Question

求数学函数题若x∈[1/27,9],求函数y=log3 x/27 · log3 3x的最大值和最小值___3是底数

Answer 1

y=log3 x/27 · log3 3x

=log3 x/3^3 · log3 3x

=1/3·log3 x ·( log3 3+log3 x)

=1/3·log3 x +1/3·log3 x ·log3 x

=1/3·(log3 x)^2+1/3·log3 x

令log3 x＝t

1/3·(log3 x)^2+1/3·log3 x=t^2/3+t/3=1/3·(t^2+1)

因为t^2>=0,所以当t=0时,1/3·(t^2+1)最小,且最小取1/3·(t^2+1)＝1/3,即log3 x＝0,x=1

当x=9时,t=log3 x=2

1/3·(t^2+1)=5/3

所以

函数y=log3 x/27 · log3 3x

的最大值和最小值分别是1／3,5／3