1.
設(shè)t=2x+1,則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.
設(shè)f(x)=ax^2+bx+c
f(2)=f(4)說明二次函數(shù)f(x)關(guān)于x=(2+4)/2=3對稱,x=3為其對稱軸
又f(2)=f(4)＞f(3),說明函數(shù)圖像開口向上,左半邊遞減,
f(5)=f(1)