find the domain of the rational expression

[CaptHaddock]

at a loss on how to do this one my teacher gave us this insanely hard take home test that is way harder than anything we have done in class...:x just giving me some steps to follow or any kind of help would be appreciated. thanks in advance!

1)

56-14y
-------
24x³-54x


2)

2m³-7m²-4m
----------------
24m⁴-62m³+70m²



3)

4m³-108
-----------------
7m⁵-91m³+252m



4)

m³-5m²-8m
--------------
28m²-m-15

 

[DrPhil]

at a loss on how to do this one my teacher gave us this insanely hard take home test that is way harder than anything we have done in class...:x just giving me some steps to follow or any kind of help would be appreciated. thanks in advance!

The domain of a rational expression must exclude any points for which the denominator is zero. For each problem, factor the denominator and find roots.

Sometimes a factor of the denominator may cancel the same factor in the numerator. That makes it a "reducible" infinity, but as long as the denominator is written to include that factor, that root must be omitted from the domain.

 

[DrPhil]

Did you TRY....did you google "domain of rational expression"?

Denominator: 24x^3 - 54x
If 24x^3 = 54x, then denominator = 0, right?

24x^3 = 54x
Divide by 6x: OOPS .. division by 0 not allowed, so what about x=0?
4x^2 = 9
x^2 = 9/4
x = 3/2

What does that tell you?

.

 

[CaptHaddock]

Did you TRY....did you google "domain of rational expression"?

Denominator: 24x^3 - 54x
If 24x^3 = 54x, then denominator = 0, right?

24x^3 = 54x
Divide by 6x:
4x^2 = 9
x^2 = 9/4
x = 3/2

What does that tell you?

wow thanks for the warm welcoming... Dr Phil

anyways to mr warm welcome yes i did try and i tried google, however it was fairly useless and none of the examples really where like my question above..

i came up with an answer of x=0,3/2,-3/2 for #1

in all your smugness you didn't even do it right yourself. GO Team Canada!

thanks...

 

[Dale10101]

Or, if somebody wants to show some leg ..

\[\begin{array}{l}
24{x^3} - 54x = 0 = > \\
6x(4{x^2} - 9) = 0 = > \\
6x({(2x)^2} - {3^2}) = 0 = > \\
6x(2x + 3)(2x - 3) = 0 = > \\
x = [0,\frac{3}{2},\frac{{ - 3}}{2}]
\end{array}\]


-heh, heh

 

[lookagain]

i came up with an answer of x=0,3/2,-3/2 for #1

CaptHaddock, your answer for the first question is wrong. \(\displaystyle \ \ \) What you gave are the values for what x cannot be.

The domain consists of all of the permissible values of x.

\[\begin{array}{l}
24{x^3} - 54x = 0 = > \\
6x(4{x^2} - 9) = 0 = > \\
6x({(2x)^2} - {3^2}) = 0 = > \\
6x(2x + 3)(2x - 3) = 0 = > \\
x = [0,\frac{3}{2},\frac{{ - 3}}{2}]
\end{array}\]

You wouldn't use brackets in the last line of that quote box.




x = 0, 3/2, -3/2



x = 0 or 3/2 or -3/2



x belongs to the set {0, 3/2, -3/2}.

 

[HallsofIvy]

Notice the difference between saying "x= [0, 3/2, -3/2]" or even "x= {0, 3/2, -3/2}" (which are wrong) and "x belongs to {0, 3/2, -3/2}" (which is correct). x is a number, not a set.

 

[Dale10101]

but

Notice the difference between saying "x= [0, 3/2, -3/2]" or even "x= {0, 3/2, -3/2}" (which are wrong) and "x belongs to {0, 3/2, -3/2}" (which is correct). x is a number, not a set.

not that I originally thought about it ... but in this case are we not actually saying that all three points are going to blow-up the original expression so that we are not indicating one or the other but rather all three as the set of points x that must be excluded from the domain?

I do take your point about saying a root of of the polynomial is 0 or 3/2 or -3/2 ...

wait, maybe I don't understand.

Had I written x = (0, 3/2, -3/2) that would be wrong because that would be a listing of the coordinates of a point. So is the problem with "x= [0, 3/2, -3/2]" that in listing roots of a polynomial it is the convention think of x as a set whose member are 0, 3/2, -3/2, rather than a list of discrete values that x might assume. Or is it that using brackets or braces are alternate forms of parentheses.

... to find the limit of point so to speak.