VCE Methods Question Thread!

[S_R_K]

There are a few reasons

Just to fix some notation, let's say that we have a function \(f : X \longrightarrow Y\) where \(X\) is the domain and \(Y\) is the co-domain.

1) In general, finding the range of a function is hard. For a function to be defined, we require that, for each "input" value \(x \in X\), there is a unique "output" value \(y \in Y\) such that \(y=f(x)\). This definition can be satisfied by a function even if the co-domain is larger than the range - ie, there exists a \(y \in Y\) such that \(y \neq f(x)\) for any \(x \in X\). For example, consider \(f : R \longrightarrow R,\;f(x)=x^2\) - all the negative numbers in the co-domain aren't in the range of the function.

Now you might say "who cares, why not just say that \(Y\) is the range, whatever that turns out to be?" One big problem is with function composition (it sounds like you're in year 11, and this is getting into Year 12 content). Function composition is a way of building up more complicated functions where the "input" of another function \(g\) is the "output" of our function \(f\) - we write, for example, \(g(f(x))\) to refer to "\(g\) composed with \(f\)". For this to work, we need the "outputs" of \(f\) to be within the domain of \(g\). This is guaranteed if the co-domain of \(f\) is a subset of the domain of \(g\), since the range is a subset of the co-domain. But if instead we say that \(Y\) is the range, and we don't know what the range is, then it may not be clear that the outputs of \(f\) are a subset of the domain of \(g\). Obviously in practice, and in specific cases, this problem is not entirely eliminated by saying that \(Y\) is the co-domain, but it does make the general issue of defining function composition more straightforward.

2) The second reason is going beyond the scope of Methods, and this is an area where Methods lies to you to avoid additional complications. You'll shortly come across the ideas of a "one-to-one" function and an "inverse" function. In VCE Methods you are told that a function \(f\) has an inverse - written as \(f^{-1}\) - if \(f\) is one-to-one. However this is not quite right. A function has an inverse only if it is one-to-one. The other requirement is that the function is "onto", which means that every value in the co-domain is the output, by that function, of some value in the domain. In other words, an "onto" function is when the co-domain = range. Importantly, not all functions are "onto", and so do not have inverses (even if they are one-to-one), and so it is important to distinguish between co-domain and range.

[Rose34]

Hello everyone,

A serious question. Why is the period of sine and cosine 2pie/n? like why not pie/n? why not 3pie/n? why does it have to be 2?

Thanks in advance.

[colline]

Hello everyone,

A serious question. Why is the period of sine and cosine 2pie/n? like why not pie/n? why not 3pie/n? why does it have to be 2?

Thanks in advance.

Hey Rose!
The period of a circular function is basically its length along the x-axis before the graph is repeated again. For both y=sin(x) and y=cos(x), the graph repeats once x=2pi.
This screenshot of y=sin(x) might explain it a bit better: https://prnt.sc/vf2ltw. See how after x=2pi, the graph repeats itself. That's basically the start of a new period.
Hope that explains it!

[Corey King]

There are a few reasons

Just to fix some notation, let's say that we have a function \(f : X \longrightarrow Y\) where \(X\) is the domain and \(Y\) is the co-domain.

1) In general, finding the range of a function is hard. For a function to be defined, we require that, for each "input" value \(x \in X\), there is a unique "output" value \(y \in Y\) such that \(y=f(x)\). This definition can be satisfied by a function even if the co-domain is larger than the range - ie, there exists a \(y \in Y\) such that \(y \neq f(x)\) for any \(x \in X\). For example, consider \(f : R \longrightarrow R,\;f(x)=x^2\) - all the negative numbers in the co-domain aren't in the range of the function.

Now you might say "who cares, why not just say that \(Y\) is the range, whatever that turns out to be?" One big problem is with function composition (it sounds like you're in year 11, and this is getting into Year 12 content). Function composition is a way of building up more complicated functions where the "input" of another function \(g\) is the "output" of our function \(f\) - we write, for example, \(g(f(x))\) to refer to "\(g\) composed with \(f\)". For this to work, we need the "outputs" of \(f\) to be within the domain of \(g\). This is guaranteed if the co-domain of \(f\) is a subset of the domain of \(g\), since the range is a subset of the co-domain. But if instead we say that \(Y\) is the range, and we don't know what the range is, then it may not be clear that the outputs of \(f\) are a subset of the domain of \(g\). Obviously in practice, and in specific cases, this problem is not entirely eliminated by saying that \(Y\) is the co-domain, but it does make the general issue of defining function composition more straightforward.

