設u=3^x,若y=(7+9u)/(u+2),求值域?

2019-05-18 12:45 am

回答 (3)

2019-05-18 5:35 pm
設u=3^x,若y=(7+9u)/(u+2),求值域?
Sol
y=(7+9u)/(u+2)
yu+2y=7+9u
yu+2y-7-9u=0
(y-9)u=7-2y
u=(2y-7)/(y-9)>0
(2y-7)(y-9)>0
3.5<y<9
2019-06-08 9:20 pm
y=(9u+18-11)/(u+2)
=9-11/(u+2)
值域為 {y∈R∣7 / 2 < y < 9} #


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