Math help (1 Viewer)

[Carrotsticks]

I see limits as an arm wrestle, or as two people pushing either side of a box.

Works like a charm and anybody can understand it.

 

[seanieg89]

I see limits as an arm wrestle, or as two people pushing either side of a box.

Works like a charm and anybody can understand it.

Haha I have to admit I don't understand limits when you put it like that.

 

[JINOUGA]

I see limits as an arm wrestle, or as two people pushing either side of a box.

Works like a charm and anybody can understand it.

carrotsticks i appreciate you are an excellent mathematician but lol that was a terrible example imo

 

[D94]

Re: Math help (Because Carrot made me do this)

How do you do that?

This:



The manipulation is subtracting 4 then adding 4. By adding/subtracting 4 I haven't changed the overall value of the function. (x + 6 - 6) = x = (x - 2 + 2). The value is still x.

This is probably getting confusing, so just stick to polynomial long division if you know it. If you don't then I can't see why a tutor would give you that question.

How do you know WHEN to use polynomial division?

It depends on the question - there's no rule stating you have to use long division. Generally, when the numerator is larger than the denominator, you should attempt to divide it to find the oblique asymptote (which again, IDK why you're being taught this or have been given a question which requires it!!!).

 

[Carrotsticks]

Haha this is how I see it, and I hope you can understand. Best explained irl though, as opposed to typed.

Suppose I have some function (for the sake of sticking in the Preliminary course, I will assume that f(x) and g(x) are polynomials).



I see it to be an arm wrestle between f(x) and g(x).

As x -> infinity, f(x) tries to pull the whole function towards infinity (assuming it's a positive polynomial ofc).

As x -> infinity, g(x) approaches infinity too. However, since he's in the bottom (so a little upset by that), he decides to troll by pulling the whole function down towards 0.

So now the question is.... who wins? f(x) or g(x)?

Suppose f(x) is some really beast guy (so a polynomial of say degree 3) and g(x) is some guy who's a little bit weaker (say a polynomial of degree 2).

Then f(x) will just destroy g(x) and will win the arm wrestle, hence the overall function approaches infinity.

However suppose f(x) is some wimp (linear function) and g(x) is a gym junkie (quartic). Then g(x) will win the arm wrestle and hence he will succeed, and the overall function will approach 0.

Now suppose f(x) and g(x) are around the same strength (both cubics), but f(x) is just a TINY bit stronger than g(x). Then they will both tie in the arm wrestle (constant limit), but f(x) will be a tiny tiny bit ahead, because he's a little bit stronger.

So suppose the limit is y=1. This means f(x) is EXACTLY the same strength as g(x) but suppose the limit is y=0.5. This means g(x) is a tiny bit stronger than f(x), but they are around the same power level (obligatory over 9000).

===============================

This mentality can be applied to other functions too apart from polynomials.

So suppose I have:



As x -> infinity, ln(x) tries to pull the function towards infinity too (though he's a pretty weak guy).

e^x on the other hand is an ABSOLUTE GUN and just destroys ln(x).

So the limit is 0, since e^x is a clear winner.

 

[deswa1]

^That's not bad actually- its pretty original. Obviously the gym junkie quartic was me and you were the wimp linear function right?

 

[seanieg89]

Oh right, you were talking about asymptotics. When you said limits = armwrestles I was thinking you were using the word in its most general sense...i.e. convergence.

In terms of which functions dominate others, an armwrestle IS a good way to think of things.

As an exercise in this concept, try to order the following contestants by how they would far in an armwrestling tournament:

x, 0.1x^2, 2x^{0.1}, log(x), sin(x), log(x^2), e^x, log(x)^2, 0.00000001x^5.

 

[Sy123]

Re: Math help (Because Carrot made me do this)

How do you know WHEN to use polynomial division?

If you still remember our 'formula' I posted to calculate horizontal asymptotes. I left out what happens when n > m because that is covered in 4U.

Basically when n > m you must use polynomial division (or otherwise) to find the oblique asymptote.

Take:



When we apply polynomial division, we divide by the bottom in order for us to get:



Where Q(x) is the quotient (the one you get on top of the polynomial division thing), and R(x) is the remainder.
Lets divide both sides by g(x)



Now lets take the limit to infinity of both sides in order to get our asymptote


We know that the degree of g(x) is higher than R(x), so when we take the limit to infinity of this, it converges to zero.
Such that


This means that, that taking the limit to infinity of the f(x)/g(x) (the one are trying to sketch), is the same as the limit to infintiy of the quotient when you divide f(x) by g(x).

