Ӗ5 ⺯?/p>
⺯ʽǸ߿c֮һ.Ҫ⺯ĻϣʃeZrng?
Ӗų
()֪f(2-cosx)=cos2x+cosx,f(x-1).
̽
[1](1)֪f(x)f(logax)= (a>0,a1,x>0),f(x)\ʽ.
(2)֪κf(x)=ax2+bx+c|f(1)|=|f(-1)|=|f(0)|=1,f(x)\ʽ.
DҪ麯еҪأֵꠦԃr֪?Ŀ.
֪RУú֪RرǶԡfúõȼתעⶨ.
˼SҪ]hĿȼת?
?1)]Ԫ;(2)ôϵ.
?1)t=logax(a>1,t>0;0
f(t)= (at-a-t)
f(x)= (ax-a-x)(a>1,x>0;0
(2)f(1)=a+b+c,f(-1)=a-b+c,f(0)=c
f(1)f(-1)f(0)ͬr1-1f(x)=2x2-1f(x)=-2x2+1f(x)=-x2-x+1f(x)=x2-x-1f(x)=-x2+x+1f(x)=x2+x-1.
[2]f(x)Rϵżx-1ry=f(x)DN(-20)б1]y=f(x)DһǶ(02)ҹ(-11)һ]f(x)\ʽDD.
DҪ麯֪ʶߡߵĻDԷֶκķҪ˼S.ˣֶκo.Ŀ. ֪RУżؼϵ߷.
˼SҪܸ]ۡۺ֪R.
뷽мSôϵʽ.
?1)x-1rf(x)=x+b
߹(-20).0=-2+bb=2f(x)=x+2.
(2)-1
߹(-11)1=a·(-1)2+2,a=-1
f(x)=-x2+2.
(3)x1rf(x)=-x+2
{֪f(x)= Dɶ.
Ӗ漰⼰ҪУ
1.ϵ֪ʽĹrôϵ;
2.Ԫݚ֪rf[g(x)]\ʽ]ԪʽϼrҲݚ;
3.η֪ĺʽýIķf(x);
⣬ڽоõۡȼתѧO.
2018ɿ2018qr/
2018ָϣ2018qָ