐨勫畾涔夛紝鏁匜(2-x)=G(x)=a(x-2)-2(x-2)^3锛屼护2-x=t锛屾墍浠(t)=2t^3-at锛屾墍浠(x)=2x^3-ax(x鈭圼-1,0])銆傚張F(X)鏄畾涔夊湪[-1,1]涓婄殑鍋跺嚱鏁锛宖(x)=f(-x)=-2x^3+ax(x鈭圼0,1])銆2)F(X)鍦╗0,1]涓婃槸澧炲嚱鏁,鎵浠...
绛旓細绛旓細鍋跺嚱鏁f(x)鍦╗1,3]涓婃槸鍑忓嚱鏁版湁鏈灏忓-1,x=3澶勫彇寰楁渶灏忓-1 鍦╢(x)鍦ㄥ尯闂碵-3,-1]涓婃槸澧炲嚱鏁版湁鏈灏忓-1,x=-3澶勫彇寰楁渶灏忓-1 鎵浠ワ細閫夋嫨D 鏈鐩冨療鍋跺嚱鏁板叧浜巠杞村绉扮殑鐭ヨ瘑鐐
绛旓細鏄殑锛鍋跺嚱鏁鏄叧浜巠杞村绉帮紝棰樼洰鏉′欢鍏充簬x=1瀵圭О鏄柊澧炲姞鐨勶紝鍗璇存槑杩欎釜鍑芥暟鍏充簬涓ゆ潯鐩寸嚎瀵圭О锛坸=0,x=1).
绛旓細f(x-1)=f(x+1),浠-1=t,涓婂紡鍙寲涓猴細f(t)=f(t+2)鎵浠ヨ鍑芥暟鏄懆鏈熶负2鐨勫嚱鏁.鍙坒(x)鏄疪涓婄殑鍋跺嚱鏁,f(x)=f(-x)褰搙鈭圼-1,0]鏃,f(x)=f(-x)=f(-x+2)=(-x+2)+1=-x+3.
绛旓細鈭礷锛-x-1锛=g锛-x锛=-g锛坸锛=-f锛坸-1锛夛紝鍙坒锛坸锛涓哄伓鍑芥暟 鈭磃锛坸+1锛=f[-锛坸+1锛塢=f锛-x-1锛夛紝浜庢槸f锛坸+1锛=-f锛坸-1锛夆埓f锛坸+1锛+f锛坸-1锛=0锛庘埓f锛2013锛+f锛2015锛=f锛2014-1锛+f锛2014+1锛=0锛屾晠閫夛細C锛
绛旓細f(2013)=f(2014-1)=g(2014)锛堟潯浠秅(x)=f(x-1)锛=-g(-2014) 锛坓鏄鍑芥暟锛=-f(-2014-1)锛堟潯浠秅(x)=f(x-1)锛=-f(-2015)=-f(2015)锛坒鏄伓鍑芥暟锛夋墍浠(2013)+f(2015)=0銆
绛旓細(1)璇曞啓鍑鍑芥暟fx鐨勫叧绯诲紡 (2)璁ㄨ鍑芥暟fx鐨勫崟璋冩 (1)瑙ｆ瀽:鈭鍑芥暟fx鏄伓鍑芥暟锛屽綋x澶т簬绛変簬0鏃.fx=x鐨勪笁鍒嗕箣涓娆℃柟 鈭磃(-x)=f(x)锛屽浘鍍忓叧浜嶻杞村绉 鍏惰В鏋愬紡涓:f(x)=x^(1/3) (x>=0)f(x)=(-x)^(1/3) (x<=0)(2)瑙ｆ瀽:x>=0锛屽崟璋冨;x<=0鍗曡皟鍑 ...
绛旓細杩欎釜鍙笉涓瀹氥璇存槑涓夌偣锛1.f(0)鍙兘娌℃湁鎰忎箟銆傚鍑芥暟銆f(x)=1/x锛岋紙琛ㄧずx鍒嗕箣涓锛夊畠鏄剧劧鏄鍑芥暟锛屼絾f(0)娌℃湁鎰忎箟銆2.鍋跺嚱鏁版椂锛宖(0)涔熷彲鑳芥槸0銆傚銆f(x)=x²鏄伓鍑芥暟锛屼笖f(0)=0 3.鍙湁褰撳鍑芥暟鐨勫畾涔夊煙涓寘鍚0鏃讹紝f(0)=0.鍥犱负 f(-x)=-f(x)灏嗐x=0浠ｅ叆銆锛...
绛旓細鏍规嵁 g锛圶锛夛紳锛峠锛圶锛夛紝鍙煡g锛圶锛変负濂囧嚱鏁帮紒fx=1/f(-x)锛涗唬鍏ユ暣鐞嗭細鍙團锛堬紞x锛夛紳[ f(-x)-1]/[f(-x) -1]*(-g(x))鏁寸悊寰楋細F锛堬紞X锛夛紳[f(x)-1]/[fx+1]*g(x)=F锛圶锛夋墍浠 F锛堬紞X锛夛紳F锛圶锛夋墍浠ュ叾涓哄伓鍑芥暟銆
绛旓細瑙ｇ敱f锛坸+1锛=f锛坸-1锛夊彇x涓簒+1寰 f锛坸+2锛=f锛坸锛夋晠鍑芥暟鐨勫懆鏈熶负2 鍒欑敱f锛坙og锛1/3)(5))=f锛-log3(5))=f锛坙og3(5))...f锛坸锛鏄伓鍑芥暟 =f锛坙og3(5)-2)...f锛坸锛夌殑鍛ㄦ湡涓2 =f锛坙og3(5)-log3锛9锛)=f锛坙og3(5/9))娉ㄦ剰鍒1/3锛5/9锛1 鏁卨og3锛1/3锛...
