calculus problem: For which non-negative values of a does the function y=(x^2+3x)/x+a have a clear discontinuity point?

[NirMoyal]

Hey,
I know the question is supposed to be quite simple but I'm getting confused with it and can't understand.
I would appreciate your help please, thanks in advance.

For which non-negative values of a does the function y=(x^2+3x)/x+a have a clear discontinuity point?

 

[stapel]

I know the question is supposed to be quite simple but I'm getting confused with it and can't understand.
I would appreciate your help please, thanks in advance.

For which non-negative values of a does the function y=(x^2+3x)/x+a have a clear discontinuity point?

What you have posted means the following:

[imath]\qquad y = \dfrac{x^2 + 3x}{x} + a[/imath]

Was this what you meant? Because, if so, I can understand your confusion; the value of [imath]a[/imath] has nothing to do with any discontinuity of the function. If not, what *did* you mean?

When you reply, please clarify at what point you are becoming confused. Thank you!

 

[mario99]

If [imath]\displaystyle y = \frac{x^2 + 3x}{x + a}[/imath], then they want you to factor the numerator and cancel some of it with the denominator.


[imath]\displaystyle y = \frac{x^2 + 3x}{x + a} = \frac{x(x + 3)}{x + a} = x[/imath]

This can be true only and only when there is a discontinuity. Then, the value of [imath]a = \ ?[/imath]

 

[Dr.Peterson]

Hey,
I know the question is supposed to be quite simple but I'm getting confused with it and can't understand.
I would appreciate your help please, thanks in advance.

For which non-negative values of a does the function y=(x^2+3x)/x+a have a clear discontinuity point?

@mario99's advice provides a good hint, assuming that what "clear discontinuity point" means is a point at which there is a discontinuity for a "bigger" reason (more visible) than a mere "hole" -- a jump or infinite discontinuity. Do you understand these concepts? That could be your difficulty.

(I don't find "clear discontinuity" to be a standard technical term, but I can easily imagine a teacher using it. Since there is a discontinuity somewhere for any value of a, I think this is what they mean.)

But in fact, there are two special values of a, not just the one he implied, and they do not produce "clear discontinuities" in this sense. You'll have to turn things around a little ...

Of course, if our understanding is correct, then you should have typed y=(x^2+3x)/(x+a).