VCE Methods Question Thread!

[EspoirTron]

Hi Monsieur Kebab, Thanks for the quick response

I have tried substituting in values and solving using simultaneous equations, but I end up having something similar to this: E-1=E-3
So I am unable to get the horizontal transformation this way
Unless I'm doing something wrong

My pleasure. Would you like to post the example question and perhaps we can work for a solution from there?

[paper-back]

Sorry, Took a while to compress the picture
The question is on the picture I've attached

[b^3]

Sorry, Took a while to compress the picture
The question is on the picture I've attached

Although it looks like it goes through , there isn't actually a curve that will have those endpoints, amplitude and period that goes through (you can force it to go through the origin but then it won't satisfy the endpoints). So it's kinda a badly drawn question.

Since we aren't given 4 points, you'll have to do some of it visually, but we'll still find the horizontal translation using a point. (Also to make the question work I'm assuming that it doesn't go through the origin).

Since it's centered vertically at we have our shift of units downwards.

Then we can use one other point, in this case to find our horizontal translation. So we substitute in the point and solve for

Technically it would then need the domain restriction, but since you're just asked to find the values you wouldn't need anymore.

EDIT: Try moving the slider to see what I mean with it not going through those points and the origin.
https://www.desmos.com/calculator/ax0x0gvtrd

[EspoirTron]

This is going to get a tad bit messy since I won't be using Latex so I hope it still makes sense.

We will begin by solving for the amplitude, a. a= (max value-min value)/2 = 2-(-10)/2 =12/2=6. So great, we know that a=6.
So now we have: y=6sin(n(x+v))+c. I am calling v epsilon here.
Well n can be found by the period of the function too. So we can see that the period is 3pi. So 3pi=2pi/n and therefore n = 2/3
So I can also see that the graph has been translated four units downwards. How? Well if the amplitude is 6 then it should go from -6 to 6, which it doesn't so that you can just see from the graph. Hence c=-4
So looking at the graph we have the points: (-pi,-4), (pi,-10), (2pi,-4) and something else which we don't need anyway because we have one unknown and three points.

Let's sub in a set of points (pick the one you would like), so we get:
-4=6sin(2/3(-pi+v))-4

Let's solve this by hand:
0=sin(2/3(-pi+v)) (divide both sides by 6)

0=sin(2/3(-pi+v))
sin^-1(0)=2/3(-pi+v)
And then 0=2/3(-pi+v)

Solving through we find that v=pi

Therefore we find that: a=6, v=pi, n=2/3 and c=-4.

And I think that is right so I hope that helps! By the way I saw in your working that you said a=-6. However, the amplitude can never be negative so just be careful there.

EDIT: BEATEN

[paper-back]

Thanks Monsieur Kebab and b^3

The horizontal translation was: E=

I substituted in the other 2 sets of points and I got the following:



But why can't we use ?

EDIT: Latex isn't working, so I put working out as attached picture

[jessss0407]

hey guys!

I need some help understanding the difference between limits, continuity and differentiability... like I have a vague idea about all of them..

i'm mainly confused about how a limit can exist in a graph that is discontinuous

also, i'm starting to learn change of bases for log at school now.. is there an easy way to understand/learn the rules?

thanks!

[paper-back]

also, i'm starting to learn change of bases for log at school now.. is there an easy way to understand/learn the rules?

Do you mean as in:
?

[EspoirTron]

hey guys!

I need some help understanding the difference between limits, continuity and differentiability... like I have a vague idea about all of them..

i'm mainly confused about how a limit can exist in a graph that is discontinuous

also, i'm starting to learn change of bases for log at school now.. is there an easy way to understand/learn the rules?

thanks!

Hey Jess,

Limits, continuity and differentiability are all very intimately connected concepts. Here is a crash course:

Let's think a function and call it f(x). In this example you are free to think of f(x) as anything you would like. A limit is a way for me to say: as I take any value of x what value does f(x) approach? So basically I am saying as I take x arbitrarily small near a particular value what does f(x) approach? So basically if I had f(x)=x then the limit as x approaches 2 would just be 2. What that is basically saying is that if I take values really close to 2, but never equal to 2, f(x) will approach 2.

