Just a quickie (1 Viewer)

[Raginsheep]

Do relations count as being continuous? So if i had x^2+y^2=4, is it continuous?

 

[Slidey]

Depends what kind of relation. Circles are continuous:
x^2+y^2=r^2.

Ellipses are.

I mean, you can extend differentiation to these things anyway.

 

[aud]

Found something in my Year 11 notes...

test for differentiability at a point
1. test for continuity
2. check .....blah

Then I have the graph y = |x|
A proof of continuity
A 'therefore, continuous at x = 0'
A proof (or disproof) of differentiability
A 'therefore, not differentiable'

So y=|x| is continuous at x=0... oops

 

[Raginsheep]

yeah.....its just that aud said that you can differentiate if the graph is continuous.

 

[aud]

yeah.....its just that aud said that you can differentiate if the graph is continuous.

? Oh... yea, that's the defition of differentiability... that, plus the point that f'(a) exists... you need both things to exist for the graph to be differentiable...? I'm lost... maybe I should just stop talking

 

[Raginsheep]

Nah, its just that you can only differentiate if its a function (for me anywayz ) and its continuous at that point.

You were going good before though

 

[Slidey]

Yea, that's what I thought, auth. Absolute values are indeed continuous.

 

[Archman]

oh ye, watch me differentiate x^2+y^2=4
2x + 2y*dy/dx = 0
dy/dx = -x/y

 

[Slidey]

I would add that since y=+sqrt(4-x^2),
dy/dx= +sqrt(4-x^2)/x

 

[Raginsheep]

Hey I said I couldn't do it and Im sure its not in the 3u course anywayz.

 

[mr EaZy]

ah ok
i did my history extn project on zero too
beleive me, you will never do dividing by zero

u really luv ur zeroes.
But Where did the Zero come from anyway?

 

[aud]

Hey I said I couldn't do it and Im sure its not in the 3u course anywayz.

Maths 2/3U Syllabus
8.1. Informal discussion of continuity
8.2 The notion of the limit of a function and the definition of continuity in terms of this notion. Continuity of f+g, f-g, fg in terms of continuity of f and g

So it looks like you need just a little knowledge... just says that "the behaviour of 1/x and |x|/x near the origin should be demonstrated, but discontinuities should not be further stressed"

 

[velox]

na 1/0 is not anything, not in maths i know anyway.

that means in any maths

 

[Xayma]

I would add that since y=+sqrt(4-x^2),
dy/dx= +sqrt(4-x^2)/x

dy/dx=± x/√(4-x²)

 

[Li0n]

u really luv ur zeroes.
But Where did the Zero come from anyway?

aparently the arabs were the first to define "zero" and before that time it was very puzzling to define 'nothing'.

 

[mojako]

f(x)=|x|/x is not continuous because you can't divide by zero, and absolute vale functions are not continuous.

So the two functions it is composed of a discontinuous:
y=|x|
w=1/x
f(x) is just y.w

I don't think anyone has corrected this.. but, absolute value is continuous!
continuous at x=a is when the following 3 values exist and are equal: (1) the value of f(a) (2) the limit of the value of f(x) as x->a^- (3) the limit of the value of f(x) as x->a⁺
----
EDIT: or more simply.. when u can sketch the curve around that point without lifting your pen off paper
----
y=|x| is continuos..
but there is a critical value at x=0. critical value is when the derivative is not defined.
having continuous function doesnt mean it must have derivative at all points.
but if it has derivative at a particular point, it must be continuous there.

It isn't continuous as

Because it changes from 1 for x>0 to -1, x<0.

But.. how can someone be expected to know that the two reasons are these two... well he can guess.. trying to make up two reasons because the Q asks for 2

 

[ngai]

Found something in my Year 11 notes...

test for differentiability at a point
1. test for continuity
2. check .....blah

Then I have the graph y = |x|
A proof of continuity
A 'therefore, continuous at x = 0'
A proof (or disproof) of differentiability
A 'therefore, not differentiable'

So y=|x| is continuous at x=0... oops

I don't think anyone has corrected this.. but, absolute value is continuous!

u thought wrong

 

[mojako]

u thought wrong

haha thanks ngai
ur very good at showing other people's mistakes, hey?

anyway I'll give the situation where critical point exists (taken off my ext2 notes):
-> if the tangent is vertical
-> at any point of discontinuity in the original function
-> at an endpoint of a finite domain (a special case of discontinuity), eg (0,0) in y=sqrt(x) or at x=1 in y=sin^-1(x) (all of which happen to have vertical tangents but if they didnt its still a critical point.. u can make functions like "f(x) = x^2 where x<=4" for example)
-> at sharp edges and cusps like in y=|x|

 

[ngai]

I cant think of any normal function which naturally has a finite domain...

sqrtx
logx

 

[mojako]

sqrtx
logx

sqrt(x) has vertical tangent x=0
log(x)... hmm... this one.. is a bit strange hahaha... I dont think this has an endpoint... (with my definition of endpoint anyway)