VCE Methods Question Thread!

[Hancock]

The stuff under the square root must be positive for the function f(x) to be defined.





and

which is the domain of f(x).

[Will T]

Will the graph of have an inverse which is a function?

is obviously not a function, but it is one-to-many.
When you reflect it about the line y = x you get the many-to-one function

So that should mean that will have an inverse which is a function.
Correct me if I am wrong please.

[Jaswinder]

a one to one?  one x value has two y values though 

[FlorianK]

a one to one?  one x value has two y values though 

[Will T]

What am I on about....
Yes so is clearly one-to-many, i.e. one x produces two y values, not one-to-one (sorry about that).

But when you reflect it about the line y=x it becomes a many-to-one function.

So I mean, its inverse is still technically a function?

[FlorianK]

What am I on about....
Yes so is clearly one-to-many, i.e. one x produces two y values, not one-to-one (sorry about that).

But when you reflect it about the line y=x it becomes a many-to-one function.

So I mean, its inverse is still technically a function?

Yes.
The inverse of a one-to-many relation is a many to one function

[pi]

Well a function has to be restricted to be one-to-one to have an "inverse" in the first place hmmm

[Jenny_2108]

What am I on about....
Yes so is clearly one-to-many, i.e. one x produces two y values, not one-to-one (sorry about that).

But when you reflect it about the line y=x it becomes a many-to-one function.

So I mean, its inverse is still technically a function?

If its not one-to-one function, the inverse function doesnt exist

When you relfect it about the line y=x, the new graph can be called many-to-one function/relation but not its inverse

[BubbleWrapMan]

They never stated y^2 = x + 1 was a function.

I don't really like how they phrased the question though. An equation isn't a function. A graph isn't a function.

y = x isn't a function. It's a straight line equation. x is a function though; it's the identity function f(x) = x.

y = x^2 isn't a function. x^2 is a function though. And y is a function of x. The equation itself, y = x^2, is a relation.

A function of x doesn't have anything to do with y, if you think about its definition. It has a domain, and it has an image for each of the domain values. It's pretty much a one-dimensional object - it's only when you let y be equal to f(x) that it has a shape on the Cartesian plane.

Let's say we have a function with a domain D and we expressed the rule as f(x) = ...

What would happen if we replaced x with y? Nothing; if we let y = 1, then f(y) would be f(1), which would be the same as if we had f(x) and evaluated f(1). So basically we have the same function, if we replace x with y. Nothing about the inputs and outputs changed. We only changed the variable, but if you think about it the variable is only an intermediate object. It's basically a dummy variable. We changed x to y, but nothing about the function changed. We can replace it with u, v, or w, and it still doesn't matter. A function is separate from the variables themselves.

But now, let's let y = f(x). It now matters what the variables are, if we are considering the graph of the function on the Cartesian plane.

And now, swap x and y. We then have x = f(y). In other words, x is a function of y. Every value of y gives a unique value of x.

Let's define f. Say its domain is R, and its rule is f(y) = y^2 - 1. It doesn't have a shape yet, so let x = f(y) = y^2 - 1. Now it has a shape on the Cartesian plane; it's a parabola.

So, in the equation y^2 = x + 1, x is a function of y. When you take the inverse relation, y becomes a function of x.

It's complete to ask whether an equation is a function. If I asked whether y = x^2 was a function, I would have be more specific - I would have to say "is y a function of x?" In which case the answer would be yes. But if I didn't specify there, you could say "no, because x = √y or -√y so it's not a function of y."

Is x^2 + y^2 = 1 a function? Well, that's still an incomplete question. I should ask if y is a function of x, or if x is a function of y. The answer to both is of course no.

What's my point here... I guess we are too accustomed to graphing y as a function of x, which leads us to believe an equation only represents a function if y is a function of x.

y^2 = x + 1 still represents a function, you just have to be more specific. x is a function of y in this case. The function itself is f(u) = u^2 - 1, which does not have an inverse, so y is not a function of x.

When you swap x and y, you get the inverse relation. In that case, x is not a function of y, but y is now a function of x.

[Will T]

Cheers for that Calvin, very enlightening.

So if the question was, does have an inverse where y is a function of x, the answer would be yes.
But if the question was, does it have an inverse which is a function, then the answer could be either yes as y is a function of x, or no as x is not a function of y.
But we sort of assume because we're talking about the Cartesian plane they're asking us whether or not y will be a function of x.

But your point stands and it's interesting, you're right when you say the language of the question lacks specificity and promotes ambiguity.

But also, I think I'd just like to assert this and have people confirm or deny:

If you have a one-to-one relation (1x for 1y) and a one-to-many relation (1x for many y) when you invert them (i.e. reflect about the line you will always get a one-to-one relation (1x for 1y) and a many-to-one relation (many x for 1y).

But anything that is either one-to-one or many-to-one can be thought of as being a function where y is a function of x, because one x input will always yield one y output.

[Will T]

Essentials Chapter 3 chapter review extended response question 2f.

I solved for and got
but apparently the graphs intersect along the line as well?
Can someone explain this because I thought that any co-ordinates of intersection between a graph and its inverse occurred along the line y=x?

[Conic]

Essentials Chapter 3 chapter review extended response question 2f.

I solved for and got
but apparently the graphs intersect along the line as well?
Can someone explain this because I thought that any co-ordinates of intersection between a graph and its inverse occurred along the line y=x?

See this thread: Inverse Functions querie

[Will T]

Ahh wow, so I'm not the first person to come across this, and I'm also not the first person to have been told by my teacher that solving against the line y=x reveals the co-ordinates of intersection.

[kenners]

hey, bit confused on transformations
VCAA 2012 Tech-active Q2e,
A transformation that maps graph to the graph of has the rule (thats the answer)
however
VCAA 2006 Q 13
A transformation that maps to has the rule

Wondering why in the 2012 question, the horizontal translation stated is "4" while the 2006 question has horizontal translation "2". Shouldnt the dilation factor in the 2012 question affect the value of the translation (like in the 2006 question)?
Could someone break up the transformations happening in each equation? thanks
thanks

[pi]

Hint: think of which graph is being transformed to which graph The questions are opposites.