VCE Methods Question Thread!

[lzxnl]

Lol, sorry if my question didn't make sense. But I was doing a question on composite functions.
The question was asking me to state whether g o f exists/defined. From what I learnt, g o f is defined if the domain of g is greater or equal to the range of f
So yeah, I've found the range of f and the domain of g now I'm wondering whether domain of g is greater/equal to range of f and how so?

That's different. They mean f(g(x)) is defined if the range of g is a subset of the domain of f. In your example, [-inf, 3] does not fit into [0, inf) or vice versa.

[T-Infinite]

That's different. They mean f(g(x)) is defined if the range of g is a subset of the domain of f. In your example, [-inf, 3] does not fit into (0, inf) or vice versa.

OMG, so it's subset! I quickly read over the textbook and it was " g o f is defined if the range of f is a subset of domain of g". The way my teacher drew the 'subset' symbol looks sooo much like the 'greater or equal' symbol. Bloody hell, so that 'greater of equal' theory doesn't exist or something? hahahah. silly me.

[soNasty]

Can someone show me how to properly calculate the range of
f(x)=squareroot(x^2-49)

The answer is R+ \ {0}
How come it's not just [0,inf)?

[SocialRhubarb]

Are you sure it's not R+ U {0}? The set R+ does not include 0, hence R+ \ {0} doesn't really make sense.

R+ U {0} would be the same as your answer.

[soNasty]

Are you sure it's not R+ U {0}? The set R+ does not include 0, hence R+ \ {0} doesn't really make sense.

R+ U {0} would be the same as your answer.

Yes it is that, sorry I have a habit of copying it wrong haha

[soNasty]

Oh it DOESNT include zero, wow I forgot that. I get it now, thanks!

[TrueTears]

so that 'greater of equal' theory doesn't exist or something? hahahah. silly me.

One must be really careful how they define things. E.g. what exactly does it mean for a set to be "larger" (or likewise, "smaller") than another set? Take an interval as an example, you may be inclined to believe that if the "length" of some real interval is greater than the length of another interval, then the former is "larger", but does this definition really work for all kinds of sets? To illustrate, consider the Cantor set which is formed by taking the closed interval and then removing the middle , denote and , continue this inductively and define , now let's see what the "length" of this set is. Since , and to construct we remove middle thirds of length , the length of is . So the Cantor set has zero length. But wait, does this mean it contains no elements? Thinking about this further, note that each endpoint is never removed, so contains at least the endpoints of each iteration. We have seemingly an impossible situation, how can a set have zero "length" yet still be non-empty? The reason is because our initial definition precludes us from seeing the larger picture.

Indeed, Cantor's discoveries lead to the notion of cardinality and cardinal numbers, I won't delve into it but you can have a read here: http://en.wikipedia.org/wiki/Cardinality and http://en.wikipedia.org/wiki/Cardinal_number.

Using the idea of countability, it can be shown that the Cantor set is a collection of points. Intuitively, we would think a collection of points is a "small" set right? Well it turns out, the Cantor set is "larger" than the entire set of natural and rational numbers, in fact it is uncountable, meaning there is no way we can list the elements in the Cantor set such that we can form a bijection with or . Thus, what seemingly appeared to be very "small" is infact so large that it's "infinitely large".

[soNasty]

How the hell does one sketch
?

There's all these turning points and stationary points of inflections, where do they begin?  How do you know when to start drawing the points of inflection within the graph?

[brightsky]

Mark in the x-intercepts x = -4, x = -1, x = 1, x = 3. Now, x = 1 and x = -4 are just normal intersections. x = 3 is a turning point. x = -1 is a stationary point of inflection. The graph is a 'positive' graph. You should be able to sketch it now.

[lzxnl]

How the hell does one sketch
?

There's all these turning points and stationary points of inflections, where do they begin?  How do you know when to start drawing the points of inflection within the graph?

This actually isn't so bad.
Firstly, let's add up the powers. It looks like the polynomial is of degree 9, which is odd, and the coefficient of the x^9 term (if expanded) seems to be positive, so when x is very negative, y is negative, and when x is very positive, so is y.

Then, mark in your intercepts. You can probably reason that as x=-4 is the first x intercept (and it's not a (x+4)^2 term), the graph will be increasing there. That means the graph goes over the x axis. Then, for the next intercept at x=-1, the graph must be decreasing, which means the graph must have turned around at some stage between x=-4 and x=-1. As (x+1)^5 is a factor, because it's an odd power greater than 1, you have a stationary point of inflection there (you sort of need to know this; look at y=x^3 and y=x^5 for instance). However, the graph is still decreasing as a result. At some stage, the graph will have to turn around again to increase to zero at x=1. The graph increases past x=1 and then has to turn around and decrease to zero at x=3. x=3 is a turning point as you can see from the (x-3)^2 term, so it turns around there and keeps increasing.

You now know where the turning points are; I haven't found where they are for you though. Do you seriously need the exact coordinates of them? I doubt it...

[Nato]

need help with this one:
The graph of has been reflected in the x-axis, shifted 3 units to the right and 1 unit up.
what is the resulting equation.

well i started off with


so from here



and for the y



and then i subbed this into the above equation


and then after rearranging i got
which doesn't seem to be correct.
 
can someone please tell me where i may have went wrong?
cheers.

[brightsky]

Below is an excerpt of a post I made a few days ago:

Alternatively, you can just remember the following:
- To dilate y = f(x) by a factor of k from the x-axis, replace y with y/k.
- To dilate y = f(x) by a factor of k from the y-axis, replace x with x/k.
- To reflect y = f(x) in the x-axis, replace y with -y.
- To reflect y = f(x) in the y-axis, replace x with -x.
- To translate y = f(x) k units in the positive direction of the x-axis, replace x with x - k.
- To translate y = f(x) k units in the positive direction of the y-axis, replace y with y - k.

So:
1. Reflect in the x-axis. y = -2(x+3)^3 - 1.
2. Translate 3 units in the positive direction of the x-axis. y = -2x^3 - 1.
3. Translate 1 unit in the positive direction of the y-axis. y = -2x^3.

[krisskross]

you went wrong there!
it should be
NO BRACKETS for -y+1




and for the y



subbing it in,



rearrange,


[edit]

Below is an excerpt of a post I made a few days ago:

So:
1. Reflect in the x-axis. y = -2(x+3)^3 - 1.
2. Translate 3 units in the positive direction of the x-axis. y = -2x^3 - 1.
3. Translate 1 unit in the positive direction of the y-axis. y = -2x^3.

Brightsky's method is way more efficient & is definitely my favourite when approaching transformations
[/edit]

[~T]

need help with this one:
The graph of has been reflected in the x-axis, shifted 3 units to the right and 1 unit up.
what is the resulting equation.

well i started off with


so from here



and for the y



and then i subbed this into the above equation


and then after rearranging i got
which doesn't seem to be correct.
 
can someone please tell me where i may have went wrong?
cheers.

Where you have:


It should be:


When we are showing transformations in this form, we don't need to put brackets around anything when we apply the translations. Thus, once we have reflected in the x-axis, our next transformations (the two translations) can just be applied like this:




Another approach is to apply each transformation to the equation in the following way:



We make y negative for a reflection in the x-axis:



We subtract 3 from x (with brackets of course) for a 3 unit shift to the right:



We subtract 1 from y for a 1 unit shift upwards:

[T-Infinite]

This may seem like a silly question but how do you find the inverse of function step by step. So I interchanged the and , which becomes . I'm not sure what to do next to make the y the subject. Someone please explain as I don't really want to use the CAS to solve it, I want to learn step by step.