Simplification of a rational polynomial function

In summary, the conversation discusses simplification of rational functions and its significance. It is shown that while simplification may seem insignificant, it can actually change the definition set and number of solutions of a function. The concept of removable discontinuity is also mentioned, which refers to a hole in a graph caused by a simplified function.
  • #1
Yh Hoo
73
0
Now if I have a function y=(ab+ac)/a, it can be further factorised, y=(a(b+c))/a. Now if we cancel off the a, we will have only y=b+c that will also give the same y-values as the original form of the function y with respect to the same x-value. This statement implies that cancellation or simplification of a rational function is NOT SIGNIFICANT.
However, when we let the y-value equal to 0, for the simplified form,y=b+c, we have only one solution, (b+c)=0. For the original form, y=(a(b+c))/a, we can have 2 solutions ,(a)(b+c)=0,so (a)=0 or (b+c)=0. This shows that simplification of a function is SIGNIFICANT!
Moreover, when we have a function, y=((x+1))/((x+1)(x-1)). Apparently, we can see that this rational polynomial function have 2 vertical asymptotes: x=1 and x=-1 in which at this line, the y-value is undefined! Now if we simplify the function to y=1/(x-1), when x=-1,y=-1/2 which is a definite value!
So , is that simplification of a function is significant or not significant??
 
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  • #2
Yh Hoo said:
Now if I have a function y=(ab+ac)/a, it can be further factorised, y=(a(b+c))/a. Now if we cancel off the a, we will have only y=b+c that will also give the same y-values as the original form of the function y with respect to the same x-value. This statement implies that cancellation or simplification of a rational function is NOT SIGNIFICANT.



This is not accurate: the original function is not defined in a, whereas the function [itex]\,y=b+c\,[/itex] is, thus they both are different functions as they don't have the same definition set.




However, when we let the y-value equal to 0, for the simplified form,y=b+c, we have only one solution, (b+c)=0. For the original form, y=(a(b+c))/a, we can have 2 solutions ,(a)(b+c)=0,so (a)=0 or (b+c)=0. This shows that simplification of a function is SIGNIFICANT!



This is wrong: a = 0 cannot be argument of the original function as was seen above. The only solution in both cases is [itex]\,b+c=0\,[/itex]





Moreover, when we have a function, y=((x+1))/((x+1)(x-1)). Apparently, we can see that this rational polynomial function have 2 vertical asymptotes: x=1 and x=-1 in which at this line, the y-value is undefined! Now if we simplify the function to y=1/(x-1), when x=-1,y=-1/2 which is a definite value! [\QUOTE]




Again, much more care is required here: it is not true that both [itex]\,x=1\,,\,x=-1\,[/itex] are vertical asymptotes: only the first

one is. The other value [itex]\,x=-1\,[/itex] is what's called a removable discontinuity , not an asymptote.

DonAntonio


So , is that simplification of a function is significant or not significant??
 
  • #3
DonAntonio said:
This is not accurate: the original function is not defined in a, whereas the function [itex]\,y=b+c\,[/itex] is, thus they both are different functions as they don't have the same definition set.

DonAntonio

Thanks for your reply. Now, Let’s look at this function again. y=((x+1))/((x+1)(x-1)) . Aren’t this function can be simplified by multiplying both the numerator and denominator by 1/((x+1) ) ?? So this y=((x+1))/((x+1)(x-1)) = 1/(x-1). Is it true?? Are this 2 function equal and equivalent? or just equivalent but not equal??

And what is actually mean by"removable discontinuity" and what is the principle of that?? For the function y=((x+1))/((x+1)(x-1)), when the x-value are very very close to -1 ,but not yet equal to-1, it can either be -1.0000...000...01 or -0.999...999...999, the function will give a y-value that is approximately equal to -0.5, but at exactly x=-1, the function is undefined! now what should be plotted on the graph of y=((x+1))/((x+1)(x-1)) when x=-1?? and sudden broken on the functional curve??
 
  • #4
Yh Hoo said:
Thanks for your reply. Now, Let’s look at this function again. y=((x+1))/((x+1)(x-1)) . Aren’t this function can be simplified by multiplying both the numerator and denominator by 1/((x+1) ) ?? So this y=((x+1))/((x+1)(x-1)) = 1/(x-1). Is it true?? Are this 2 function equal and equivalent? or just equivalent but not equal??



Ok, what is true is [itex]\displaystyle{\,y=\frac{x+1}{(x+1)(x-1)}=\left\{\begin{array}{cc}\frac{1}{x-1}\,&\,,x\neq -1\\UNDEFINED\,&\,,x=-1\end{array}\right.}[/itex]



And what is actually mean by"removable discontinuity" and what is the principle of that??



If you haven't yet studied then your confusion is understandable so wait a little.





For the function y=((x+1))/((x+1)(x-1)), when the x-value are very very close to -1 ,but not yet equal to-1, it can either be -1.0000...000...01 or -0.999...999...999, the function will give a y-value that is approximately equal to -0.5, but at exactly x=-1, the function is undefined! now what should be plotted on the graph of y=((x+1))/((x+1)(x-1)) when x=-1?? and sudden broken on the functional curve??

DonAntonio
 
  • #5

What is a rational polynomial function?

A rational polynomial function is a mathematical expression that can be written as the ratio of two polynomials, where the numerator and denominator are both polynomials with real coefficients.

Why is simplification of a rational polynomial function important?

Simplification of a rational polynomial function is important because it allows us to reduce a complex expression into a simpler form, making it easier to work with and understand.

How do you simplify a rational polynomial function?

To simplify a rational polynomial function, we use the rules of algebra to combine like terms, factor out common factors, and cancel out common factors in the numerator and denominator.

What are the steps to simplify a rational polynomial function?

The steps to simplify a rational polynomial function are:1. Factor both the numerator and denominator2. Cancel out common factors in the numerator and denominator3. Simplify any remaining expressions in the numerator and denominator4. Write the simplified expression as a single fraction

What are some common mistakes to avoid when simplifying a rational polynomial function?

Some common mistakes to avoid when simplifying a rational polynomial function include forgetting to factor out common factors, cancelling out incorrect terms, and not simplifying the expression fully. It is important to double check your work and simplify as much as possible to avoid any errors.

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