Thank you for your help.
However, I meant that the (x-1) and (x+2) are both under the square root sign.
Sorry for the confusion.
VCE Methods Questions Thread
No worries, in that case we have a whole new situation. The same principle applies in that x+2 cannot equal zero and everything inside the bracket must be 0 or greater. To be honest, I would convert (x-1)/(x+2) into a partial fraction which would yield 1 - 3/(x+2) which can be obtained through long division. If you graph that you get a hyperbola (I think it's called that). You just gotta find when the graph is greater than or equal to zero. I would strongly suggest graphing it and you'll come to realise that from -2<x<1 the graph is negative. Thus the domain of that function would be the direct opposite, aka your aforementioned answer.
You'll find graphing stuff is always gonna help you and you need to learn and get used to do doing it. Hope that Helps.
PizzaMaster Nice to meet a fellow pizza lover
- Edited
Can a one-to-one function have a many-to one inverse function?
For example:
f(x) = 2x^ 3/4 + 1
has an inverse:
f-1(x) = [(x-1) / 2]^ 4/3
How come?
PizzaMaster
If you have f(x) = x0.5
Then f-1(x) = x2, but restricted to the domain R+ u {0}, which means the inverse is also 1-1. The negative branch of x2 isn't a reflection of y = x0.5 in the line y = x, so it's not part of the inverse function
Ohh so you mean that the range of the original is the domain of the inverse?
And so that's why
f(x) = 2x^ 3/4 + 1
does have an inverse:
f-1(x) = [(x-1) / 2]^ 4/3
BUT with domain [1, infinity)
How do we know whether the question wants us to give the dilation from x axis or y axis when given both the original and transformed function.
For example:
f(x)= 1/x2
f1(x)= 5/x2
I got dilation by a factor of 1/5 from the y axis but the answers say factor of 5 from the x axis. Is my answer still valid?
Correct that's why you need to restrict the domain of the original function
find values of a and b such that a(x+2)+b(x+3)=18x+8 for all values of x
if someone could help that would be great
chemistry1111 Expand then factorising LHS gives (a+b)x + (2a + 3b) = 18x + 8 --> by comparing coefficients, you obtain a+b = 18 and 2a+3b =8, solve simultaneously for a and b, giving a=46 and b=-28
thank you so much that makes sense
also this question
For the polynomial P(x)=(a+1)x3 + (b-7)x2 +c+5, find values for a,b and c if P(x) has a degree 2, a leading coefficient of 3 and the constant term is -1
when factorizing polynomials in methods unit 3/4, what method do most people use vcaa prefer, long division or equating coefficients
Do we need to buy the new Methods 3/4 book for the new study design (2023)?
I feel like the textbook questions are the same...
Can anyone plz help with this question (part b, c and d in particular):
PizzaMaster
Howdy, I did QCE Methods, so if I mess this up, I’m really sorry.
For b) may I know what the answers are? I may be able to work backwards
I didn’t do d) because I wasn’t that confident in doing the question.
Hope this helps,
-jinx_58
Hi jinx_58,
Thank you so much for solving!
The answer to part b is Domain = [0, 6]; Range = [0, 9/2]
The answer to part c is 9/2