What is the Point Discontinuity Problem in Rational Functions?

[neuro.akn]

Homework Statement

Find the value of p at which the discontinuity would occur.

Homework Equations

f(x) = x^2 - 6x + 9 / x - p

The Attempt at a Solution

Able to solve if p has an assigned numerical value, but help is needed for determining the value of p at which the discontinuity would occur. Any help is appreciated. Thank you.

 

[pasmith]

Factorize $x^2 - 6x + 9$.

 

[neuro.akn]

Okay. Thanks; once I have done that, how would I solve for p?

 

[pasmith]

If you haven't yet found the answer to your other problem, I would suggest you do that and then come back to this problem, at which point the solution will hopefully be obvious.

In any event I would recommend re-reading the material on which these problems are based, since you don't seem to be fully comfortable with it.

 

[lendav_rott]

A discontinuity occurs when you cannot determine the function value at a certain argument(x) value. There's an operation only Chuck Norris can do..or so they say.

X-p , you know that there is one value that it can't have since it's in the denominator.
You also know that the function at 1st glance Could be 0 when X=?? according to the numerator.

What's the 1 value that cannot be P? Everything else can. Think of a hyperbole.

 

[Ray Vickson]

Homework Statement

Find the value of p at which the discontinuity would occur.

Homework Equations

f(x) = x^2 - 6x + 9 / x - p

The Attempt at a Solution

Able to solve if p has an assigned numerical value, but help is needed for determining the value of p at which the discontinuity would occur. Any help is appreciated. Thank you.

As in your other posting: you need parentheses! If I read what you wrote using standard rules for mathematical expressions, I would see
$$f(x) = x^2 - 6x + \frac{9}{x} - p.$$

 

[neuro.akn]

Ray Vickson, (x^2 - 6x + 9) / (x - p)
Thank you all for your help.

 

[neuro.akn]

The thing is, we have not gone over this material at all.

 

[neuro.akn]

I have solved the problem. Thank you everyone.