2) The second reason is going beyond the scope of Methods, and this is an area where Methods lies to you to avoid additional complications. You'll shortly come across the ideas of a "one-to-one" function and an "inverse" function. In VCE Methods you are told that a function \(f\) has an inverse - written as \(f^{-1}\) - if \(f\) is one-to-one. However this is not quite right. A function has an inverse only if it is one-to-one. The other requirement is that the function is "onto", which means that every value in the co-domain is the output, by that function, of some value in the domain. In other words, an "onto" function is when the co-domain = range. Importantly, not all functions are "onto", and so do not have inverses (even if they are one-to-one), and so it is important to distinguish between co-domain and range.

That definitely clears things up
Additionally, thanks for letting me know about inverse functions. I knew that in subjects like chemistry we were ''lied'' to with simplifications, but honestly I assumed that wasn't the case in math. Interesting and good to know.

[Corey King]

Hey guys
I'm a bit confused about when to add in the plus-minus sign when finding the root of a term.
I'm doing inverse functions at the moment.

So this is an example question, in which they include the plus-minus sign after taking the root:
https://gyazo.com/433483a81d5a041da59afb7e04e5f336

And then for Q3 b), they take the root of both sides in a similar situation but don't include the +- sign.
https://gyazo.com/46060bea8db1b117c63d8e8d3a72c9df
https://gyazo.com/f67ebab745a734bcd2078376033a1053


Is it just a textbook error for question 3 b, or am I missing something?

Many thanks
Corey

[eloisegrace]

Hey guys
I'm a bit confused about when to add in the plus-minus sign when finding the root of a term.
I'm doing inverse functions at the moment.

So this is an example question, in which they include the plus-minus sign after taking the root:
https://gyazo.com/433483a81d5a041da59afb7e04e5f336

And then for Q3 b), they take the root of both sides in a similar situation but don't include the +- sign.
https://gyazo.com/46060bea8db1b117c63d8e8d3a72c9df
https://gyazo.com/f67ebab745a734bcd2078376033a1053


Is it just a textbook error for question 3 b, or am I missing something?

Many thanks
Corey

It has to do with the domain of the inverse function. As the domain of the inverse function is the range of the original function, you use this to determine if it is the positive or negative square root.

I hope this makes sense

[Corey King]

It has to do with the domain of the inverse function. As the domain of the inverse function is the range of the original function, you use this to determine if it is the positive or negative square root.

I hope this makes sense

Oh nice! Thanks
I'm still a bit confused though. I thought that the domain and the relation were kind of separated?
So that you can have multiple domains apply to the same relation.
I think something like this was brought up back near the chapter on hybrid functions?

[AlannahBottino]

I have a few questions:

First, how do you find the domain and range of composite functions?

Then, when there is a question where you have to find the derivative, and hence find the antiderivative of a different function, how does that work?

Also antiderivatives of composite functions and fraction ones like (3/(x-2)^2) or even (3/(x-2))?

Thanks 

[keltingmeith]

Oh nice! Thanks
I'm still a bit confused though. I thought that the domain and the relation were kind of separated?
So that you can have multiple domains apply to the same relation.
I think something like this was brought up back near the chapter on hybrid functions?

Domains and relationships are separated, but they're together if they're in the form of the function. For example, here's three functions:

f: [0,1]->R, f(x)=x
g: [0,2]->R, f(x)=x
h: [0,2]->R, f(x)=x^2

Even though the first and second function have the same relationship, they have different domains, and so they are different functions. Even though the second and third functions have the same domain, they have different relationships, so they are different functions. So in that sense - yes, they are separated. However, for the function g, that domain is now paired with that relationship - so if I wanted to talk about the inverse function of g, g^-1, then I need to consider the domain that's with it. But also, the inverses of f and g here are boring, so let's talk about the inverse of h.

For h, its inverse function h^-1 must have a range of [0,2]. Because the domain of a function becomes the range of its inverse. Since the range must be [0,2], that means I know that I must take the positive square-root - because if I take the negative square-root, the range will also be negative.

I have a few questions:

First, how do you find the domain and range of composite functions?

Then, when there is a question where you have to find the derivative, and hence find the antiderivative of a different function, how does that work?

Also antiderivatives of composite functions and fraction ones like (3/(x-2)^2) or even (3/(x-2))?

Thanks 

For the first one, in methods we assume a function composition can only happen if the range of the inner function is a subset of the domain of the outer function. This isn't strictly true in maths, but methods likes to take liberties with facts, and it does make our lives slightly easier if it is true. This means the domain of the composite function is the domain of the inner function - because, think about it. If all of the outputs, the range, of the inner function can enter the outer function, then what is a list of numbers that give you all of the outputs of that inner function? It's the domain of the inner function.

