How do you find the domain of #g(x)=6/(9-5x)#?

2 Answers
Aug 3, 2018

Answer:

#x inRR,x!=9/5#

Explanation:

The denominator of g(x) cannot be zero as this would make g(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be.

#"solve "9-5x=0rArrx=9/5larrcolor(red)"excluded value"#

#"domain is "x inRR,x!=9/5#

#(-oo,9/5)uu(9/5,oo)larrcolor(blue)"in interval notation"#
graph{6/(9-5x) [-10, 10, -5, 5]}

Aug 3, 2018

Answer:

#x inRR, x!=9/5#

Explanation:

The only thing that will make #g(x)# undefined is when the denominator is zero, so let's set it to zero to find any excluded values in the domain.

#-5x+9=0=>-5x=-9=>x=9/5#

The value #x=9/5# is not included in our domain, so we can say

#x inRR, x!=9/5#

We can even see this graphically, as we have a vertical asymptote at #x=9/5#.

graph{6/(9-5x) [-10, 10, -5, 5]}

Hope this helps!