Redefining discontinous functions

[Ribbin]

I've had to redefine discontinuous functions before, but all of a sudden I get questions with more than just x as a variable, so I have no idea what to do with it. Help would be greatly appreciated, here it is:

Find the value of p for which f(x) would have a removable discontinuity and then define the functions so that f(x) = x^2 - 6x + 9 / (x - p) is continuous for all values of x.

I don't understand it. If you need to find how it's discontinuous and both of the numbers on the bottom are variables, how do I do that? So lost.

 

[stapel]

The "p" isn't a variable; it is some constant.

You have to figure out the value for "p", so that this rational function can be reduced to a linear function (expect for x = p), just as you did back in algebra, when you graphed certain rationals as straight lines except for the one hole.

You're just being asked to find the value of "x = p" where the hole in the line is. Factor, and figure out what you'd have to cancel, just like back in algebra.

Eliz.

P.S. Welcome to FreeMathHelp!

 

[Ribbin]

That's my main problem though. I'm not sure on how to factor it with the constants in it. All the questions I had for the homework didn't have constants and all of a sudden they throw this at me. If you could show me how I start to factor it I would be very grateful. Thanks for the help!

 

[stapel]

That's my main problem though. I'm not sure on how to factor it with the constants in it.

I'm not sure what you mean...? How would you factor x² - 6x + 9 without using the 6 and the 9...?

Or did you not learn how to factor quadratics back in algebra...?

Please reply with clarification. Thank you.

Eliz.

 

[Ribbin]

Well I understand you get (x - 3)(x - 3) on the top, but then what do I do at the bottom?

 

[stapel]

Well I understand you get (x - 3)(x - 3) on the top, but then what do I do at the bottom?

As suggested before, think back on when you graphed rational functions in algebra. There were times when a factor in the denominator "cancelled out", leaving you with a hole in the graph.

Now figure out what "p" should be to allow for this cancellation, leaving you with the graph of a straight line (except for at x = p). That is, figure out what the denominator should be, in order for the discontinuity (the zero in the denominator) to be removable (cancellable).

Eliz.