VCE Methods Question Thread!

[Maths Forever]

TT is getting at this point:

You have assumed that a limit of a function as x approaches b must be the function evaluated at that point if it exists. This is only actually true for a certain class of function called 'continuous functions'. It so happens that all polynomials, exponentials, logs, trig functions and sums/products of these are continuous over their maximal domains and so are quotients, assuming the denominator isn't 0. However, this is something that needs to be proved (not in VCE) and you can't just assume that the limit of x as x approaches 1 is 1. You need a rigorous working definition of a limit and then you need to work with it.

Fortunately this isn't covered in Methods so you don't have to worry about it. But at the same time, it's not healthy for your maths learning to just parrot limit theorems and calculations without understanding what's going on with them. Find a rigorous proof of the Sandwich theorem, or indeed l'Hopital's rule (which I'm sure you've seen); that'll be quite useful for your learning.

My main goal is to assist any student studying VCE maths methods, using the proofs which I know exist, and are part of the course I know well. I look forward to learning the derivations. Good luck with your future learning too lzxnl!

[Maths Forever]

My hybrid function example has nothing to do with whether is undefined or defined as , it was to illustrate a fallacy in your approach.
Yes there is.
Yes I do, but this has nothing to do with my example.

Perhaps an even simpler example to illustrate why your approach is incorrect is as follows. Define:



Again I ask, what is ? Using your approach:

1) g(x) is not a product function.
2) g(x) is defined at x=0.

Hence .

Also the above example illustrates why this statement is false.
Clearly at , but there is an limit to evaluate, i.e., , whether it exists or not is not of the concern.

I think I know what you mean now. g(x) is discontinuous at x = 0.

Limit as x approaches 0 for g(x) = x + 3 (from left hand side) = 0 + 3 = 3

Limit as x approaches 0 for g(x) = x + 3 (from right hand side) = 0 + 3 = 3

Hence the limit as x approaches 0 for your hybrid function is equal to 3, regardless of the discontinuity at x = 0.

[lzxnl]

Actually the limit is still 3. It exists; as the function is x + 3 for all nonzero x, as x approaches 0 from either direction, it behaves like x + 3 and thus approaches 3. The exact value as x = 0 of the function is irrelevant.

[SE_JM]

Can someone please explain to me:
Find the maximal domain of
h(x) = ((x-1)/(x+2))^1/2

I'm getting pretty confused. I also remember there was a thread about finding maximal domains but i forgot where it is now.. Does anyone know?
Thank you

[TrueTears]

I think I know what you mean now. g(x) is discontinuous at x = 0.

Limit as x approaches 0 for g(x) = x + 3 (from left hand side) = 0 + 3 = 3

Limit as x approaches 0 for g(x) = x + 3 (from right hand side) = 0 + 3 = 3

Hence the limit as x approaches 0 for your hybrid function is equal to 3, regardless of the discontinuity at x = 0.

Correct

Hence, the dangers of substituting in values when the function is not continuous.

[keltingmeith]

Can someone please explain to me:
Find the maximal domain of
h(x) = ((x-1)/(x+2))^1/2

I'm getting pretty confused. I also remember there was a thread about finding maximal domains but i forgot where it is now.. Does anyone know?
Thank you

Simply look for problem areas - for example, you can't take the square root of a negative number (note that the domain of the square root function is undefined for negative x), so we want everything inside of it to be positive or zero:



Now, multiplying x+2 across poses a problem, because if the number is negative then we have to change the inequality. So, we have two situations - either x>=-2 (so x+2 is positive/zero) or x<-2 (so x+2 is negative).

For x>=-2:

The intersection of these is x>=1, so we take x>=1 as one set that the function is defined.

For x<-2:

The intersection of these is x<-2, so we take x<-2 as another set that the function is defined.

Finally, the maximal domain comes from the union of these two sets - so, the maximal domain is
You can, of course, supplement this with a graph to make sure you haven't messed up (checking with the graph of (x-1)/(x+2), we see that it is positive over that domain)

[Maths Forever]

Correct

Hence, the dangers of substituting in values when the function is not continuous.

Great! I'm glad we sorted it out! 

Also, where do you find all of those mathematical symbols on ATAR notes?

[keltingmeith]

Great! I'm glad we sorted it out! 

Also, where do you find all of those mathematical symbols on ATAR notes?

LaTeX typesetting. Judging by your name, I'm guessing you'll be studying maths at uni? Because LaTeX is just beautiful for writing up your maths assignments.

[Maths Forever]

LaTeX typesetting. Judging by your name, I'm guessing you'll be studying maths at uni? Because LaTeX is just beautiful for writing up your maths assignments.

Alright thanks. I'll have a look into LaTeX! Yes, I will be studying maths, so it should be quite useful. 

[cosine]

Can someone please help me with 6a, I keep getting the gradient of the chord PQ as -2+h but answer says its just h

Thanks

[Maths Forever]

Can someone please help me with 6a, I keep getting the gradient of the chord PQ as -2+h but answer says its just h

Thanks

6. a) f(x) = x(x+2) = x^2 + 2x

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

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

Chord Gradient: (h^2 - 1 - (-1) ) / (-1 + h - (-1) ) = (h^2) / h = h

I hope this helps!

(b) f'(x) = 2x + 2

At point P, x = -1, f(x) = -1, f'(-1) = 2(-1) + 2 = -2 + 2 = 0

Hence the gradient at point P is 0.

[IntelxD]

Can someone please help me with 6a, I keep getting the gradient of the chord PQ as -2+h but answer says its just h

Thanks

Substitute -1 into f(x) to get the y-coordinate of the point:

f(-1)=-1(-1+2)=-1

Substitute -1+h into f(x) to get the y-coordinate of the point:
 
f(-1+h)=(-1+h)((-1+h)+2)
            =(h-1)(h+1)
            =(h^2-1) Recall difference of two squares.

Our points are (-1,-1) and (-1+h,h^2-1). Now we can use (y2-y1)/(x2-x1) to obtain the gradient in terms of h.

Gradient = (h^2-1)-(-1)/(-1+h)-(-1)
             = (h^2 -1 + 1)/(h-1+1)
             =h^2/h
             =h

Edit: Someone beat me to it, hope this still helps.

[cosine]

Oh, thank you people!

Also, what is the graphical difference of a tangent and a chord. This is my underatanding:

Chord: a straight line which connects two points together which we can work out the gradient by m=rise/run
Tangent: the line created at a specific point?

Thanks!

[keltingmeith]

Oh, thank you people!

Also, what is the graphical difference of a tangent and a chord. This is my underatanding:

Chord: a straight line which connects two points together which we can work out the gradient by m=rise/run
Tangent: the line created at a specific point?

Thanks!

Your definition of a chord is good.
A tangent is a line that just touches (as opposed to intersecting) at a single point.

[cosine]

Thanks Euler!!

Quick question, is finding the derivative the shorter way to find the gradient of a function? Say f(x)= 3x + 1

The way to graphically see the gradient is by rise/run, but if we find the derivative f'(x) = 3 could this work as well?

And if it does work, then why dont we just always use the derivative methods to find out the gradient? Thanksss