時(shí)間:2024-05-13
x1、x2∈(-∞0]且x1<x2則
f(x1)-f(x2)=(2x1+2-x1)-(2x2+2-x2)=(2x1-2x2)+( 1 2x1 - 1 2x2 )= (2x1-2x2)(2x12x2-1) ? 2x12x2
∵x1<x2<0∴0<2x1<2x2<1∴2x12x2>0∴2x1-2x2<0∴2x12x2-1<0
∴f(x1)-f(x2)>0即f(x1)>f(x2)
(x)=2x+2ax+b且f(1)=52f(2)=174.(1)求a、b;(2)
(x)=2x+2ax+b且f(1)= 5 2 f(2)= 17 4 .
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