Take the above example where the oblique asymptote is x+2.
When x=100, the oblique will be 100+2
When x=10000, the oblique will be 10000+2

In very essence, the function is approaching Q(x) which is our oblique asymptote.

So it's like a rainbow right? because you can see a rainbow but you can never 'touch' it or something.

Whatever works for you, your analogy doesnt seem to have logical flaws (but remember some asymptotes CAN be crossed, only horizontal/oblique ones CAN (doesnt mean they have to))

what's that? :s

The function as it approaches infinity, it approaches the asymptote closer and closer. For something to display asymptote behaviour, a function must approach it at infinity.

 

[Fawun]

Re: Math help (Because Carrot made me do this)

This:



The manipulation is subtracting 4 then adding 4. By adding/subtracting 4 I haven't changed the overall value of the function. (x + 6 - 6) = x = (x - 2 + 2). The value is still x.

This is probably getting confusing, so just stick to polynomial long division if you know it. If you don't then I can't see why a tutor would give you that question.


It depends on the question - there's no rule stating you have to use long division. Generally, when the numerator is larger than the denominator, you should attempt to divide it to find the oblique asymptote (which again, IDK why you're being taught this or have been given a question which requires it!!!).

I have no idea why she gave me this question but okay thanks! Your way of manipulating it is confusing me so i'm just going to stick with polynomial long division (which I can do btw)

Oh right, you were talking about asymptotics. When you said limits = armwrestles I was thinking you were using the word in its most general sense...i.e. convergence.

In terms of which functions dominate others, an armwrestle IS a good way to think of things.

As an exercise in this concept, try to order the following contestants by how they would far in an armwrestling tournament:

x, 0.1x^2, 2x^{0.1}, log(x), sin(x), log(x^2), e^x, log(x)^2, 0.00000001x^5.

??

If you still remember our 'formula' I posted to calculate horizontal asymptotes. I left out what happens when n > m because that is covered in 4U.

Basically when n > m you must use polynomial division (or otherwise) to find the oblique asymptote.

Take:



When we apply polynomial division, we divide by the bottom in order for us to get:



Where Q(x) is the quotient (the one you get on top of the polynomial division thing), and R(x) is the remainder.
Lets divide both sides by g(x)



Now lets take the limit to infinity of both sides in order to get our asymptote


We know that the degree of g(x) is higher than R(x), so when we take the limit to infinity of this, it converges to zero.
Such that


This means that, that taking the limit to infinity of the f(x)/g(x) (the one are trying to sketch), is the same as the limit to infintiy of the quotient when you divide f(x) by g(x).

Take the above example where the oblique asymptote is x+2.
When x=100, the oblique will be 100+2
When x=10000, the oblique will be 10000+2

In very essence, the function is approaching Q(x) which is our oblique asymptote.



Whatever works for you, your analogy doesnt seem to have logical flaws (but remember some asymptotes CAN be crossed, only horizontal/oblique ones CAN (doesnt mean they have to))



The function as it approaches infinity, it approaches the asymptote closer and closer. For something to display asymptote behaviour, a function must approach it at infinity.

Okay thanks! I kinda have an idea of how this works. So if I were to do this question:

Question 5A,

Is this correct so far?







Therefore it's an even function.

Vertical asymptote:





No Solution? is this right?

Horizontal Asymptote:









x-intercepts:





y-intercept:





So how do I graph it?

Is this the part where I sub in numbers?

 

[JINOUGA]

Re-check ur x intercept

 

[Sy123]

Re: Math help (Because Carrot made me do this)

Okay thanks! I kinda have an idea of how this works. So if I were to do this question:

Question 5A,

Is this correct so far?







Therefore it's an even function.

Vertical asymptote:





No Solution? is this right?

Horizontal Asymptote:









x-intercepts:





y-intercept:





So how do I graph it?

Is this the part where I sub in numbers?

Yes that is exactly right, since we have no vertical asymptote however, there are no points to test that are around the vertical asymptote (because there are none).
We have already 'subbed' in infinity (you found the limit).

Remember earlier how I said safe path/smart path?

Safe Path (doesnt take much longer at all)
Lets take the limit of negative infinity, we find that we get the same y=0.