Finding the range of the composition is slightly more complicated. First, figure out the range of the outer function. Let's say I have the function composition f(g(x)). Well, to make my life easier, I'm going to call g(x)=u. So, now I'm trying to graph f(u). How would you find the range of f(u)? Well, I would figure out all the numbers u can be, and I would call that the domain. Then I would graph f(u), and see what y-values it has. Okay, so how can I figure out all of the numbers that u can be? Well, u=g(x), so surely all the numbers that u can be is simply all of the outputs of g(x) - so, the domain of u is the same as the range of g(x). So, to find the range of f(g(x)), all I need to do is:

1. find the range of g(x)
2. put those numbers into the domain of f(u), and then
3. the range of f(u) is going to be the same as the range of f(g(x)).

Simple 3 step solution!

Next - you're asking about integration by inspection. The trick to these questions is to know that the derivative sign and the anti-derivative sign sort of cancel each other out. Let's say I'm told to find the derivative of x*sin(x). This would be:

\[
\frac{d}{dx} x\sin(x)=x\cos(x) +\sin(x)
\]

Simple product rule, hopefully you can see how I made it - let me know if you can't, I'll be much more explicit. Next, the question asks me to find an antiderivative of x*cos(x). Well, I don't know how to do that! But, what if I just take the top expression, and take the antiderivative of both sides? This would give me:

\[
\int\frac{d}{dx} x\sin(x)\: dx=\int x\cos(x) +\sin(x)\:dx
\]

Okay, so that integral sign on the left cancels out with the derivative sign (but I need to make sure I put a +C the moment one of these sides doesn't have an integral sign!), and I can split up the one on the right like so:

\[
x\sin(x)+C=\int x\cos(x)\:dx +\int\sin(x)\:dx
\]

And now, I can solve for the integral of x*cos(x):

\[
x\sin(x)+C=\int x\cos(x)\:dx -\cos(x)\\
x\sin(x)+\cos(x)+C=\int x\cos(x)\: dx
\]

Which is exactly what the question wanted me to find! So, those questions are all about recognising the derivatice and anti-derivative signs "cancel out", and then just some basic algebra manipulation.

Finally, antiderivatives of reciprocal functions. In methods, these can only be in the form of \(\int\frac{a}{(bx+c)^n}\:dx\) Integrating these isn't too hard - for example, you should also know how to do \(\int a(bx+c)^n\:dx\) All you do is add one to the power, divide by the new power, and divide by the derivative of what's in the brackets. It's kind of like the chain rule, but in reverse. So, how do you do that if it's a fraction? Well, flip the fraction, then add 1 to the power, divide by the power, and divide by the derivative of the bracket. Here's some examples:

\[
\int 7(2x-3)^3\:dx = \frac{7}{4\times 2}(2x-3)^4 +C\\
\int\frac{2}{(5x-1)^{12}}\:dx=\int 2(5x-1)^{-12}\: dx=\frac{2}{-11\times 5}(5x-1)^{-11}+C
\]

(normally you would simplify the denominator - I haven't in this case, so it's super clear where all the numbers are coming from)

Finally, there is one exception to this rule - and that's when the power of the denominator is 1. For example, \(\int\frac{1}{x-2}\:dx\). This one actually becomes a log to the base e - more specifically, \(\int\frac{1}{x-2}\:dx=\log_e(x-2)+C\). And the reason for this is because if you differentiate a logarithm, you get 1/the number inside the logarithm. You should remember this, and it is on your formula sheet:

\[
\frac{d}{dx}\log_e(2x-1)=\frac{2}{2x-1}\\
\frac{d}{dx}\log_e(5-6x)=\frac{-6}{5-6x}=\frac{6}{6x-5}
\]

Since we know that integration is the opposite of differentiation, it should then make sense if I wanted to integrate something of the form \(\frac{a}{bx+c}\), then it should become a logarithm. However, it is important to note this is only true for when the number inside the logarithm is positive, since you can't take the logarithm of a negative number

Let me know if I've confused you with any of these points.

[fun_jirachi]

Hey

In general, for a function f(g(x)), the domain is all x in the domain of g such that g(x) is in the domain of f. The range is all such values of f(x) that the domain maps to.

Example:
\(f(x) = \sqrt(x), g(x) = x^2, f(g(x)) = \sqrt{x^2}\)  -  the domain of g is all positive reals, while the range of f is all positive reals. We have also that the domain of f is all reals, while the range of g is all positive reals. (This is just the absolute value function!) Since the range of the 'inside' is the same as the domain of the 'outside', we don't need to do anything - just keep the domain of f, and the range of g for f(g(x)).

