MHB What is the domain of the rational function f(x) = 2/(x - 3)?

[mathdad]

Find the domain of
f(x) = 2/(x - 3).

1. Are we looking for the domain of f or f(x)?

2. Solution

Set x - 3 = 0 and solve for x.

x - 3 = 0

x - 3 + 3 = 3

x = 3

Let D = domain

D = ALL REAL NUMBERS except for x = 3.

Yes?

P.S. Does x = 3 mean there is a hole at the point (3, 0) for this function? If so, what is the name of the hole? Asymptote?

 

[rootbarb]

Domain is the input or f(a number). OR the X-axis on the graph
Range is the output or f(x)=a number OR the Y-axis on the graph
Let's say your given the point (1,2)
f(1) would be input
f(x)=2 the outputAs for the second question I thought it would include 3 or 3 would be the endpoint. I'm not positive about that.

 

[mathdad]

Domain is the input or f(a number). OR the X-axis on the graph
Range is the output or f(x)=a number OR the Y-axis on the graph
Let's say your given the point (1,2)
f(1) would be input
f(x)=2 the outputAs for the second question I thought it would include 3 or 3 would be the endpoint. I'm not positive about that.

Is my answer correct?

 

[kaliprasad]

Is my answer correct?

yes it is correct

 

[mathdad]

I will post more rational function problems when I get to that particular chapter.

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yes it is correct

Does x = 3 for this question mean there is a hole at the point
(3, 0)? If so, what is the name of the hole? Asymptote?

 

[kaliprasad]

I will post more rational function problems when I get to that particular chapter.

- - - Updated - - -
Does x = 3 for this question mean there is a hole at the point
(3, 0)? If so, what is the name of the hole? Asymptote?

x =3 is vetical asymtote but it is not a hole as x = 3 denomiator is zero but numerator is not

it is a hole when numerator is zero as well.

 

[mathdad]

x =3 is vetical asymtote but it is not a hole as x = 3 denomiator is zero but numerator is not

it is a hole when numerator is zero as well.

Let A = a number

Are you saying that A/0 is a vertical asymptote?

Are you also saying that 0/A is a hole in the graph of the function?

 

[skeeter]

Let A = a non-zero number

Are you saying that A/0 is a vertical asymptote? the value of x that makes the denominator = 0 and the numerator not equal to 0 is the location of a vertical asymptote.

Are you also saying that 0/A is a hole in the graph of the function? the value of x that makes both the denominator and numerator = 0 is the location of a point discontinuity.

consider $f(x) = \dfrac{x(x+1)(x-2)}{(x-1)(x-2)}$

$f(x)$ has vertical asymptote $x=1$ and a point discontinuity ("hole") at $x=2$.

 

[mathdad]

(Non-zero number)/0 = vertical asymptote

(number that makes top zero)/(number that makes bottom zero) = hole in the graph of the function at a point (x, y).

That HOLE is called a POINT DISCONTINUITY.

TRUE?