- #1

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However, is it merely just looking for a term that cancels out, or is it more specifically a term that cancels out

*makes the numerator equal to zero?*

__and__Thanks in advance.

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- Thread starter hgducharme
- Start date

- #1

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However, is it merely just looking for a term that cancels out, or is it more specifically a term that cancels out

Thanks in advance.

- #2

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- #3

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Something like this

Edit: Actually, this picture might have just answered my question.

So we have the equation: [itex]\frac{x^2-1} {x-1}[/itex]

which reduces to: [itex]\frac{(x+1)(x-1)} {(x-1)}[/itex]

The [itex](x-1)[/itex] terms will both cancel out, but that still leaves the numerator as a non-zero value. Thus, maybe it's merely just a term that cancels out that causes a hole. In this case, the term is [itex](x-1) = (x = 1)[/itex] which corresponds with the hole in the graph at [itex]x = 1[/itex]

Last edited:

- #4

Stephen Tashi

Science Advisor

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One can define funtions in complicated ways using if...then rules as well as algebraic expressions. Suppose we only consider a function defined by the ratio of polynomial factors. (i.e. its is a single fraction , not a sum of several fractions).

Functions like [itex] f(x) =\frac{(x^2 + 1) (2x + 3)} { (x^2 + 1)} [/itex] don't have a hole in their graph because (in the real number system) there is no value of [itex] x [/itex] that would make the denominator zero. So the fact that the numerator and denominator have a common factor does not always imply the function has a hole in its graph. Looking for terms that cancel out, doesn't automatically locate a hole in the graph.

By contrast, the graph of the function [itex] g(x) = \frac{(x^2 -1 )(2x + 3)}{(x^2 - 1) } [/itex] has a holes when [itex] x = 1 [/itex] and [itex] x = -1 [/itex].

The graph of the function [itex] h(x) = \frac{ 6 + (x^2 )}{(x^2-1) } [/itex] doesn't exist at the values [itex] x = 1 [/itex] and [itex] x = -1 [/itex] because those values make the denominator zero. Since [itex] x^2 -1 [/itex] is not a common factor, the fraction cannot be reduced. The graph does not have a hole at any finite point (x,y). Values of [itex] x [/itex] that make the denominator zero cause the graph not to exist, even if there are no common factors in the fraction.

- #5

- 14

- 0

One can define funtions in complicated ways using if...then rules as well as algebraic expressions. Suppose we only consider a function defined by the ratio of polynomial factors. (i.e. its is a single fraction , not a sum of several fractions).

Functions like [itex] f(x) =\frac{(x^2 + 1) (2x + 3)} { (x^2 + 1)} [/itex] don't have a hole in their graph because (in the real number system) there is no value of [itex] x [/itex] that would make the denominator zero. So the fact that the numerator and denominator have a common factor does not always imply the function has a hole in its graph. Looking for terms that cancel out, doesn't automatically locate a hole in the graph.

By contrast, the graph of the function [itex] g(x) = \frac{(x^2 -1 )(2x + 3)}{(x^2 - 1) } [/itex] has a holes when [itex] x = 1 [/itex] and [itex] x = -1 [/itex].

The graph of the function [itex] h(x) = \frac{ 6 + (x^2 )}{(x^2-1) } [/itex] doesn't exist at the values [itex] x = 1 [/itex] and [itex] x = -1 [/itex] because those values make the denominator zero. Since [itex] x^2 -1 [/itex] is not a common factor, the fraction cannot be reduced. The graph does not have a hole at any finite point (x,y). Values of [itex] x [/itex] that make the denominator zero cause the graph not to exist, even if there are no common factors in the fraction.

Thank you, this helped!

- #6

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