Smart Path:
Recognise that it is an even function, therefore symterical about the y-axis, therefore the limit to negative infintiy should be the same as the one to infinity.

We have some pieces of information:
Horizontal asymptote y=0.
At infinity it approaches our asymptote
At negative infinity it approaches our asymptote
Our graph goes through (0,1) (intercept)
Our graph is symetrical

Now assume nothing crazy happens, do whatever you think you have to do in order to get the graph from one asymptote up and through the intercept down again back to the asymptote, you will find that you are quite accurate.

EDIT: It does not have ANY x-intercept, because if y=0
You will find, 0=1

 

[Fawun]

Re: Math help (Because Carrot made me do this)

Yes that is exactly right, since we have no vertical asymptote however, there are no points to test that are around the vertical asymptote (because there are none).
We have already 'subbed' in infinity (you found the limit).

Remember earlier how I said safe path/smart path?

Safe Path (doesnt take much longer at all)
Lets take the limit of negative infinity, we find that we get the same y=0.

Smart Path:
Recognise that it is an even function, therefore symterical about the y-axis, therefore the limit to negative infintiy should be the same as the one to infinity.

We have some pieces of information:
Horizontal asymptote y=0.
At infinity it approaches our asymptote
At negative infinity it approaches our asymptote
Our graph goes through (0,1) (intercept)
Our graph is symetrical

Now assume nothing crazy happens, do whatever you think you have to do in order to get the graph from one asymptote up and through the intercept down again back to the asymptote, you will find that you are quite accurate.

EDIT: It does not have ANY x-intercept, because if y=0
You will find, 0=1

How do you do that? :s

 

[Sy123]

We know how to do this, what we are doing here is we arent finding the limit, we are simply transforming the question in a way where we can use our limit rule:



This is what we need to apply, but we are this time applying negative infinity to 1/x^2
But x^2 is positive no matter what, so doing (negative infinity)^2 is the same thing as (infinity)^2

So our limit turns out the same.

It is actually better in this case to just sub in numbers, and see where the function approaches.

Taking limits isnt necessary, just sub in numbers (sub in -1000, and +1000), see where the function approaches and go on from there in sketching.

I need to go study eco yearly now, so Ill explain this in detail tomorrow.

 

[Fawun]

We know how to do this, what we are doing here is we arent finding the limit, we are simply transforming the question in a way where we can use our limit rule:



This is what we need to apply, but we are this time applying negative infinity to 1/x^2
But x^2 is positive no matter what, so doing (negative infinity)^2 is the same thing as (infinity)^2

So our limit turns out the same.

It is actually better in this case to just sub in numbers, and see where the function approaches.

Taking limits isnt necessary, just sub in numbers (sub in -1000, and +1000), see where the function approaches and go on from there in sketching.

I need to go study eco yearly now, so Ill explain this in detail tomorrow.

So basically what you're trying to say is negative infinity will always be the same as positive infinity since negative infinity squared is the same thing as positive infinity squared?

Also, I can't be bothered latexing this so I just took a photo but what happens if my x-intercept is the same as my vertical asymptote? and you said before that it can never cross a vertical asymptote only a horizontal one :s

inb4 i get posts from carrot and everyone else mocking my handwriting.

 

[Shazer2]

inb4 your handwriting is amazing.

 

[Carrotsticks]

inb4 i get posts from carrot and everyone else mocking my handwriting.

horizontol osymptote

 

[Fawun]

inb4 your handwriting is amazing.

lol thanks

horizontol osymptote

stop dat

 

[Sy123]

So basically what you're trying to say is negative infinity will always be the same as positive infinity since negative infinity squared is the same thing as positive infinity squared?

Also, I can't be bothered latexing this so I just took a photo but what happens if my x-intercept is the same as my vertical asymptote? and you said before that it can never cross a vertical asymptote only a horizontal one :s

inb4 i get posts from carrot and everyone else mocking my handwriting.

There is a reason why x=-2 is not a vertical asymptote, its because we can simplify our fraction a bit:



So it cant be an asymptote, because we have a cancellation going on. BUT this doesnt mean that x=-2 is a valid point for this, where x=-2 there will be a open circle on the function to indicate that it is discontinuous.

