(1)因为f(x-2)=ax2-(a-3)x+(a-2),则f(x)=a(x+2)2-(a-3)(x+2)+(a-2)又因为过点(-2,-3),代入-3=a-2 a=-1所以 f(x)=-x2+1(2)g(x)=f(f(x))=f(-x2+1)=-(-x2+1)2+1=2x2-x4又因为F(x)=Pg(x)-4 f(x) 所以F(x)=P(2x2-x4)+4x2-4=-Px4+(2P+4)x2-4设x2=y(y>=0),so F(x)=-Py2+(2P+4)y-4F(x)在(- , f(2))上是增函数,在(f(2),0)上是减函数,且f(2)=3,so F(x)在(- , f(2)),即X在(-,3)上...
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