已知函数f(x)=loga^x,记g(x)=f(x)[f(x)+f(2)-1],若g(x)在区间[1/2,2]上是增函数
2个回答

f(x)=log a^x = (lg a) * x

g(x)=f(x) [ f(x) + f(2) - 1] = (lga)² x² + (2lga -1)lga * x

g '(x) = 2 (lga)² x + (2lga -1)lga

g '(x) ≥ 0 => x + (2lga -1) / (2lga) ≥ 0 => x ≥ 1/(2lga) -1

[1/2,2] => 1/(2lga) -1 ≤ 1/2 => 1/lga ≤ 3

(1) a>1,lga ≥ 1/3 => a ≥ 10^(1/3)

(2) 0

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