Simplification of a rational polynomial function

[Yh Hoo]

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??

 

[DonAntonio]

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 $\,y=b+c\,$ 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 $\,b+c=0\,$

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 $\,x=1\,,\,x=-1\,$ are vertical asymptotes: only the first

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

DonAntonio

So , is that simplification of a function is significant or not significant??

 

[Yh Hoo]

This is not accurate: the original function is not defined in a, whereas the function $\,y=b+c\,$ 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??

 

[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??

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

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

 

[BloodyFrozen]

The removable discontinuity mentioned by DonAntonio is a hole in a graph in simpler terms (taught in Alg II of PreCalc).

You can search it up or read about it at the link below.

http://www.purplemath.com/modules/grphrtnl4.htm