You can sketch y=1/x-2 by itself, but you have to consider the open circle.
So you ask, where do I put the open circle, is it (-2,0)?
No its not because although it might have an intercept, it doesn't have one.

Because we know that:



When we put y=0



But this has no solution since x=/=-2
Therefore there is no x-intercept

But where is our open circle?

We can find it by taking the limit of the function as it approaches x=-2, and see what happens:



So our open circle is in fact at (-2, 1/4 )
We then have the pieces of information:

1) Horizontal Asymptote y=0
2) Vertical Asymptote x=2
3) Open circle (-2, 1/4)
4) Approaches horizontal asymptote from above at positive infinity
5) Approaches horizontal asymptote from below at negative infinity

To find out 4 and 5, we can just sub in points at neg/pos infinity.

For number 5 you will get -0.0000123
For number 4 you will get +0.0002313
Then we graph.

 

[Fawun]

You can sketch y=1/x-2 by itself, but you have to consider the open circle.
So you ask, where do I put the open circle, is it (-2,0)?

Where did you get -2 from?

But where is our open circle?

We can find it by taking the limit of the function as it approaches x=-2, and see what happens:



So our open circle is in fact at (-2, 1/4 )

So whenever you want to find where the open circle is, you take the limit of the function?

To find out 4 and 5, we can just sub in points at neg/pos infinity.

For number 5 you will get -0.0000123
For number 4 you will get +0.0002313
Then we graph.

Sub in where though? :s

Also, how would you do this question?

1. Shade in the region represented by the following inequalities. Show the coordinates of any axis intercepts and points of intersection in your diagrams.

a)

I don't get it :s I know how to find the point of intersection and stuff by using simultaneous equations but I don't know how to graph them or how to even start...

 

[Sy123]

Where did you get -2 from?

Look at the function again:



The graph cannot be at x=-2 AND x=2.
The point is, at x=2 there is an asymptote, but x=-2 there is no asymptote because it cancels out with the numerator, the only alternative is an open circle.

So whenever you want to find where the open circle is, you take the limit of the function?

To find the y-coordinate of the open circle take the limit of the x value (in this case the x value is -2, so we take the limit of the function to x=-2

Sub in where though? :s

We sub it into our original function, just sub in large numbers, and see if they are negative or positive numbers.
For example if I sub in x=1000, and I get the output of y=-0.0000...
Then it approaches positive infinity below the y-axis (because it is a negative y at a very large positive x)

Also, how would you do this question?

1. Shade in the region represented by the following inequalities. Show the coordinates of any axis intercepts and points of intersection in your diagrams.

a)

I don't get it :s I know how to find the point of intersection and stuff by using simultaneous equations but I don't know how to graph them or how to even start...

Lets look at our first function, it has an x^2 and a y^2, because we are not doing 4U maths, we can assume that this is a circle graph.

So lets put the circle equation in the form:



Where (h,k) is the centre and r radius.

Looking at the equation given to us, lets utilise completing the square in order to get it into a recognisable form:



What do we need to factorise... x because there is an x AND an x^2 there, to get it into its own bracket we must add 4 to both sides to get a perfect square of x on the LHS



Now we have manipulated our equation to get a perfect square of x:



Here we have the circle equation in a recognisable form

Our centre is (2,0) and radius is 2.

Graph it on the Cartesian axes.

Now for the second graph it is simply a straight line:



It is a line going through the centre of gradient
To make graping this straight line easy, just plot (0,0) and the point (We just sub in x=1 to make plotting easier)

So now we arrive at this, we have a circle and a straight line that intersects it, we arent done yet though, we have inequalities to shade.

Take the circle graph again:

It is:



The easiest way to know which region to shade, is to pick a point, any point (that DOESNT lie on the curve), and test it.

lets take the point 1,1 and sub it in:

(1-2)^2+1^2=2



Is this true? Yes it is, that means the point (1,1) is valid for this inequality, that means the region which is bounded by the curve that contains (1,1) must be shaded.

Lets do the same for the second inequality, sub in (1,1)



This is INCORRECT, therefore the region that is bounded by the curve that contains (1,1), IS NOT SHADED. We shade the other region, so the other side of the line.

NOTE: Our inequality sign for the SECOND one is >, not
That means we have to make our line dotted to symbolise this inequality.

Now we have two shaded regions, the valid region that is our answer, is the region where they both overlap. This is the region that is our answer, and you should shade it in a differnet color or something idk.