Finding the domain of rational expressions HELP!! PLEASE!!!! #DESPERATE

dancer6266

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Jan 7, 2013
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I cannot figure out how to find the domain of this specific problem: 5x-7
------
4
I know you're supposed to make the denominator equal to 0, but 0 doesn't = 4 so is it undefined? Any help would be appreciated! Thanks:D
 
dancer6266,

Note that \(\displaystyle y=\frac{5x-7}{4}\) is a monotonically increasing function.

The corresponding inverse function
\(\displaystyle x=\frac{4y+7}{5}\) is also monotonically increasing.

For every real value 'x', there is a corresponding real value y and vice-versa.

Hence the domain (x values) is the set of real numbers.

Cheers,
Sai.


 
I cannot figure out how to find the domain of this specific problem: 5x-7
------
4
I know you're supposed to make the denominator equal to 0, but 0 doesn't = 4 so is it undefined? Any help would be appreciated! Thanks:D
Internet explorer does not respect spaces so it would be better to write something like (5x- 7)/4. Even better would be f(x)= (5x- 7)/4.

Please, please, do not just memorize formulas or methods without understanding them! You are "supposed to make the denominator equal to 0" to do what? "so is it undefined?". Is what "undefined"? You are apparently asked to find the domain of this functions which means all points at which the function is defined.

You are "supposed to make the denominator equal to 0" because you cannot divide by 0- if, for some x, the denominator is 0, you cannot do that division and so the function is not defined there. That point is not in the domain.

Yes, for this function, there is NO value of x that makes the denominator 4. That means that the function is defined for all x.

(Because the denominator does NOT depend on x, this is not, strictly speaking, a "rational function". It is, rather, a linear polynomial.)
 
Last edited:
Internet explorer does not respect spaces so it would be better to write something like (5x- 7)/4. Even better would be f(x)= (5x- 7)/4.


You are "supposed to make the denominator equal to 0" because you cannot divide by 0- if, for some x, the denominator is 0,
you cannot do that division and so the function is not defined there. That point is not in the domain.

lookagain's point #1) Yes, for this function, there is NO value of x that makes the denominator 4.
That means that the function is defined for all x.


(Because the denominator does NOT depend on x, this is not, strictly speaking, a "rational function".
lookagain's point #2)It is, rather, a linear polynomial.)


Point #1) For that function, because there is no value of x that makes
the denominator 0 (as opposed to 4), then that function is defined for all real x.


Otherwise, look at these two contrasting examples:


\(\displaystyle g(x) \ = \ \dfrac{5x - 7}{x + 4}\)

x = 0 makes the denominator 4, and this function is not defined for all real x.


versus


\(\displaystyle h(x) \ = \ \dfrac{5x - 7}{x^2 + 4}, \)

x = 0 makes the denominator 4, but this function is defined for all real x.


- - - - - -- - - - - -- - -


Point # 2)

dancer6266,

know that if you are given any function equal to a polynomial solely in terms of the variable x
(and that may include any non-zero constant terms), then that function is defined for all real x.
 
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