How do you find the domain of #f(x)=(3x-1)/(x^2+9)#?

2 Answers
Apr 28, 2018

Answer:

The domain of #f(x)# is #RR#

Explanation:

The function is

#f(x)=(3x-1)/(x^2+9)#

#AA x in RR," the denominator is " x^2+9>0# and is #!=0#

Therefore,

The domain of #f(x)# is #RR# or in interval notation #x in (-oo, +oo)#

graph{(3x-1)/(x^2+9) [-7.02, 7.03, -3.51, 3.51]}

Apr 28, 2018

Answer:

#x in(-oo,oo)#

Explanation:

#"the denominator of "f(x)" cannot be zero as this would"#
#"make "f(x)" undefined"#

#x^2+9" is always positive for all "x inRR#

#rArr"domain "x in(-oo,oo)#
graph{(3x-1)/(x^2+9) [-10, 10, -5, 5]}