How do you find the domain and range of #5/(4-9x)#?

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Answer 1 · 2016-06-30T13:12:30

The domain is made of all real numbers except #4/9#
The range is made of all numbers except 0

you can find the domain by excluding x values that make null the denominator:

#4-9x!=0#

that's

#x!=4/9#

Let's find the inverse function through these simple steps:

let

#y=f(x)=5/(4-9x)#

#4y-9xy=5 and x!=4/9#

#9xy=4y-5#

#x=(4y-5)/(9y)#

so the range of y is the domain:

#y!=0#