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

Answer 1

Eva Abramson

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

The role is

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

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

Consequently,

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

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

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Answer 2

Ethan Adkins

#x in(-oo,oo)#

#"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]}

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Answer 3

Eva Abramson

To find the domain of ( f(x) = \frac{3x - 1}{x^2 + 9} ), we need to determine the values of ( x ) for which the function is defined. The only restriction is that the denominator cannot be equal to zero. Therefore, the domain of the function is all real numbers except the values that make the denominator zero. In this case, ( x^2 + 9 ) can never be zero, so the domain of ( f(x) ) is all real numbers.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.