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求函数g(x)=lgx²+2x+3/x+1在[1,2]上的值域
人气:459 ℃ 时间:2020-03-21 21:40:56
解答
g(x)=lg【(x²+2x+3)/(x+1)】分母不为零,x≠-1g(x)=lg【(x²+2x+3)/(x+1)】= lg【(x+1)(x+2)】= lg【(x+2)】单调增,g(1) = lg(1+2) = lg3,g(2) = lg(2+2) = 2lg2在[1,2]上的值域【lg3,2lg2】...
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