Continuous Extension to a Point (Factoring Question)

In summary, the problem is asking for a way to define f(6) so that the function is continuous at s=6. The solution involves finding the limit of the function as s approaches 6 and factoring out a linear term from the numerator and denominator to cancel out a zero-denominator.
  • #1
BraedenP
96
0

Homework Statement



Define f(6) in a way that extends
gif.latex?f(s)=\frac{s^3-216}{s^2-36}.gif
to be continuous at s=6.

Homework Equations



None. Only limits are required.

The Attempt at a Solution



In order to figure out which point needs to be added to the function, I have to find the limit of this function as s->6. This will result in a zero-denominator, however, so I need to factor [PLAIN]http://latex.codecogs.com/gif.latex?s^2-36 out of the numerator, so that I can cancel them and use the resulting polynomial to calculate the limit.

Basically, I just need help factoring [PLAIN]http://latex.codecogs.com/gif.latex?s^2-36 out of [URL]http://latex.codecogs.com/gif.latex?s^3-216;[/URL] I can solve the rest of the question easily.

Help would be appreciated.

Thanks,
Braeden
 
Last edited by a moderator:
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  • #2
BraedenP said:

Homework Statement



Define f(6) in a way that extends
gif.latex?f(s)=\frac{s^3-216}{s^2-36}.gif
to be continuous at s=6.

Homework Equations



None. Only limits are required.

The Attempt at a Solution



In order to figure out which point needs to be added to the function, I have to find the limit of this function as s->6. This will result in a zero-denominator, however, so I need to factor [PLAIN]http://latex.codecogs.com/gif.latex?s^2-36 out of the numerator, so that I can cancel them and use the resulting polynomial to calculate the limit.

Basically, I just need help factoring [PLAIN]http://latex.codecogs.com/gif.latex?s^2-36 out of [URL]http://latex.codecogs.com/gif.latex?s^3-216;[/URL] I can solve the rest of the question easily.
The numerator doesn't have a factor of s2 - 36. There is a linear factor that can be pulled out, however.
 
Last edited by a moderator:
  • #3
Mark44 said:
The numerator doesn't have a factor of s2 - 36. There is a linear factor that can be pulled out, however.

Yeah, I figured it out, thanks. I'm not quite sure what I was trying to do. I ended up factoring (x-6) out of the numerator and denominator and cancelling those.
 
  • #4
Hopefully, you factored out s - 6 rather than x - 6.
 
  • #5
Mark44 said:
Hopefully, you factored out s - 6 rather than x - 6.

Oh yeah.. sorry. I factored (s-6)
 

FAQ: Continuous Extension to a Point (Factoring Question)

What is Continuous Extension to a Point?

Continuous Extension to a Point, also known as Factoring, is a mathematical process that involves breaking down a polynomial equation into smaller parts. This allows for easier analysis and solving of the equation.

Why is Continuous Extension to a Point important?

Continuous Extension to a Point is important because it allows for the simplification of complex polynomial equations, making them easier to solve. It is also used in various fields of science, such as physics and engineering, for modeling and analyzing real-world problems.

How do you perform Continuous Extension to a Point?

To perform Continuous Extension to a Point, you need to first identify common factors in the polynomial equation. Then, use the distributive property to break down the equation into smaller parts. Continue this process until the equation is fully factored.

What are the benefits of using Continuous Extension to a Point?

Using Continuous Extension to a Point can help simplify complex equations and make them easier to solve. It also allows for a better understanding of the relationships between the variables in the equation. Additionally, factoring can help identify the roots of an equation and find solutions to real-world problems.

Are there any limitations to Continuous Extension to a Point?

While Continuous Extension to a Point is a useful tool, it may not always be possible to fully factor an equation. Some equations may not have any common factors or may have factors that are not easily identifiable. In these cases, other methods may need to be used to solve the equation.

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