f(x)=x的平方-x+b 且f(log以2為底的a的對數)=blog2[f(a)]=2(a不等于1) 求f(log以2為底的x的對數)的最小值及對應的x的值
熱心網友
f(log以2為底的a的對數)=bf(log以2為底的a的對數)= [log2(a)]^2 - log2(a) + b =b所以 [log2(a)]^2 - log2(a) = 0log2(a) * [log2(a) -1] = 0log2(a)=0 或 log2(a) = 1a = 1 或 a =2a=1 根據題目 舍去。a=2log2[f(a)]=2f(a) = 4a^2 - a + b =4a=2代入所以 b = 2f(log以2為底的x的對數)= [log2(x)]^2 - log2(x) + b= [log2(x) - 1/2)^2 + 7/4當 log2(x) - 1/2 = 0 即 x=√2 時,上式取最小值 7/4
熱心網友
答的有條理,精彩!