设函数f(x)=x^2,x<=1 f(x)=ax+b,x>1 (这两个式子在一个大根号内）在点x=1处可导，试确定a和b的值。 设函数f(x)=x^2/x-2,f(-x+1)=
\u51fd\u6570 f(x)=ax^2,x>1;bx+1,x<=1 \u5728x=1 \u5904\u8fde\u7eed\uff0c\u53ef\u5bfc\uff0c\u6c42a,b \u7684\u503c\uff1b1. \u6839\u636e\u8fde\u7eed\u77e5\u9053x=1\u65f6\u4e24\u4e2a\u5f0f\u5b50\u76f8\u7b49\uff0c\u6240\u4ee5a=b+1
2. f'(x)=2ax , x>1
b , x<=1
\u6839\u636e\u53ef\u5bfc\u77e5\u9053\u4e24\u5f0f\u5728x=1\u65f6\u76f8\u7b49\u30022a=b
\u4e24\u5f0f\u8054\u7acb a=-1 b=-2
f(-x+1)
=(-x+1)²/[(-x+1)-2]
=(1-x)²/[-(x+1)]
=-(x²-2x+1)/(x+1)
简单分析一下即可，详情如图所示
x<=1时，f'(x)=2x
x>1时，f'(x)=a
x=1可导，则必连续，得：f(1)=f(1-)=f(1+),即：2=a+b
左右导数须相等，得：f(1)=f'(-1)=f'(1+),即2=a
因此有a=2, b=0
绛旓細s(x)=x^2 x鈭圼2k蟺,2(k+1)蟺)s(x)=2蟺^2 x=2(k+1)蟺 s(4蟺)=(4蟺)^2=16蟺^2
绛旓細璇佹槑锛氳 x = y^2,f(y)=f(y^2), ===> f(x)= f(x^(1/2))浠荤粰x 澶т簬0锛 涓嶇瓑浜1, f锛坸) = f(x^(1/2))= f(x^(1/4))=...=f(x^(1/2^n))=...鍥犱负 x, x^(1/2), ..., x^(1/2^n), ... ---> 1 鏍规嵁杩炵画鎬э紝 浜庢槸 f锛1锛=lim f(x^(...
绛旓細f(1/e)=1/e²-2*(-1)=2+1/e²f(e)=e²-2>2+1/e²鎵浠max(x)=f(e)=e²-2 (3).f(x)=X^2-x-a 鍗筹細x²-2lnx=X^2-x-a 2lnx=x+a 浠(x)=2lnx-x g'(x)=2/x-1 浠'(x)>0锛屽緱锛0<x<2 浠'(x)<0锛屽緱锛歺>2,x<0(...
绛旓細f(x)=x^2-x+(1/2)=(x-(1/2))^2+(1/4)瀵圭О杞翠负x=1/2 鑰岋細n鈭堟鏁存暟 鎵浠ワ細鍦╗n,n+1]鍗曡皟閫掑 f(n)=n^2-n+(1/2),鏈灏忔暣鏁板=n^2-n+1 f(n+1)=(n+1)^2-(n+1)+(1/2),鏈澶ф暣鏁板=(n+1)^2-(n+1)=n^2+n 鏁存暟鐨勪釜鏁=(n^2+n)-(n^2-n+1)+1=...
绛旓細涓よ竟瀵箈姹傚 2x=1-f(2x)路(2x)'2f(2x)=-2x+1 f(2x)=-x+(1/2)浠2x=t,鍒檟=t/2 f(t)=-(t/2)+(1/2)鎵浠f(x)=-(x/2)+(1/2)
绛旓細鎶x鍒嗗埆鎹㈡垚2锛-2锛宎锛-a锛屽垎鍒畻鍑虹粨鏋滃氨琛屼簡锛屾病浠涔堥毦鐨勫晩銆
绛旓細濂鍑芥暟 鎵浠(3)=-f(-3)=-2^(-3)=-1/8
绛旓細1銆乫'(x)锛1+4x+3x^2锛3(x+1)(x+1/3)锛岄┗鐐规槸-1/3鍜-1f''(x)锛4+6xf''(-1/3)锛0锛屾墍浠f(x)鍦▁锛-1/3澶勫彇寰楁瀬灏忓糵(-1/3)锛-4/27锛沠''(-1)锛0锛屾墍浠(x)鍦▁锛-1澶勫彇寰楁瀬澶у糵(-1)锛0 2銆佺敱鍗曡皟鎬锛屽嚱鏁鍦╗a,0]涓婄殑鏈灏忓煎彧鑳藉湪f(a)锛宖(0)锛宖(-...
绛旓細鎶f(x)=x^2-1浠ｅ叆f(x/a)-4a^2f(x)-f(x-1)<=4f(a)骞跺寲绠鍙緱 2x+3 鈮 (4a^2 + 1 - 1/(a^2))*x^2, 鍥犱负x >0, 鎵浠4a^2 + 1 - 1/(a^2) 鈮 (2x+3)/(x^2)瀵逛换鎰弜鈭圼1锛+鈭)閮芥垚绔嬨傚洜涓(2x+3)/(x^2) 鈮5锛屽綋x=1鏃剁瓑鍙锋垚绔嬶紝鎵浠4a^2 + 1 - 1/...
绛旓細y'=lim螖x鈫抣im螖x鈫0螖y[f(x+螖x)-f(x)]/螖x f鈥(-1)=limx鈫-1[f(x)-f(-1)/(x+1)]=limx鈫-1(10x^2-10)/(x+1)=limx鈫-1(10x-10)=-20 ^f(x)=10x^3-x f'(x)=(10x^3)-(x)'=10脳3脳x^(3-1)-1 =30x²-1 f'(-1)=30脳(-1)²-1 ...
