题目
f(x2)-f(x1) |
x2-x1 |
A.f(3)<f(-2)<f(1) | B.f(1)<f(-2)<f(3) | C.f(-2)<f(1)<f(3) | D.f(3)<f(1)<f(-2) |
答案
f(x2)-f(x1) |
x2-x1 |
∴函数在(-∞,0)上单调递减
∵函数是偶函数,∴函数在(0,+∞)上单调递增
∴f(1)<f(2)<f(3)
∴f(1)<f(-2)<f(3)
故选B.
f(x2)-f(x1) |
x2-x1 |
A.f(3)<f(-2)<f(1) | B.f(1)<f(-2)<f(3) | C.f(-2)<f(1)<f(3) | D.f(3)<f(1)<f(-2) |
f(x2)-f(x1) |
x2-x1 |