函数,已知f(2x)=x²-1,则f(2)=? 怎么解,求方法。并求第六题的解析。 已知函数f(2x+1)=x^2+2x,求函数f(x-1)的解...
\u82e5\u51fd\u6570f(2x +1)=x^2-2x,\u5219f(x)=?f(2x +1)=x^2-2x
f\uff082x+1\uff09=1/4\uff084x²+4x+1\uff09-1/4-3x
=1/4\uff082x+1\uff09²-1/4-3/2\uff082x+1\uff09+3/2
\u4ee42x+1=a
\u5219 f\uff082x+1\uff09=f\uff08a\uff09=1/4a²-3/2a+5/4
f\uff08x\uff09=1/4x²-3/2x+5/4
\u5c31\u662f\u5c06\u53f3\u4fa7\u53d8\u6210\u542b\u67092x+1\u7684\u5f0f\u5b50\uff0c\u7136\u540e\u66ff\u6362\u5c31\u53ef\u4ee5\u4e86
\u4ee42x+1=t\uff0c\u5219x=(t-1)/2
\u539f\u5f0f\u5316\u4e3a\uff1af(t)=(t-1)²/4+t-1
\u6240\u4ee5\uff1af(x-1)=(x-2)²/4+x-2=x²/4-1
\u795d\u4f60\u5f00\u5fc3\uff01\u5e0c\u671b\u80fd\u5e2e\u5230\u4f60\uff0c\u5982\u679c\u4e0d\u61c2\uff0c\u8bf7Hi\u6211\uff0c\u795d\u5b66\u4e60\u8fdb\u6b65\uff01
请采纳
6.把(1,0)代入得 ,A不满足条件
7.f(2)=f(2×1),即x=1时,代入得 f(2)=1²-1=0
8.2x²-1=7, 2x²=8 x²=4 a=正负2
9.f(x-1)=3x-2=3(x-1)+1
f(x)=3x+1
10.正比例函数,则f(x)=kx
f(2)=2k=3,f(-2)=-2k=-3
f(2)=f(2×1)=1²-1=0
第六题,把x=1 代进去,A:y=2所以选A
如图
绛旓細f(t)=(1/2t)鐨勫钩鏂+1/2t=1/4t^2+1/2t 鎵浠,f(x)=(1/2x)^2+1/2x=1/4x^2+1/2x f(x+1)=1/4(x+1)^2+1/2(x+1)=1/4x^2+x+3/4
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绛旓細鏄痻=1/2 鐢辨潯浠跺彲鐭 f'(1)=0 瀵筬(2x)姹傚 [f(2x)]'=2f'(2x)浠ゅ叾=0 2f'(2x)=0 瑙e緱x=1/2 鎴栫敤鐗规畩鍑芥暟 f(x)=(x-1)^2 f(2x)=(2x-1)^2 =4x^2-4x+1 -b/2a=1/2