This reasoning gets more problematic when they're not the same, and this is where a lot of people mess up. It's important to consider both
Example:
\(f(x) = \sqrt(x), g(x) = -x^2, g(f(x)) = -(\sqrt{x})^2\)
Domain of f = all positive reals
Range of f = all positive reals
Domain of g = all reals
Range of g = all negative reals

Here, we can see that the range of f and the domain of g are not the same - we need to 'adapt' the domain of f such that the range of values produced by f is a subset of the domain of g - g's range will be the range of numbers that this g maps this subset to. Try this with this function and other functions to get your head around it more

Second question:
Example:
\(\text{Find } \frac{d}{dx} x\ln x, \text{ and hence find } \int_0^1 \pi \ln x \ dx\).

Finding the derivative is pretty standard product rule - \(\frac{d}{dx} x\ln x = \ln x + 1\).
To find the integral, we manipulate it into a better from so it matches the right hand side from our first part - as such
\(\int_0^1 \pi \ln x \ dx = \int_0^1 \pi (\ln x + 1 - 1) \ dx = \pi \left(\int_0^1 \ln x + 1 \ dx \int_0^1 1 \ dx\right)\). Then by the Fundamental Theorem of Calculus, we can use the first part's LHS as the antiderivative for the left part of our current RHS (a bit of a reductive explanation, i know ). A lot of the tricks you're likely to come across when manipulating stuff include multiplying and dividing by the same number, and adding and subtracting the same number, etc.

As a generalisation, the antiderivative of something of the form \(\frac{a}{bx+c}\) is \(\frac{a}{b} \ln (bx+c) + C\). Something of the form \(\frac{a}{(bx+c)^n}\), where n is a positive number greater than 1, is best converted to a form like \(a(bx+c)^{-n}\), where we can use the reverse chain rule.

Hope this answers your questions adequately

[keltingmeith]

oops, looks like you got in just as I finished editing, sorry jirachi Some things I wanted to say about the answer, though:

(This is just the absolute value function!)

Very frustratingly, the absolute value function is stripped from VCE methods

This reasoning gets more problematic when they're not the same, and this is where a lot of people mess up. It's important to consider both

Extremely important to consider! But, VCE gets around this by pretending that the composition just doesn't exist when the range of the inner function is not a subset of the domain of the outer.

Sorry for having to drill holes in your honestly, very lovely explanations, but unfortunately VCE is butchering maths a bit

[Corey King]

Domains and relationships are separated, but they're together if they're in the form of the function. For example, here's three functions:

f: [0,1]->R, f(x)=x
g: [0,2]->R, f(x)=x
h: [0,2]->R, f(x)=x^2

Even though the first and second function have the same relationship, they have different domains, and so they are different functions. Even though the second and third functions have the same domain, they have different relationships, so they are different functions. So in that sense - yes, they are separated. However, for the function g, that domain is now paired with that relationship - so if I wanted to talk about the inverse function of g, g^-1, then I need to consider the domain that's with it. But also, the inverses of f and g here are boring, so let's talk about the inverse of h.

For h, its inverse function h^-1 must have a range of [0,2]. Because the domain of a function becomes the range of its inverse. Since the range must be [0,2], that means I know that I must take the positive square-root - because if I take the negative square-root, the range will also be negative.

Ah, makes sense All cleared up.

[Corey King]

I'm stuck on this question about function applications (final question of chapter).

I'm stuck at question b) ii - Find the allowable values of x.
https://gyazo.com/b38f9fb23d45081e919ca7b9723d0ae4
I understand why x must be < 24, but I can't see how they infer 24 - 2x < 0 in the solved answers.
https://gyazo.com/5ceb1349ab91e3e7478074236fce7922

[keltingmeith]

I'm stuck on this question about function applications (final question of chapter).

I'm stuck at question b) ii - Find the allowable values of x.
https://gyazo.com/b38f9fb23d45081e919ca7b9723d0ae4
I understand why x must be < 24, but I can't see how they infer 24 - 2x < 0 in the solved answers.
https://gyazo.com/5ceb1349ab91e3e7478074236fce7922

See attached picture. If y-x is positive, then the little overhang bit I've put there exists. If y-x=0, it doesn't exist, and y=x. However, if y-x is negative, then you'd have a negative length, and that's not allowed. Hence, y-x>0. And since y=24-x, this means that y-x=24-x-x=24-2x>0, and gives you that x<12.

[Sherlock.Holmes]

How much do you need in each exam for a 25 ss?