VCE Methods Question Thread!

[snowisawesome]

I know i've asked this before, but can someone please explain how to tell if y = 2+((1/x-3)) is a one-to-one function?
I've sketched it, but it still doesn't quite make sense


Also, if x^2 + y^2 = 1
then y^2 = 1-x^2,
then would y = +square root of(1-x^2), or would y = +-square root of (1-x^2)

Thanks

[JamesMaths]

I know i've asked this before, but can someone please explain how to tell if y = 2+((1/x-3)) is a one-to-one function?
I've sketched it, but it still doesn't quite make sense


Also, if x^2 + y^2 = 1
then y^2 = 1-x^2,
then would y = +square root of(1-x^2), or would y = +-square root of (1-x^2)

Thanks

Could I put my solution here?

Thanks
James.

[Lear]

I know i've asked this before, but can someone please explain how to tell if y = 2+((1/x-3)) is a one-to-one function?
I've sketched it, but it still doesn't quite make sense

This visualization might help you (attached). So if a function is one to one, there must be a unique y value to each x value. To check this we use a vertical line test and horizontal line test where we draw any arbitrary vertical/horizontal line to see if each x,y point is unique. In this picture I have graphed your function in red, the vertical lines in blue and horizontal in red. Notice how none of these lines intersect the graph More than once
From this we can conclude that each y value has a unique x value and each x value has a unique y value.

[snowisawesome]

Thanks Lear and JamesMaths for the help

Does anyone know approximately what sac and exam scores i need to get 35 raw in methods?

[Lear]

Thanks Lear and JamesMaths for the help

Does anyone know approximately what sac and exam scores i need to get 35 raw in methods?

I would highly suggest giving it your absolute best instead of playing a numbers game to see how much you need for x study score. You still have a lot of time and I don't believe limiting yourself to a 35 is ideal, you can definitely do better!

[snowisawesome]

Does anyone know how you would find the inverse of y = 9x-x^3

Also
Mathsquest 12 methods
exercise 3.6
question 3
consider the function f:(-infinity, 2) → R, f(x) = -1/(x-2)^2
Fully define the inverse, f^-1
I got until y-2 = +-√(-1/x)
but then the answers said it's only y-2 = -√(-1/x) since x belongs to (-infinity, 0)
Can someone please explain how they got this?

Thanks

[Bri MT]

Does anyone know how you would find the inverse of y = 9x-x^3
Thanks

For finding any inverse function
1. Remember that the equation must be one-to-one
2. Swap the x's with y's and the y with an x
3. Rearrange to have the equation in the form y =



Remember that the squareroot of any negative number isn't real.
To have it be the square root of a positive number the negative 1 must be divided by a negative

Also remember that the domain of the original is the range of its inverse

[snowisawesome]

For finding any inverse function
1. Remember that the equation must be one-to-one
2. Swap the x's with y's and the y with an x
3. Rearrange to have the equation in the form y =



Remember that the squareroot of any negative number isn't real.
To have it be the square root of a positive number the negative 1 must be divided by a negative

Also remember that the domain of the original is the range of its inverse

I'm confused how to find the inverse since it's a cubic function (y = 9x-x^3)
Could you please explain step by step?

[VanillaRice]

Does anyone know how you would find the inverse of y = 9x-x^3

Thanks

Think we discussed this question awhile back. There's no way of doing this by hand. If allowed a CAS, you can simply solve the equation x = = 9y-y^3 for y, and then sketch it.

[snowisawesome]

Think we discussed this question awhile back. There's no way of doing this by hand. If allowed a CAS, you can simply solve the equation x = = 9y-y^3 for y, and then sketch it.

Thanks
I was allowed a CAS for the question, just wanted to know how to do it by hand, in case it shows up in exam 1?

[VanillaRice]

Thanks
I was allowed a CAS for the question, just wanted to know how to do it by hand, in case it shows up in exam 1?

It won't. See what the equation is in your CAS and you'll see why 

[snowisawesome]

For finding any inverse function
1. Remember that the equation must be one-to-one
2. Swap the x's with y's and the y with an x
3. Rearrange to have the equation in the form y =



Remember that the squareroot of any negative number isn't real.
To have it be the square root of a positive number the negative 1 must be divided by a negative

Also remember that the domain of the original is the range of its inverse

Would it be possible for you to explain the bold part in a bit more detail please?

Also, would I be correct in saying that if -y^2=x^2-16
then y^2 = -x^2+16

Also,
f:[-4,0]  → R, f(x) = √(16-x^2)
I found my inverse to be y = +-√(16-x^2)
but since the domain of the original function is [-4,0], the inverse is only y = -√(16-x^2) according to the answer
Could someone please explain why this is the case?
And it said that he inverse is a one-to-one function with domain =[0,4] and range = [-4,0]
Can someone please explain this as well?
Thanks

[DBA-144]

Would it be possible for you to explain the bold part in a bit more detail please?

Also, would I be correct in saying that if -y^2=x^2-16
then y^2 = -x^2+16

Also,
f:[-4,0]  → R, f(x) = √(16-x^2)
I found my inverse to be y = +-√(16-x^2)
but since the domain of the original function is [-4,0], the inverse is only y = -√(16-x^2) according to the answer
Could someone please explain why this is the case?
And it said that he inverse is a one-to-one function with domain =[0,4] and range = [-4,0]
Can someone please explain this as well?
Thanks

So i guess i could have a go at this.

For the inverse u swap x and y values. So what used to be the domain becomes the range. And vice versa. 
You cant have a negative square root. Eg what is the square root of -4? There is not one ; no 2 same numbers multiply to make a negative number.
By multiplying the negative 1 by negative number what happens? Well what happens when -1 is divided by -1? They become positive and the reason we do this is because we CANT HAVE A NEGATIVE  SQUARE ROOT. 

For ur other question,  domain of function is range of inverse. So range is what uv given above. Domain of inverse is range of function. Find the range of the function. U should get the domain of the inverse which uv listed above. 

It has minus in front of square root sign because the domain is from -4 to 0. Remember this transformation is in the y axis?

[Mattjbr2]

How would you guys go about answering part -d- of this question? Page 61 of the year 12 Cambridge book. The worked solutions ignore it.
If you graph r in terms of x, the range is R. However, it makes sense to say that r can't realistically (in the real world) be negative nor could it be zero. So would the answer be R+? As in, (0,infinity)?

[Bell9565]

How would you guys go about answering part -d- of this question? Page 61 of the year 12 Cambridge book. The worked solutions ignore it.
If you graph r in terms of x, the range is R. However, it makes sense to say that r can't realistically (in the real world) be negative nor could it be zero. So would the answer be R+? As in, (0,infinity)?

I'm not 100% certain but if that point where the r intersects with x, im pretty sure that would have to be less that 3 so it would definitly be restricted between 0 and 3 but I don't have my cas here so i'm not exactly sure how you'd go about it but I'd assumes it would be something regarding the angle at the top right corner being less than 90 degrees or something.
I could be wrong but logically that makes sense