Stupid Teachers

[Dan Cardin]

The problem i was given something like this -

f(X) =
{x^2 + 3, x > 1
ax^2 + bx2, x < 1}

^that being a piecewise function.

That's not exactly what the problem was, but the important parts are still preserved. Now the question was "At which values of a and b is the function continuous?"

Naturally, my response is, "x is never equal to 1, therefore the function can never be continuous, regardless of the values of a and b."

Now, in my definition that I was given for "continuous function", it went something like the dictionary definition of
Of or relating to a line or curve that extends without a break or irregularity.Obviously it has a break. But she tells me
But the function is continuous elsewhere and I respond with my above arguement, so she tells me
just because the function doesnt specify a value at x =1 , doesnt mean that the graph doesnt exist there

Well, i still completely disagree, can anyone tell me why I am wrong?

[3Garrett3]

X doesn't have to be 1, but say x is .5 and a is 2. It would make the overall = 1.

It wont fit in the equations she gave you, but I'm quite sure that is how it works.

[Cartmann]

you need 3 lines in order to get to the 3 unknown numbers. You only have 2

[Runaway]

you need 3 lines in order to get to the 3 unknown numbers. You only have 2

@Cartmann: ...what?

anyways.

seeing as the line would be drawn with open-ended circles on 1, a piecewise function like you mentioned would technically be continuous. It depends on how you look at it. There might not be a value for x = 1, but there are values for x = .99 and x = 1.01. Therefore, when the graph is drawn it is continuous, but through algebra it doesn't necessarily have to be continuous.

[Dan Cardin]

X doesn't have to be 1, but say x is .5 and a is 2. It would make the overall = 1.

It wont fit in the equations she gave you, but I'm quite sure that is how it works.

that would make y = 1.

As for the drawing of it, it was meant to be algebraic.

[mixster]

Is it possibly a case of 0.9999... being equal to 1? Although theoretically never 1, x becomes infinitely closer to 1. One of those peculiar things people try to convince us are true despite being nonsense yet still making sense.

[Nava2]

Its a piecewise function, 1 is not in the domain of X therefore it does not exist.

Technically its a removable discontinuity, but its still discontinuous at x = 1.

The graph is not continuous at f(1) because f(1) does not exist. (Directly taken from my calc textbook.)

[senrath]

Is it possibly a case of 0.9999... being equal to 1? Although theoretically never 1, x becomes infinitely closer to 1. One of those peculiar things people try to convince us are true despite being nonsense yet still making sense.

Psst, 0.999... is exactly equal to 1. Nothing theoretical about it.

[Dan Cardin]

Its a piecewise function, 1 is not in the domain of X therefore it does not exist.

Technically its a removable discontinuity, but its still discontinuous at x = 1.

The graph is not continuous at f(1) because f(1) does not exist. (Directly taken from my calc textbook.)
The question didnt ask whether it is continuous at f(1), it asked where a and b made the function continuous. The above, sort of proves it not being a continuous function, but she didnt dispute that f(1) wasnt continuous, so i felt i should make the distinction.

[senrath]

The question didnt ask whether it is continuous at f(1), it asked where a and b made the function continuous. The above, sort of proves it not being a continuous function, but she didnt dispute that f(1) wasnt continuous, so i felt i should make the distinction.

While it's not continuous at 1, the notation indicates that 1 is not within the range, and is ignored completely. Therefore, whether or not it's continuous at 1 is irrelevant to the problem.

[Dan Cardin]

i dont see how! a continuous function is defined everywhere(continuous at every point).

[Nava2]

What you can say is that f(x) is continuous on intervals.

[Dan Cardin]

i dont think that would answer the question.

[boberman]

Dan, I'm going to have to side with you. The function is never continuous because 1 is undefined. No value of a or b will change that.