Recall that the domain of a rational function A2 is all values of x except those values for which Q(x) = 0. Q(x) Solve 6x - 9 = 0 for x. 6x - 9 = 0 X = х+ 1 Therefore the domain of f(x) = is all real numbers except x = 6x - 9 x + 1 x + 1 Because it is a rational function, f(x) = is continuous over its domain. Thus, f(x) = is discontinuous at 6х- 9 бх - 9 the point x = since this is the value at which f(x) is not defined.
Recall that the domain of a rational function A2 is all values of x except those values for which Q(x) = 0. Q(x) Solve 6x - 9 = 0 for x. 6x - 9 = 0 X = х+ 1 Therefore the domain of f(x) = is all real numbers except x = 6x - 9 x + 1 x + 1 Because it is a rational function, f(x) = is continuous over its domain. Thus, f(x) = is discontinuous at 6х- 9 бх - 9 the point x = since this is the value at which f(x) is not defined.
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section4.5: Rational Functions
Problem 54E
Related questions
Question
![Recall that the domain of a rational function 2 is all values of x except those values for which Q(x) = 0.
Q(x)
Solve 6x – 9 = 0 for x.
6x – 9 = 0
X =
x + 1
Therefore the domain of f(x) =
is all real numbers except x =
бх — 9
x + 1
x + 1
Because it is a rational function, f(x) =
is continuous over its domain. Thus, f(x)
is discontinuous at
бх — 9
бх — 9
the point x =
since this is the value at which f(x) is not defined.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6a453cf0-4fa7-465e-ab7f-471dd1e543ba%2Ff6cd4172-e299-4024-a45e-d905de569e46%2Fc6sjwr_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Recall that the domain of a rational function 2 is all values of x except those values for which Q(x) = 0.
Q(x)
Solve 6x – 9 = 0 for x.
6x – 9 = 0
X =
x + 1
Therefore the domain of f(x) =
is all real numbers except x =
бх — 9
x + 1
x + 1
Because it is a rational function, f(x) =
is continuous over its domain. Thus, f(x)
is discontinuous at
бх — 9
бх — 9
the point x =
since this is the value at which f(x) is not defined.
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