Question

求函数y=

lg x x2-7x+12

的定义域．

Answer 1

依题意，令lg

x

x2-7x+12

≥0，
即lg

x

x2-7x+12

≥lg1．
于是有  

x

x2-7x+12

≥1?

x

x2-7x+12

-1≥0?

x2-8x-12

x2-7x+12

≤0?

(x-2)(x-6)

(x-3)(x-4)

≤0?

(x-2)(x-6)(x-3)(x-4)≤0 (x-3)(x-4)≠0

?x∈[2，3)∪(4，6]．