So continuity. A continuous function is just any function that is defined for all values within its given domain, or over all reals. For a function to be continuous at a point the partial limits must approach the same value for that number. So if again I have f(x) as I approach a particular value of x from the left-hand side I will get a value. As I approach from the right-hand side I will also get a value. For the function to be continuous at this point both these left-hand and right-hand limits must give me the same numerical value. A function is not continuous at jumps, endpoints or anywhere with hollow circles really.

Differentiability. I won't go too much into this, but basically when I am finding the derivative I am actually find the rate at which my function, f(x), is changing. This has a very intimate connection with limits and unfortunately because I am horrible with Latex I can't derive anything for you! However, understand your limits and everything in calculus will become a lot more easier!

On a side note: a partial limit does exist for an end point, but it isn't continuous there since the partial limits on both sides  don't equal each other.

Edit: I want to be more vigorous.

So the latter part of your question asked why a limit can exist if the function is discontinuous? Well let's think about Mr f(x) again. Let's say f(x) is defined over all reals excluding 3. Well intuitively f(3) doesn't exist. However, remember a limit is telling me what value f(x) approaches as I take x arbitrarily small close to 3. So if I ask you to find the limit as f(x) approaches 3, it exists - or, more precisely, one of the partial limits will assume a value the other partial limit does not. So I would be approaching a value when I approach 3 but f(x) would never equal that value. Think of it in this way: f(x) is a happy lad, but as he approaches 3 he gets scared because he sees the girl he likes so he tries going up to her but he gets so close to telling her but he just doesn't do it! So the poor guy never sees what is going on at x=3.

I hope my cheesy example shed some light on the topic

[soNasty]

Can someone show me how to do question 8 of the 2013 VCAA exam 1 please?

[paper-back]

Is there a method for getting the range of a composite function?
E.g. we know that domain of f(g(x)) is the domain of g(x)
Is there a rule similar to this to get the range of f(g(x))?

Thanks

[EspoirTron]

Can someone show me how to do question 8 of the 2013 VCAA exam 1 please?

Okay so E(X) = integrand of xf(x) from 0 to 2. Since 0 won't give us anything we'll say E(x)= integrand (x multiplied by pi/4cos(pix/4).
Well from the information they gave us: pi4cos(pix/4) = d/dx(xsin(pi/4)-sin(pix/4)
So it follows that E(X)= integrand (d/dx(xsin(pi/4))-(sin(pix/4)) with upper bound = 2 and lower bound = 0.
Well if I take the integral of the derivative I am essentially just going to be left with my original function, so integrand(d/dx(xsin(pi/4))=xsin(pi/4)

So it follows that E(X)=xsin(pi/4)-(-4/picos(pi/4)) evaluated between 0 and 2.

Substituting our values now we will find that E(X)= 2-4/pi

Is there a method for getting the range of a composite function?
E.g. we know that domain of f(g(x)) is the domain of g(x)
Is there a rule similar to this to get the range of f(g(x))?

Thanks

The best way to identify the range is by looking at a sketch of the graph - imo.

[paper-back]

Can we assume that the range of f(g(x)) is the range of f(x). Or does this not work

[alchemy]

Can we assume that the range of f(g(x)) is the range of f(x). Or does this not work

Doesn't work mate

[EspoirTron]

Can we assume that the range of f(g(x)) is the range of f(x). Or does this not work

Well I would not make that assumption. IIRM for f(g(x)) to exist the range of the inner function must be a subset of the domain of the outer function; that is rang is a subset of domf. We say that the domain of the given composite function is the domain of g(x). If I was to say that f(x) was the range I would be making the assumption that the domain of f=domain of g, which isn't necessarily true.

If you know the domain of the function and you know it is function you are familiar with, e.g f(g(x))= x^3 then the range is clear, so that is okay. But if you get something like f(g(x))=ln(x) times 15^x/cos(x) times mod(x) then the range isn't very clear. By the way, I'm not even sure the function I just listed exists, I just put random things together to accentuate the fact

[ETTH96]

Hey guys can someone help me out with this?

let f(x)=1/x  -3 and g(x) =-ax. The values of a for which the graphs of y=f(x) and y=g(x) have two unique intersection points are?

a) a<9/4
b) a>9/4
c) a>4/9
d) a<4/9
e) -infinity <a <0 or 0<a <9/4

How do i go about doing this? Thanks