Question

函数f(x)=x^2-bx+c满足f(-1+x)=f(2-x)且f(0)=2,则f(b^|x| )与f(c^|x|)的大小关系为？

Answer 1

将两个条件代入分别可得出b=1和c=2，由于f(x)在x>1/2是增函数，易知f(b^|x| )<=f(c^|x|)

Answer 2

f(-1+x)=x^2-(2+b)x+1+c;
f(2-x)=x^2+(b-4)x+4-2b+c;
f(-1+x)=f(2-x) → b=1;
f(0)=2 → c=2;
f(x)=x^2-x+2;
f(b^|x|)=f(1)=2;
f(c^|x|)=f(2^|x|);
2^|x|≥1 → f(2^|x|)≥f(1) → f(c^|x|)≥f(b^|x|) .