Maths Methods 3/4 Help Thread 2011

[b^3]

Yes smooth is the same gradient on either side of the point that is between the two functions that make up the hybrid fucntion.
A continuous function has no breaks in it, i.e. there is no open circles in it's domain.

e.g.
This is continous since there is not a 'break' at x=0

This is not continous since there is a 'break' at x=0.

[dc302]

could someone please help clarify --> so if a question states that the hybrid function is
a) continuous
b) smooth
 
what these indicate, I know "smooth" means same gradient? I think I came across a question that had smooth but not continuous?

Note that a smooth function is always continuous on its domain.

[dc302]

Yes smooth is the same gradient on either side of the point that is between the two functions that make up the hybrid fucntion.
A continuous function has no breaks in it, i.e. there is no open circles in it's domain.

e.g.
This is continous since there is not a 'break' at x=0

This is not continous since there is a 'break' at x=0.

The second example is in fact continuous on its domain. I think you are confused with a 'break' in the domain with a 'break' in the range. A discontinuous function is something like:

f(x) =   1 for x>0
            0 for x=0
            -1 for x<0

Then the discontinuity is at x = 0.

[abeybaby]

definition of continuity:

f is continuous at x=a if:
 lim       exists and is equal to f(a).  that is, f is approachable from the left, and from the right of x=a, and the same value is attained from both the
x->a     left and right limits, and is equal to f(a).

'smooth' means that the gradient as approached from the left of x=a is equal to the gradient as approached from the right of x=a.
         

[b^3]

definition of continuity:

f is continuous at x=a if:
 lim       exists and is equal to f(a).  that is, f is approachable from the left, and from the right of x=a, and the same value is attained from both the
x->a     left and right limits, and is equal to f(a).

'smooth' means that the gradient as approached from the left of x=a is equal to the gradient as approached from the right of x=a.

I just want to double check this, isn't there three conditions that must be satisfied for it no be continuous, not two?
From Methods Essentials textbook 1/2 (keep in mind the textbook isn't always right, its just "right" at vce level).

A function is continuous at the point x=a if the following three condiionts are met:
1. is defined at
2. exsits
3.

Now the second two conditions are met but my point here is that we don't have a f(a) for the limits to be equal to do we, since it is defined for x>0, x<0 not x=0? I may be wrong, I just want to make sure though.

Also your example is a better example, it shows it more clearly.

[dc302]

definition of continuity:

f is continuous at x=a if:
 lim       exists and is equal to f(a).  that is, f is approachable from the left, and from the right of x=a, and the same value is attained from both the
x->a     left and right limits, and is equal to f(a).

'smooth' means that the gradient as approached from the left of x=a is equal to the gradient as approached from the right of x=a.

I just want to double check this, isn't there three conditions that must be satisfied for it no be continuous, not two?
From Methods Essentials textbook 1/2 (keep in mind the textbook isn't always right, its just "right" at vce level).

A function is continuous at the point x=a if the following three condiionts are met:
1. is defined at
2. exsits
3.

Now the second two conditions are met but my point here is that we don't have a f(a) for the limits to be equal to do we, since it is defined for x>0, x<0 not x=0? I may be wrong, I just want to make sure though.

Also your example is a better example, it shows it more clearly.

Yes, your example has a discontinuous domain, but it is actually still a continuous function, since those conditions are met everywhere where f(x) is defined.


edit: in fact, the 2 conditions abes said are enough, since asserting that f(a) = lim f(x) also asserts that f(a) exists (which is what 1 says). So you only need conditions 2, and 3. Keep in mind this definition is only for functions from the real line to the real line though.

[b^3]

definition of continuity:

f is continuous at x=a if:
 lim       exists and is equal to f(a).  that is, f is approachable from the left, and from the right of x=a, and the same value is attained from both the
x->a     left and right limits, and is equal to f(a).

'smooth' means that the gradient as approached from the left of x=a is equal to the gradient as approached from the right of x=a.

I just want to double check this, isn't there three conditions that must be satisfied for it no be continuous, not two?
From Methods Essentials textbook 1/2 (keep in mind the textbook isn't always right, its just "right" at vce level).

A function is continuous at the point x=a if the following three condiionts are met:
1. is defined at
2. exsits
3.

Now the second two conditions are met but my point here is that we don't have a f(a) for the limits to be equal to do we, since it is defined for x>0, x<0 not x=0? I may be wrong, I just want to make sure though.

Also your example is a better example, it shows it more clearly.

Yes, your example has a discontinuous domain, but it is actually still a continuous function, since those conditions are met everywhere where f(x) is defined.

Ok thanks, that makes sense now.

[dc302]

Damn you beat me, but i wanted to add that:

In fact, the 2 conditions abes said are enough, since asserting that f(a) = lim f(x) also asserts that f(a) exists (which is what 1 says). So you only need conditions 2, and 3. Keep in mind this definition is only for functions from the real line to the real line though

[dc302]

edit: double posted..

[nacho]

with transition matrices, if you want to find the probability of soemthig occurring n days later/n weeks later
u just set T to T^(n-1)
Correct?

[Greatness]

with transition matrices, if you want to find the probability of soemthig occurring n days later/n weeks later
u just set T to T^(n-1)
Correct?

Yep when you have a given condition.

[luken93]

Quick question, what's the best way to set out domain of a function?

1) Df = (-1, 2) u (2, inf)
2) x c (-1, 2) u (2, inf)
3) {x: x c (-1, 2) u (2, inf) }

Just don't wanna get tripped up on this sort of thing!

[burbs]

Surely those are all acceptable considering VCAA themselves alternate between them?

[nacho]

Quick question, what's the best way to set out domain of a function?

1) Df = (-1, 2) u (2, inf)
2) x c (-1, 2) u (2, inf)
3) {x: x c (-1, 2) u (2, inf) }

Just don't wanna get tripped up on this sort of thing!

i use option 1 or 2, 3 is ... i dont see why one would use that, but it's good to familiarise yourself with it, i guess.

[dc302]

Quick question, what's the best way to set out domain of a function?

1) Df = (-1, 2) u (2, inf)
2) x c (-1, 2) u (2, inf)
3) {x: x c (-1, 2) u (2, inf) }

Just don't wanna get tripped up on this sort of thing!

i use option 1 or 2, 3 is ... i dont see why one would use that, but it's good to familiarise yourself with it, i guess.

Technically, the only correct solution in those 3 options, is option 3, because 3 gives you a set (which is what a domain is), whereas 1 gives you a condition on no variable, and 2 gives you only 1 number (the number x). However, I think VCAA accepts them all anyway so it shouldn't matter.

edit: also, I assume your 'c's are meant to be epsilons? Or whatever you call the E like symbol.. haha