Question

函数)1.已知f(2x+1)=x2-2x,求f(√2￣) (x2为x的二次方,(√2￣为根号二)2.f(x)是二次函数,

Answer 1

1.

设t=2x+1,则x=(t-1)/2

f(t)=[(t-1)/2]^2-2*(t-1)/2

=t^2/4-t+1/4-t+1

=t^2/4-2t+1

f(x)=x^2/4-2x+1

f((√2￣ )=(√2￣)^2/4-2(√2￣)+1=3/2-2√2￣

2.

设f(x)=ax^2+bx+c

f(2)=f(4)说明二次函数f(x)关于x=(2+4)/2=3对称,x=3为其对称轴

又f(2)=f(4)＞f(3),说明函数图像开口向上,左半边递减,

f(5)=f(1)