General

[quote=radkai]@Mathematician: Right ^_^[/quote]

Okay, Thanks. I think I'm getting this whole continuity thing now. UwU

@Mathematician: Right ^_^

[quote=radkai]Can't do ln 0 bro. x > -1. Point of discontinuity would be wrong in this case. How can it be discontinuous when it wasn't even continuous to being with? Better way of describing would be domain.[/quote]

So there would be no points of discontinuity right? Since its a continuous graph?

i dont know too bad

[quote=ChewyC]ln(x+1), x has to more than or equal to -1 cause you cant get negative ln.[/quote]

Can't do ln 0 bro. x > -1. Point of discontinuity would be wrong in this case. How can it be discontinuous when it wasn't even continuous to being with? Better way of describing would be domain.

Cot(x) is defined over spaced out multiples of (0<x<pi/2), but there's no way to write that in interval form. You can write out the [i]range[/i] of Cot(x) as (-∞,∞), but there's no way you can express the domain in interval form.

As for the second question:
[quote=ChewyC]ln(x+1), x has to more than or equal to -1 cause you cant get negative ln.[/quote]
He has it correct. ln(x+1) is asymptotic to the line x=-1, so there are points of discontinuity when x ≤ -1.

[quote=ImNoMerchant]Is this Calculus? or perhaps Pre-Calculus?[/quote]
It looks like algebra II to me.

Is this Calculus? or perhaps Pre-Calculus?

[quote=FTWJayChou]cot (x) = 1/tanx = cos x/ sin x
That would mean, cot (x) is discontinuous on the points where sin (x) = 0 because you can't have (cos x)/ 0.

I'm not sure how your teacher exactly wants it but it should be something similar to:

D: {xER || x =/= +/- n*pi}[/quote]

Oh haha silly me I put the wrong function. I meant to ask for the discontinuities of ln(x+1)
Also thanks for the notion thing.

lol ...... im in 10th grade